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./Probability_Practice.tex:218: Package amsmath Error: Erroneous nesting of equation structures;
|
||||
(amsmath) trying to recover with `aligned'.
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||||
|
||||
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|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.218 \end{align}
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||||
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||||
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|
||||
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|
||||
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||||
./Probability_Practice.tex:218: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.218 \end{align}
|
||||
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||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
./Probability_Practice.tex:218: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.218 \end{align}
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||||
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||||
I've inserted a begin-math/end-math symbol since I think
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||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./Probability_Practice.tex:218: Undefined control sequence.
|
||||
\math@cr@@@ ...t@ {&\kern -\alignsep@ }\fi \next@
|
||||
\cr
|
||||
l.218 \end{align}
|
||||
|
||||
The control sequence at the end of the top line
|
||||
of your error message was never \def'ed. If you have
|
||||
misspelled it (e.g., `\hobx'), type `I' and the correct
|
||||
spelling (e.g., `I\hbox'). Otherwise just continue,
|
||||
and I'll forget about whatever was undefined.
|
||||
|
||||
./Probability_Practice.tex:218: Missing { inserted.
|
||||
<inserted text>
|
||||
{
|
||||
l.218 \end{align}
|
||||
|
||||
I've put in what seems to be necessary to fix
|
||||
the current column of the current alignment.
|
||||
Try to go on, since this might almost work.
|
||||
|
||||
[4]
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||||
<inserted text>
|
||||
$
|
||||
l.236 P(\text{cancelled}^
|
||||
c) = 0.98\\
|
||||
I've inserted a begin-math/end-math symbol since I think
|
||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./Probability_Practice.tex:238: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.238
|
||||
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||||
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./Probability_Practice.tex:254: Package amsmath Error: Erroneous nesting of equation structures;
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(amsmath) trying to recover with `aligned'.
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||||
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||||
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|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.254 \end{align}
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||||
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||||
[5]
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||||
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||||
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||||
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|
||||
Type H <return> for immediate help.
|
||||
...
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||||
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||||
l.394 \end{align}
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||||
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||||
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||||
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|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.405 \end{align}
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||||
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||||
(amsmath) trying to recover with `aligned'.
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||||
|
||||
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|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.418 \end{align}
|
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||||
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|
||||
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|
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|
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[8]
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||||
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||||
(amsmath) trying to recover with `aligned'.
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||||
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||||
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|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.433 \end{align}
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||||
(amsmath) trying to recover with `aligned'.
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||||
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||||
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|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.442 \end{align}
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||||
[9]
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[12]
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[13] [14]
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[15] [16] (./Probability_Practice.aux) )
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\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Probability Practice Exercises}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
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|
||||
|
||||
% For partial derivatives
|
||||
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|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A jar contains four coins: a nickel (5 cents), a dime (10 cents), a quarter (25 cents), and a half-dollar (50 cents).
|
||||
Three coins are randomly selected without replacement from the jar.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item List all the possible outcomes in sample space S.
|
||||
\item What is the probabilty that the selection will contain the half-dollar?
|
||||
\item What is the probability that the total amount drawn will equal 65 cents or less?
|
||||
\item Which interpretations of probability are you applying?
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$S = \{(5,10,25), (5,25,50), (10, 25,50), (5,10,50)\}$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{half dollar}) = \frac{3}{4}$$
|
||||
|
||||
\part
|
||||
|
||||
Sum of the outcomes:
|
||||
$$S = \{40, 80, 85, 65\}$$
|
||||
|
||||
$$P(\text{total} \leq 65) = \frac{2}{4} = \frac{1}{2}$$
|
||||
|
||||
\part
|
||||
|
||||
Classical interpretation.
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
In a genetics experiment, the researcher mated two Drosophila fruit flies
|
||||
and observed the traits of 300 offsprint. The results are shown in the table
|
||||
|
||||
\includegraphics[width=\textwidth]{q2.png}
|
||||
One of these offspring is randomly selected and observed for the two genetic traits
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that the fly has normal eye color and normal wing size?
|
||||
\item What is the probability that the fly has vermillion eyes?
|
||||
\item What is the probability that the fly has either vermillion eyes or miniature wings, or both?
|
||||
\item Which interpretations of probability are you applying?
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{Normal Eye Color} \cap \text{Normal Wing Size}) = \frac{140}{300} = \frac{7}{15} = 0.46667$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{Vermillion Eyes}) = \frac{3+151}{300} = \frac{77}{150} = 0.513$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(\text{vermillion eyes } \cup \text{ miniature wings})&= P(\text {vermillion eyes}) + P(\text{miniature wings}) - P(\text{vermillion eyes} \cap \text{miniature wings})\\
|
||||
&= \frac{154}{300} + \frac{157}{300} - \frac{151}{300}
|
||||
&= \frac{160}{300}
|
||||
&= \frac{8}{15}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
Relative frequency interpretation.
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
City residents were surgeyed recently to determine readership of newspapers.
|
||||
50\% of the residents read the morning paper, 60\% read the evening paper and
|
||||
20\% read both newspapers. Find the probability that a resident selected
|
||||
at random reads the morning or evening paper
|
||||
|
||||
\solution
|
||||
|
||||
Given: \\
|
||||
$P(\text{morning}) = 0.5$\\
|
||||
$P(\text{evening}) = 0.6$\\
|
||||
$P(\text{both}) = 0.2 = P(\text{morning } \cap \text{ evening})$\\
|
||||
|
||||
To find: $P(\text{morning or evening})$\\
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(\text{morning} \cup \text{evening}) &= P(\text{morning}) + P(\text{evening}) - P(\text{morning} \cap \text{evening})$\\
|
||||
&= 0.5 + 0.6 - 0.2\\
|
||||
&= 0.9
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
The experience of a train passenger is that the train is cancelled with probability 0.02
|
||||
When it does run, it has probability of 0.9 of being on time.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that the train runs on time
|
||||
\item Given that the train is not on time, find the probability that it has been cancelled
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
Given:\\
|
||||
P(\text{cancelled}) = 0.02\\
|
||||
P(\text{cancelled}^c) = 0.98\\
|
||||
P(\text{on time} | \text{cancelled}^c) = 0.9\\
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{on time}) = 0.98 \cdot 0.9 = 0.882$$
|
||||
|
||||
\part
|
||||
|
||||
From part (a):
|
||||
$$P(\text{on time}^c) = 1 - 0.882 = 0.118$$\\
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(\text{cancelled} | \text{on time}^c) &= \frac{P(\text{cancelled} \cap \text{on time}^c)}{P(\text{on time}^c)}\\
|
||||
&= \frac{0.02\cdot 1}{0.02 \cdot 1 + 0.98 \cdot 0.1}\\
|
||||
&= 0.1694915\\
|
||||
&\approx 0.1694
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
Draw the probability tree diagram to visualise.\\
|
||||
Note: Once the train is cancelled, there is \textbf{NO WAY} the train can be on time,\\
|
||||
Therefore, $P(\text{on time} | \text{cancelled}) = 0$ and $P(\text{on time}^c | \text{cancelled}) = 1$.
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
In a group of students 60\% are female and 40\% are male.
|
||||
A third of the female students study french but only a quarter of the male students study french
|
||||
\\
|
||||
A student is chosen at random from the group.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Show that the probability that the chosen student is female and studies french is 0.2
|
||||
\item Calculate the probability that the chosen student studies french
|
||||
\item Given that the chosen student does study french, calculate the conditional probability that the chosen student is female
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
Let f: female\\
|
||||
fr: studies french
|
||||
\\\\
|
||||
Given:
|
||||
$P(f) = 0.6$\\
|
||||
$P(f^c) = 0.4$\\
|
||||
$P(fr) = 0.3$\\
|
||||
$P(fr^c) = 0.7$\\
|
||||
$P(fr | f) = 0.3$\\
|
||||
$P(fr^c | f) = 0.7$\\
|
||||
$P(fr | f^c) = 0.25$\\
|
||||
$P(fr^c | f^c) = 0.75$\\
|
||||
|
||||
Draw the probability tree to help solve the problem
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
P(f \cap fr) = P(f) \cdot P(fr | f) = 0.6 \cdot \frac{1}{3} = 0.2
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(fr) = 0.2 + (0.4 \cdot 0.25) = 0.2 + 0.1 = 0.3$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(f | fr) = \frac{P(f \cap fr)}{P(fr)} = \frac{0.2}{0.3}$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Of the patients in an emergency unit, 30\% have a sports injury. Pass records
|
||||
for this emergency unit suggest that :\\
|
||||
|
||||
A patient with sports injury has a probability of 0.2 of being admitted to hospital\\
|
||||
\\
|
||||
A patient who doesn’t have a sports injury has a probability of 0.4 of being admitted to
|
||||
hospital\\
|
||||
\\
|
||||
Now, a patient is chosen at random from those in the emergency unit.\\
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that the chosen patient has a sports injury
|
||||
\item What is the probability that the chosen patient has a sports injury and is admitted to hospital
|
||||
\item Show that the probability that the chosen patient is admitted to hospital is 0.34
|
||||
\item Given that the chosen patient is admitted to hospital, find the conditional probability that the patient has a sports injury
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
Given:\\
|
||||
$P(\text{sports injury}) = 0.3$\\
|
||||
$P(\text{sports injury}^c) = 0.7$\\
|
||||
$P(\text{admitted} | \text{sports injury}) = 0.2$\\
|
||||
$P(\text{admitted} | \text{sports injury}^c) = 0.4$\\
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{sports injury}) = 0.3$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{sports injury} \cap \text{admitted}) = P(\text{sports injury}) \cdot P(\text{admitted} | \text{sports injury}) = 0.3 \cdot 0.2 = 0.06$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{admitted}) = P(\text{sports injury} \cap \text{admitted}) + P(\text{sports injury}^c \cap \text{admitted}) = 0.3 \cdot 0.2 + 0.7 \cdot 0.4 = 0.34$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{sports injury} | \text{admitted}) = \frac{P(\text{sports injury} \cap \text{admitted})}{P(\text{admitted})} = \frac{0.06}{0.34} = 0.17647 \approx 0.18$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
Darren and Charles are friends who often go to the cinema together. On such visits there
|
||||
is a probability of 0.4 that Darren will buy popcorn.
|
||||
The probability that Charles will buy
|
||||
popcorn is 0.7 if Darren buys popcorn and 0.35 if he doesn’t.
|
||||
|
||||
When Darren and Charles visit the cinema together:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that both buy popcorn
|
||||
\item Show that the probability that neither buys popcorn is 0.39
|
||||
\item Find the probability that exactly one of them buys popcorn
|
||||
\end{enumerate}
|
||||
Sarah sometimes join Darren and Charles on their cinema visits. On these occasions,
|
||||
the probability that sarah buys popcorn is 0.55 if both
|
||||
of charles and darren buy popcorn and 0.25 if exactly one of darren and charles buys popcorn
|
||||
|
||||
When Darren, Sarah and Charles visit the cinema together:
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that the three of them buy popcorn
|
||||
\item Charles and sarah buy popcorn but Daren Doesn't
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
Given:\\
|
||||
$P(\text{Darren buys popcorn}) = 0.4$\\
|
||||
$P(\text{Charles buys popcorn} | \text{Darren buys popcorn}) = 0.7$\\
|
||||
$P(\text{Charles buys popcorn} | \text{Darren doesn't buy popcorn}) = 0.35$\\
|
||||
$P(\text{Darren doesn't buy popcorn}) = 0.6$\\
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
& P(\text{Darren buys popcorn} \cap \text{Charles buys popcorn}) \\
|
||||
&= P(\text{Darren buys popcorn}) \cdot P(\text{Charles buys popcorn} | \text{Darren buys popcorn})\\
|
||||
&= 0.4 \cdot 0.7\\
|
||||
&= 0.28
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
& P(\text{Darren doesn't buy popcorn} \cap \text{Charles doesn't buy popcorn}) \\
|
||||
&= P(\text{Darren doesn't buy popcorn}) \cdot P(\text{Charles doesn't buy popcorn} | \text{Darren doesn't buy popcorn})\\
|
||||
&= 0.6 \cdot 0.65\\
|
||||
&= 0.39
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
& P(\text{Darren buys popcorn} \cap \text{Charles doesn't buy popcorn} \text{ or } \text{Charles buys popcorn} \cap \text{Darren doesn't buy popcorn}) \\
|
||||
&= P(\text{Darren buys popcorn}) \cdot P(\text{Charles doesn't buy popcorn} | \text{Darren buys popcorn}) \\
|
||||
&+ P(\text{Charles buys popcorn}) \cdot P(\text{Darren doesn't buy popcorn} | \text{Charles buys popcorn})\\
|
||||
&= 0.4 \cdot 0.3 + 0.6 \cdot 0.35\\
|
||||
&= 0.33
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
Given:
|
||||
$P(\text{Sarah buys popcorn} | \text{Darren buys popcorn} \cap \text{Charles buys popcorn}) = 0.55$\\
|
||||
$P(\text{Sarah buys popcorn} | \text{Darren buys popcorn} \cap \text{Charles doesn't buy popcorn}) = 0.25$\\
|
||||
$P(\text{Sarah buys popcorn} | \text{Darren doesn't buy popcorn} \cap \text{Charles buys popcorn}) = 0.25$\\
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(\text{3 buy popcorn}) &= P(\text{Darren buys popcorn} \cap \text{Charles buys popcorn} \cap \text{Sarah buys popcorn}) \\
|
||||
&= 0.4 \cdot 0.7 \cdot 0.55\\
|
||||
&= 0.154
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(\text{Charles buys popcorn} \cap \text{Sarah buys popcorn} \cap \text{Darren doesn't buy popcorn}) &= 0.6 \cdot 0.35 \cdot 0.25\\
|
||||
&= 0.0525
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
|
||||
Following a flood, 120 tins were recovered from Lee’s corner shop.
|
||||
Unfortunately, the water had washed off all the labels.
|
||||
Of the tins, 50 contained pet food, 20 contained peas, 35
|
||||
contained beans and the rest contained soup.
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Lee selects a tin at random, what is the probability that it contains soup?
|
||||
\item Lee selects a tin at random, what is the probability that it doesn’t contain pet food
|
||||
\item Lee selects two tins at random (without replacement), what is the probability that both
|
||||
contains peas
|
||||
\item Lee selects two tins at random (without replacement), what is the probability that one
|
||||
contains pet food and the other contains peas
|
||||
\item Lee selects 3 tins at random (without replacement). What is the probability that one
|
||||
contains pet food, one contains peas and one contains beans.
|
||||
\item Find the probability that Lee will have to open more than two tins before he finds one
|
||||
that doesn’t contain pet food.
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
Soup left: 120 - 50 - 20 - 35 = 15\\
|
||||
|
||||
$$P(\text{soup}) = \frac{15}{120} = 0.125$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{pet food}^c) = \frac{120-50}{120} = 0.58333$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{peas} \cap \text{peas}) = \frac{20}{120} \cdot \frac{19}{119} = 0.00254$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{pet food} \cap \text{peas}) = \frac{50}{120} \cdot \frac{20}{119} \cdot 2 = 0.140056$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{pet food} \cap \text{peas} \cap \text{beans}) = \frac{50}{120} \cdot \frac{20}{119} \cdot \frac{35}{118} \cdot 3! = 0.12462612164 \approx 0.125$$
|
||||
|
||||
\part
|
||||
|
||||
What are the chances that it's just
|
||||
|
||||
$$\frac{50}{120} \cdot \frac{49}{119} = 0.1715686 \approx 0.172$$
|
||||
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A school employs 75 teachers. The following table summarizs their length of service at the school, classified by gender.
|
||||
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
& $<$ 3 years & 3 years to 8 years & $>$ 8 years\\
|
||||
\hline
|
||||
Female & 12 & 20 & 13\\
|
||||
\hline
|
||||
Male & 8 & 15 & 7\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that a randomly selected teacher is a female
|
||||
\item Find the probability that a randomly selected teacher is female given that the teacher has more than 8 years service
|
||||
\item Find the probability that a randomly selected teacher is female given that the teacher has less than 3 years service
|
||||
\item State, giving a reason, whether or not the event of selecting a female teacher is independent of the event of selecting a teacher with less than 3 years service
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{female}) = \frac{12+20+13}{75} = \frac{45}{75} = 0.6$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{female} | >\text{8 years}) = \frac{P(\text{female} \cap >\text{8 years})}{P(>\text{8 years})} = \frac{\frac{13}{75}}{\frac{20}{75}} = 0.65$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{female}|< \text{3 years}) = \frac{P(\text{female} \cap < \text{3 years})}{P(< \text{3 years})} = \frac{\frac{12}{75}}{\frac{20}{75}} = 0.6$$
|
||||
|
||||
\part
|
||||
|
||||
Test for independence: $$P(\text{female}| < \text{3 years}) = P(\text{female})$$\\
|
||||
Since both are 0.6, the events are independent.
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
A group of students bought, in total, 25 items of clothing at two shops: Mango and
|
||||
Iora. The following table shows how many tops, jeans and sweaters were
|
||||
bought at each of the two shops:\\
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
& Top & Jeans & Sweaters\\
|
||||
\hline
|
||||
Mango & 3 & 7 & 5\\
|
||||
\hline
|
||||
Iora & 2 & 5 & 3\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\
|
||||
One item of clothing is chosen at random from these 25 items:\\
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that the chosen item is a top?
|
||||
\item What is the probability that the chosen item was bought from Iora?
|
||||
\item What is the probability that the chosen item is a top and was bought from Iora?
|
||||
\item State with a reason whether the events (chosen item is a top) and (chosen item was bought from iora) are independent.
|
||||
\item Given that the chosen item is not a top, find the conditional probability that it was
|
||||
\end{enumerate}
|
||||
\solution
|
||||
|
||||
\begin{tabular}{|c|c|c|c|c|}
|
||||
\hline
|
||||
& Top & Jeans & Sweaters & Total\\
|
||||
\hline
|
||||
Mango & 3 & 7 & 5 & 15\\
|
||||
\hline
|
||||
Iora & 2 & 5 & 3 & 10\\
|
||||
\hline
|
||||
Total & 5 & 12 & 8 & 25\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\\\
|
||||
\part
|
||||
|
||||
$$P(\text{top}) = \frac{5}{25} = 0.2$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{Iora}) = \frac{10}{25} = 0.4$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{top} \cap \text{Iora}) = \frac{2}{25} = 0.08$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{top} | \text{Iora}) = \frac{P(\text{top} \cap \text{Iora})}{P(\text{Iora})} = \frac{0.08}{0.4} = 0.2$$
|
||||
|
||||
Test for independence: $$P(\text{top}) = P(\text{top} | \text{Iora})$$\\
|
||||
Since both are 0.2, the events are independent.
|
||||
\\\\
|
||||
\part
|
||||
|
||||
$$P(\text{Iora} | \text{top}^c) = \frac{P(\text{Iora} \cap \text{top}^c)}{P(\text{top}^c)} = \frac{0.4 \cdot 0.8}{0.8} = 0.4$$
|
||||
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A volunteer for the Health Literacy Center was investigating the attitudes of students towards
|
||||
smoking on campus. A random sample of 730 students from all four grade levels was taken.
|
||||
Each student was given the statement "Smoking is dangerous for your health" and asked
|
||||
whether they agreed, had no opinion or disagreed. The following contingency table summarizes the results.
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Complete the following table knowing that
|
||||
\begin{enumerate}
|
||||
\item 387 Male students participated
|
||||
\item 277 students agreed and 97 students had no opinion
|
||||
\item 154 from those who agreed were female
|
||||
\item Only 36 from the males who participated had no opinion
|
||||
\end{enumerate}
|
||||
\item Among the participants, what is the probability that a student agreed P (A)?
|
||||
\item Show whether the events of students agreeing(A) and being a female student (F) are independent
|
||||
\item What is the probability that a student is a female given that the student had no opinion $P(F|N)$? What is the probability that a student disagreed, given that the student is Male $P(D|M)$?
|
||||
\item Use Bayes' Theorem to calculate the probability $P(M|A)$
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
|
||||
\part
|
||||
|
||||
\begin{tabular}{|c|c|c|c|c|}
|
||||
\hline
|
||||
& Agree (A) & No Opinion (N) & Disagree (D) & Totals\\
|
||||
\hline
|
||||
Male (M) & 123 & 36 & 228 & 387 \\
|
||||
\hline
|
||||
Female (F) & 154 &61 &128 &343 \\
|
||||
\hline
|
||||
Totals &277 & 97 & 356 &730 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\\\
|
||||
\part
|
||||
|
||||
$$P(A) = \frac{277}{730} = 0.38$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(A|F) = \frac{P(A \cap F)}{P(F)} = \frac{0.21}{0.47} = 0.45$$\\
|
||||
Test for independence: $$P(A) = P(A|F)$$\\
|
||||
Since $P(A) = 0.38$ and $P(A|F) = 0.45$, the events are not independent.
|
||||
|
||||
\part
|
||||
|
||||
$$P(F|N) = \frac{P(F \cap N)}{P(N)} = \frac{\frac{61}{730}}{\frac{97}{730}} = 0.63$$
|
||||
|
||||
$$P(D|M) = \frac{P(D \cap M)}{P(M)} = \frac{\frac{228}{730}}{\frac{387}{730}} = 0.59$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(M|A) = \frac{P(M \cap A)}{P(A)} = \frac{\frac{123}{730}}{\frac{277}{730}} = 0.44$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Letter A is sent by first-class post and has a probability of 0.9 of being delivered next day. Letter B is sent by second-class post and has a probability of only 0.3 if being delivered the next day.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that both letters are delivered the next day
|
||||
\item Find the probability that neither letter is delivered the next day
|
||||
\item Find the probability that at least one of the letters is delivered the next day
|
||||
\item Given that at least one of the letters is delivered the next day, find the probability that letter A is delivered the next day
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
|
||||
Draw the Probability tree diagram to solve the problem.
|
||||
|
||||
\part
|
||||
|
||||
$$P(A \cap B) = P(A) \cdot P(B) = 0.9 \cdot 0.3 = 0.27$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(A^c \cap B^c) = P(A^c) \cdot P(B^c) = 0.1 \cdot 0.7 = 0.07$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.9 + 0.3 - 0.27 = 0.93$$
|
||||
OR, Using a tree diagram:\\
|
||||
$$P(A \cup B) = 0.9 \cdot 0.3 + 0.9 \cdot 0.7 + 0.1 \cdot 0.3 = 0.93$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(a|\text{at least 1 letter}) = \frac{P(a \cap \text{at least 1 letter})}{P(\text{at least 1 letter})} = \frac{0.9 \cdot 0.3 + 0.9 \cdot 0.7}{0.93} = 0.9677419 \approx 0.9677$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\textbf{ANSWERS}
|
||||
\begin{enumerate}
|
||||
\item (b) $\frac{3}{4}$, $c = \frac{1}{2}$, classical interpretation
|
||||
\item $\frac{7}{15}$, $\frac{77}{150}$, $\frac{8}{15}$, Relative frequency interpretation
|
||||
\item 0.9
|
||||
\item 0.882, 0.1694
|
||||
\item 0.2,0.3, 0.2/0.3
|
||||
\item 0.3, 0.06, 0.34, 0.17647
|
||||
\item 0.28, 0.39, 0.33,0.154, 0.0525
|
||||
\item 0.125, 0.0583, 0.0266, 0.140, 0.125, 0.172
|
||||
\item 0.6, 0.65, 0.6, independent
|
||||
\item $\frac{1}{5}$, $\frac{2}{5}$, $\frac{2}{25}$, independent, $\frac{2}{5}$
|
||||
\item B)P(A) = $\frac{377}{730}$, C) not Independent, D) $\frac{61}{97}, \frac{228}{387}$, E) 0.44404
|
||||
\item A) 0.27, B)0.07, C)0.93, D)0.9677
|
||||
\end{enumerate}
|
||||
|
||||
11a)
|
||||
|
||||
\begin{tabular}{|c|c|c|c|c|}
|
||||
\hline
|
||||
& Agree (A) & No Opinion (N) & Disagree (D) & Totals\\
|
||||
\hline
|
||||
Male (M) & 123 & 36 & 228 & 387 \\
|
||||
\hline
|
||||
Female (F) & 154 &61 &128 &343 \\
|
||||
\hline
|
||||
Totals &277 & 97 & 356 &730 \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\end{homeworkProblem}
|
||||
|
||||
\end{document}
|
||||
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|
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|
|
@ -0,0 +1,264 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
\usepackage{multicol}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ Save\ My\ Grades}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\def\therefore{\boldsymbol{\text{ }
|
||||
\leavevmode
|
||||
\lower0.4ex\hbox{$\cdot$}
|
||||
\kern-.5em\raise0.7ex\hbox{$\cdot$}
|
||||
\kern-0.55em\lower0.4ex\hbox{$\cdot$}
|
||||
\thinspace\text{ }}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Remove vector $u=(-1,3,-4,2)$ from vector $v=(-2,2,2,5,6)$\\
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Note the difference in the order between remove from and project onto.\\
|
||||
Project vector $u=(-1,3,-4,2)$ onto vector $v=(3,-3,-1,1)$\\
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x+4y+2z &=5.5\\
|
||||
-5x-22y-5z &= -45.5\\
|
||||
2x+4z+14z &=-25
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Show as an itermediate step the augmented matrix when for the first time the zeroth coulmn became a one-hot vector after performing transformations
|
||||
\item Show as an intermediate step the augmented matrix when for the first time the augmented matrix is in row echelon form.
|
||||
\item Write the set of all solutions as a single vector or a combination of vectors, None if there is no solution
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x+3y-5z &=2.75\\
|
||||
3x+12y-13z &=-9.75\\
|
||||
-4x-6z+25z &=-46.25 \text{(might be -6y)}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Show as an itermediate step the augmented matrix when for the first time the zeroth coulmn became a one-hot vector after performing transformations
|
||||
\item Show as an intermediate step the augmented matrix when for the first time the augmented matrix is in row echelon form.
|
||||
\item Write the set of all solutions as a single vector or a combination of vectors, None if there is no solution
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inverse of
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0 &= \begin{bmatrix}9&-2\\3&-4\end{bmatrix}\\
|
||||
A_1 &= \begin{bmatrix}10&3\\8&4\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
Use these inverses to Solve
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0x &= \begin{bmatrix}1\\-2\end{bmatrix}\\
|
||||
A_1x &= \begin{bmatrix}-7\\4\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
\begin{itemize}
|
||||
\item Are they invertible?
|
||||
\item Which of them has full rank? WHich one of them has lower rank and which one?
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the determinant of
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}3&-1&4\\5&2.5&3\\1&8&-6\end{bmatrix}\\
|
||||
A &= \begin{bmatrix}3&-2&0.5\\2.5&-3&1\\3&2&4\end{bmatrix}\\
|
||||
A &= \begin{bmatrix}2&-2&2\\8&3&-2\\10&-4.5&5\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
What is the determinant of this matrix? Write it as a polynomial in $c$.\\
|
||||
For what value $c$ the matrix is not invertible?
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}6&-3&c\\5&2&2\\-2&-6&-2\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute and apply the Householder matrix which makes transforms the first column of $A$ to a multipile of the first one-hot vector for
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}8&1&2\\4&-1&3\\-8&4&2\end{bmatrix}
|
||||
$$
|
||||
and for (Subtracting is nicer)\\
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}3&-4&3\\\sqrt{2}&6&4\\\sqrt{5}&3&2\end{bmatrix}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\end{document}
|
||||
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|
|
@ -0,0 +1,498 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
\usepackage{multicol}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Revision\ Tutorial\ Solutions}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\def\therefore{\boldsymbol{\text{ }
|
||||
\leavevmode
|
||||
\lower0.4ex\hbox{$\cdot$}
|
||||
\kern-.5em\raise0.7ex\hbox{$\cdot$}
|
||||
\kern-0.55em\lower0.4ex\hbox{$\cdot$}
|
||||
\thinspace\text{ }}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Remove vector $u=(-1,3,-4,2)$ from vector $v=(-2,2,2.5,6)$\\
|
||||
|
||||
\answer
|
||||
|
||||
$$v_{new} = v - \frac{v\cdot u}{u\cdot u}u$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
u\cdot u &= (-1,3,-4,2)\cdot (-1,3,-4,2) = 30\\
|
||||
v\cdot u &= (-2,2,2.5,6)\cdot (-1,3,-4,2) = 10\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
v_{new} &= (-2,2,2.5,6) - \frac{10}{30}(-1,3,-4,2) \\
|
||||
&= (-2,2,2.5,6) +(\frac{1}{3},-1,\frac{4}{3}, -\frac{2}{3})\\
|
||||
&= (-\frac{5}{3},1,\frac{13}{6},\frac{16}{3})
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
You can check:
|
||||
$$
|
||||
v_{new}\cdot u = (-\frac{5}{3},1,\frac{13}{6},\frac{16}{3})\cdot (-1,3,-4,2) = 0
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Note the difference in the order between remove from and project onto.\\
|
||||
Project vector $u=(-1,-3,-4,2)$ onto vector $v=(3,-3,-1,1)$\\
|
||||
|
||||
\answer
|
||||
|
||||
$$u_{new} = \frac{u\cdot v}{v\cdot v}v$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
u_{new} &= \frac{(-1,-3,-4,2)\cdot (3,-3,-1,1)}{(3,-3,-1,1)\cdot (3,-3,-1,1)}(3,-3,-1,1)\\
|
||||
&= \frac{12}{20}(3,-3,-1,1)\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x+4y+2z &=5.5\\
|
||||
-5x-22y-5z &= -45.5\\
|
||||
2x+4y+14z &=-25
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Show as an itermediate step the augmented matrix when for the first time the zeroth coulmn became a one-hot vector after performing transformations
|
||||
\item Show as an intermediate step the augmented matrix when for the first time the augmented matrix is in row echelon form.
|
||||
\item Write the set of all solutions as a single vector or a combination of vectors, None if there is no solution
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
&\left[
|
||||
\begin{array}{ccc|c}
|
||||
1 & 4 & 2 & 5.5\\
|
||||
-5 & -22 & -5 & -45.5\\
|
||||
2 & 4 & 14 & -25
|
||||
\end{array} \right] \\
|
||||
\rho_2+5\rho_1, \rho_3-2\rho_1 &\rightarrow \left[
|
||||
\begin{array}{ccc|c}
|
||||
1 & 4 & 2 & 5.5\\
|
||||
0 & -2 & 5 & -18\\
|
||||
0 & -4 & 10 & -36
|
||||
\end{array} \right] \\
|
||||
\rho_3-2\rho_2 &\rightarrow \left[
|
||||
\begin{array}{ccc|c}
|
||||
1 & 4 & 2 & 5.5\\
|
||||
0 & -2 & 5 & -18\\
|
||||
0 & 0 & 0 & 0
|
||||
\end{array} \right] \\
|
||||
-\frac{1}{2}\rho_2&\rightarrow \left[
|
||||
\begin{array}{ccc|c}
|
||||
1 & 4 & 2 & 5.5\\
|
||||
0 & 1 & -2.5 & 9\\
|
||||
0 & 0 & 0 & 0
|
||||
\end{array} \right] \\
|
||||
\rho_1+4\rho_2 &\rightarrow \left[
|
||||
\begin{array}{ccc|c}
|
||||
1 & 0 & 12 & -30.5\\
|
||||
0 & 1 & -2.5 & 9\\
|
||||
0 & 0 & 0 & 0
|
||||
\end{array} \right] \\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x + 12z &= -30.5\\
|
||||
y - 2.5z &= 9\\\\
|
||||
\therefore x &= -30.5 - 12z\\
|
||||
\therefore y &= 9 + 2.5z\\
|
||||
\therefore z &= 0 + 1z\\\\
|
||||
\begin{bmatrix} x\\ y\\ z \end{bmatrix} &= \begin{bmatrix}
|
||||
-30.5\\9\\0\end{bmatrix} + \begin{bmatrix} -12\\2.5\\1\end{bmatrix}z
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x+3y-5z &=2.75\\
|
||||
3x+12y-13z &=-9.75\\
|
||||
-4x-6y+25z &=-46.25
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Show as an itermediate step the augmented matrix when for the first time the zeroth coulmn became a one-hot vector after performing transformations
|
||||
\item Show as an intermediate step the augmented matrix when for the first time the augmented matrix is in row echelon form.
|
||||
\item Write the set of all solutions as a single vector or a combination of vectors, None if there is no solution
|
||||
\end{itemize}
|
||||
|
||||
\answer
|
||||
|
||||
|
||||
$$
|
||||
\left[ \begin{array}{ccc|c}
|
||||
1 & 3 & -5 & 2.75\\
|
||||
3 & 12 & -13 & -9.75\\
|
||||
-4 & -6 & 25 & -46.25
|
||||
\end{array} \right] \\
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\rho_2-3\rho_1, \rho_3+4\rho_1 &\rightarrow \left[ \begin{array}{ccc|c}
|
||||
1 & 3 & -5 & 2.75\\
|
||||
0 & 3 & 2 & -18\\
|
||||
0 & 6 & 5 & -35.25
|
||||
\end{array} \right] \\
|
||||
\rho_3-2\rho_2 &\rightarrow \left[ \begin{array}{ccc|c}
|
||||
1 & 3 & -5 & 2.75\\
|
||||
0 & 3 & 2 & -18\\
|
||||
0 & 0 & 1 & 0.75
|
||||
\end{array} \right] \\
|
||||
\frac{1}{3}\rho_2-2\rho_1 &\rightarrow \left[ \begin{array}{ccc|c}
|
||||
1 & 3 & -5 & 2.75\\
|
||||
0 & 1 & 0 & -6.5\\
|
||||
0 & 0 & 1 & 0.75
|
||||
\end{array} \right] \\
|
||||
\rho_1-3\rho_2+5\rho_1 &\rightarrow \left[ \begin{array}{ccc|c}
|
||||
1 & 0 & 0 & 26\\
|
||||
0 & 1 & 0 & -6.5\\
|
||||
0 & 0 & 1 & 0.75
|
||||
\end{array} \right] \\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\text{first time one hot:}& \left [ \begin{array}{ccc|c}
|
||||
1 & 3&-5 & 2.75\\
|
||||
0 & 3 & 2 & -18\\
|
||||
0 & 6 & 5 & -35.25
|
||||
\end{array} \right ] \\
|
||||
\text{first time row echelon form:}& \left [ \begin{array}{ccc|c}
|
||||
1&3&-5&2.75\\
|
||||
0&1&\frac{2}{3}&6\\
|
||||
0&0&1&0.75
|
||||
\end{array} \right ] \\
|
||||
\text{Reduced row echelon form:}& \left[ \begin{array}{ccc|c}
|
||||
1&0&0&26\\
|
||||
0&1&0&-6.5\\
|
||||
0&0&1&0.75
|
||||
\end{array} \right ] \\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inverse of
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0 &= \begin{bmatrix}9&-2\\3&-4\end{bmatrix}\\
|
||||
A_1 &= \begin{bmatrix}10&3\\8&4\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
Use these inverses to Solve
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0x &= \begin{bmatrix}1\\-2\end{bmatrix}\\
|
||||
A_1x &= \begin{bmatrix}-7\\4\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0 &= \begin{bmatrix}9&-2\\3&-4\end{bmatrix}\\
|
||||
\det(A_0) &= 9\cdot(-4) - (-2)\cdot3 =-36+6 = -30\\
|
||||
A_0^{-1} &= \frac{1}{-30}\begin{bmatrix}-4&2\\-3&-9\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\\frac{1}{10}&-\frac{3}{10}\end{bmatrix}\\
|
||||
x &= A_0^{-1}\begin{bmatrix}1\\-2\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\\frac{1}{10}&-\frac{3}{10}\end{bmatrix}\begin{bmatrix}1\\-2\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{2}{15}+\frac{2}{15}\\\frac{1}{10}+\frac{6}{10}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{4}{15}\\\frac{7}{10}\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A_1 &= \begin{bmatrix}10&3\\8&4\end{bmatrix}\\
|
||||
\det(A_1) &= 10\cdot4 - 3\cdot8 = 40-24 = 16\\
|
||||
A_1^{-1} &= \frac{1}{16}\begin{bmatrix}4&-3\\-8&10\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{1}{4}&-\frac{3}{16}\\\frac{-1}{2}&\frac{5}{8}\end{bmatrix}\\
|
||||
x &= \begin{bmatrix}\frac{1}{4}&-\frac{3}{16}\\\frac{-1}{2}&\frac{5}{8}\end{bmatrix} \begin{bmatrix}-7\\4\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-\frac{10}{4}\\6\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the determinant of
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}3&-1&4\\5&2.5&3\\1&8&-6\end{bmatrix}\\
|
||||
A &= \begin{bmatrix}3&-2&0.5\\2.5&-3&1\\3&2&4\end{bmatrix}\\
|
||||
A &= \begin{bmatrix}2&-2&2\\8&3&-2\\10&-4.5&5\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Are they invertible?
|
||||
\item Which of them has full rank? Which one of them has lower rank and which one?
|
||||
\end{itemize}
|
||||
|
||||
\answer
|
||||
|
||||
\begin{multicols}{3}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}3&-1&4\\5&2.5&3\\1&8&-6\end{bmatrix}\\
|
||||
\det{A} &= 0 (\text{not invertible})\\
|
||||
\text{rank} &= 2\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}3&-2&0.5\\2.5&-3&1\\3&2&4\end{bmatrix}\\
|
||||
\det{A} &= -21 (\text{invertible})\\
|
||||
\text{rank} &= 3\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}2&-2&2\\8&3&-2\\10&-4.5&5\end{bmatrix}\\
|
||||
\det{A} &= 0 (\text{not invertible})\\
|
||||
\text{rank} &= 2\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
What is the determinant of this matrix? Write it as a polynomial in $c$.\\
|
||||
For what value $c$ the matrix is not invertible?
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}6&-3&c\\5&2&2\\-2&-6&-2\end{bmatrix}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 30-30c+4c\\
|
||||
&= 30-26c\\
|
||||
\det(A) &= 0\\
|
||||
\therefore c &= \frac{30}{26}\\
|
||||
\end{align*}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\textbf{Compute and apply} the Householder matrix which makes transforms the first column of $A$ to a multipile of the first one-hot vector for
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}8&1&2\\4&-1&3\\-8&4&2\end{bmatrix}
|
||||
$$
|
||||
and for (Subtracting is nicer)\\
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}3&-4&3\\\sqrt{2}&6&4\\\sqrt{5}&3&2\end{bmatrix}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x &=\begin{bmatrix}8\\4\\-8\end{bmatrix}\\
|
||||
u &= \begin{bmatrix}8\\4\\-8\end{bmatrix} \pm \lVert \begin{bmatrix}8,4,-8\end{bmatrix}\rVert \begin{bmatrix}1\\0\\0\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-4\\4\\-8\end{bmatrix} \text{ Subtract}\\
|
||||
H_u &= I - \frac{2}{\lVert [-4,4,-8]\rVert_2^2}\begin{bmatrix}8\\4\\-8\end{bmatrix}\begin{bmatrix}8&4&8\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\-\frac{2}{3}&\frac{2}{3}&-\frac{1}{3}\end{bmatrix}\\
|
||||
HA &= \begin{bmatrix}12 &-2.44 & 1\\0 & 2.33 & 4\\0&-2.7&0\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x &= \begin{bmatrix}3\\\sqrt{2}\\ \sqrt{5}\end{bmatrix}\\
|
||||
u &= \begin{bmatrix}3\\\sqrt{2}\\ \sqrt{5}\end{bmatrix} \pm \lVert \begin{bmatrix}3,\sqrt{2},\sqrt{5}\end{bmatrix}\rVert \begin{bmatrix}1\\0\\0\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-1\\\sqrt{2}\\ \sqrt{5}\end{bmatrix} \text{ Subtract}\\
|
||||
H_u &= I - \frac{2}{\lVert [-1,\sqrt{2},\sqrt{5}]\rVert_2^2}\begin{bmatrix}-1\\\sqrt{2}\\\sqrt{5}\end{bmatrix}\begin{bmatrix}-1&\sqrt{2}&\sqrt{5}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{3}{4}&\frac{\sqrt{2}}{4}&\frac{\sqrt{5}}{4}\\\frac{\sqrt{2}}{4}&\frac{1}{2}&-\frac{\sqrt{2}\sqrt{5}}{4}\\\frac{\sqrt{5}}{4}&-\frac{\sqrt{5}\sqrt{2}}{4}&-\frac{1}{4}\end{bmatrix}\\
|
||||
HA &= \begin{bmatrix}4&0.8&4.78\\0&-0.79&1.48\\0&-7.73&-1.99\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\end{document}
|
||||
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|
|
@ -0,0 +1,348 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
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|
||||
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|
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#9}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute $A^\top A$ for
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
2&-3\\
|
||||
-6&-9
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
0&1&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inverse of
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 = \begin{bmatrix}
|
||||
-2&-3\\
|
||||
-6&-4
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_1 = \begin{bmatrix}
|
||||
3&1\\
|
||||
-2&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Use this inverses to solve
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Note: it is not common to solve $Ax=b$ using matrix inversion.\\
|
||||
|
||||
Reasons:
|
||||
\begin{itemize}
|
||||
\item $Ax=b$ can be solvable when $A$ is not invertible.
|
||||
\item It is often slower / more costly
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the determinant of
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
2&-4&3\\
|
||||
1&1&1\\
|
||||
3&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&0&1\\
|
||||
2&3&2\\
|
||||
-1&3&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&3\\
|
||||
-5&-10&-15\\
|
||||
6&12&18
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Are they Invertible?
|
||||
\item Which of them has full rank? Which of them has lower rank and which one?
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
For what value $a$ the matrix is not invertible?
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&1\\
|
||||
2&a&4\\
|
||||
-3&1&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute and apply the Householder matrix which makes transforms the first column of
|
||||
to a multple of the first one-hot vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$ for
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
2&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
and for (hint: here subtracting is nicer)
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&-4&3\\
|
||||
3&1&1\\
|
||||
\sqrt{6}&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Verify that
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&0&0\\
|
||||
0&\cos(\alpha)&-\sin(\alpha)\\
|
||||
0&\sin(\alpha)&\cos(\alpha)
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Satisfies being an orthogonal matrix.
|
||||
|
||||
\solution
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
What is the cosine of the angle between
|
||||
|
||||
$$
|
||||
\left( 6,-6,-4,\sqrt{12} \right), \left( 6,4,2,\sqrt{25} \right)?
|
||||
$$
|
||||
|
||||
\solution
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Another $3\times3$ affine system
|
||||
|
||||
\begin{itemize}
|
||||
\item show the intermediate result when the first column is the one hot vector $\begin{bmatrix}1&0&0\end{bmatrix}$ for the first time
|
||||
\item show that the intermediate result when the matrix has row echelon form for the first time
|
||||
\item get the solution
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
2x-3y+2z &=-4\\
|
||||
7x+4.5y-1z &=16\\
|
||||
4x+3y+z &=2
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\end{document}
|
||||
Binary file not shown.
Binary file not shown.
|
|
@ -0,0 +1,705 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
\usepackage{multicol}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#10}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute $A^\top A$ for
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
2&-3\\
|
||||
-6&-9
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
0&1&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}
|
||||
2&-3\\
|
||||
-6&-9
|
||||
\end{bmatrix}\\
|
||||
A^\top &= \begin{bmatrix}
|
||||
2&-6\\
|
||||
-3&-9
|
||||
\end{bmatrix}\\
|
||||
A^\top A &= \begin{bmatrix}
|
||||
2&-6\\
|
||||
-3&-9
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2&-3\\
|
||||
-6&-9
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
40&48\\
|
||||
48&100
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}
|
||||
0&1&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}\\
|
||||
A^\top &= \begin{bmatrix}
|
||||
0&2&4\\
|
||||
1&0&3\\
|
||||
-1&2&1
|
||||
\end{bmatrix}\\
|
||||
A^\top A &= \begin{bmatrix}
|
||||
0&2&4\\
|
||||
1&0&3\\
|
||||
-1&2&1
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
0&1&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
20&12&8\\
|
||||
12&10&2\\
|
||||
8&2&6
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inverse of
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 = \begin{bmatrix}
|
||||
-2&-3\\
|
||||
-6&-4
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_1 = \begin{bmatrix}
|
||||
3&1\\
|
||||
-2&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Use this inverses to solve
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Note: it is not common to solve $Ax=b$ using matrix inversion.\\
|
||||
|
||||
Reasons:
|
||||
\begin{itemize}
|
||||
\item $Ax=b$ can be solvable when $A$ is not invertible.
|
||||
\item It is often slower / more costly
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A^{-1} &\implies \left[ \begin{array}{rr|rr}
|
||||
-2&-3&1&0\\
|
||||
-6&-4&0&1
|
||||
\end{array}\right]\\
|
||||
-\frac{1}{2}\rho_1, -\frac{1}{4}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{3}{2}&-\frac{1}{2}&0\\
|
||||
\frac{3}{2}&1&0&-\frac{1}{4}
|
||||
\end{array}\right]\\
|
||||
\frac{3}{2}\rho_1 - \rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{3}{2}&-\frac{1}{2}&0\\
|
||||
0&\frac{5}{4}&-\frac{3}{4}&\frac{1}{4}
|
||||
\end{array}\right]\\
|
||||
\frac{4}{5} \rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{3}{2}&-\frac{1}{2}&0\\
|
||||
0&1&-\frac{3}{5}&\frac{1}{5}
|
||||
\end{array}\right]\\
|
||||
\rho_1 - \frac{3}{2}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&0&\frac{2}{5}&-\frac{3}{10}\\
|
||||
0&1&-\frac{3}{5}&\frac{1}{5}
|
||||
\end{array}\right]\\\\
|
||||
A^{-1} &= \begin{bmatrix}
|
||||
\frac{2}{5}&-\frac{3}{10}\\
|
||||
\frac{3}{5}&\frac{1}{10}
|
||||
\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
A_0^{-1} A_0 x &= A_0^{-1} \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
x &= \begin{bmatrix}
|
||||
\frac{2}{5}\\
|
||||
\frac{3}{5}
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
0.9\\
|
||||
-1.6
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\pagebreak
|
||||
|
||||
|
||||
\part
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A^{-1} &\implies \left[ \begin{array}{rr|rr}
|
||||
3&1&1&0\\
|
||||
-2&2&0&1
|
||||
\end{array}\right]\\
|
||||
\rho_2 + \rho_1 &\implies \left[ \begin{array}{rr|rr}
|
||||
3&1&1&0\\
|
||||
1&3&1&1
|
||||
\end{array}\right]\\
|
||||
\frac{1}{3}\rho_1 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{1}{3}&\frac{1}{3}&0\\
|
||||
1&3&1&1
|
||||
\end{array}\right]\\
|
||||
\rho_1-\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{1}{3}&\frac{1}{3}&0\\
|
||||
0&1&\frac{1}{4}&\frac{3}{8}
|
||||
\end{array}\right]\\
|
||||
-\frac{3}{8}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{1}{3}&\frac{1}{3}&0\\
|
||||
0&1&\frac{1}{4}&\frac{3}{8}
|
||||
\end{array}\right]\\
|
||||
\rho_1 - \frac{1}{3}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&0&\frac{1}{4}&-\frac{1}{8}\\
|
||||
0&1&\frac{1}{4}&\frac{3}{8}
|
||||
\end{array}\right]\\
|
||||
A^{-1} &= \begin{bmatrix}
|
||||
\frac{1}{4}&-\frac{1}{8}\\
|
||||
\frac{1}{4}&\frac{3}{8}
|
||||
\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0 x &= \begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}\\
|
||||
A_0^{-1} A_0 x &= A_0^{-1} \begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}\\
|
||||
x &= \begin{bmatrix}
|
||||
\frac{1}{4}&-\frac{1}{8}\\
|
||||
\frac{1}{4}&\frac{3}{8}
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
\frac{9}{8}\\
|
||||
-\frac{11}{8}
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the determinant of
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
2&-4&3\\
|
||||
1&1&1\\
|
||||
3&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&0&1\\
|
||||
2&3&2\\
|
||||
-1&3&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&3\\
|
||||
-5&-10&-15\\
|
||||
6&12&18
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Are they Invertible?
|
||||
\item Which of them has full rank? Which of them has lower rank and which one?
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 0+16-6-(0+6+4)\\
|
||||
&= 0
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is not invertible.
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
2&-4&3\\
|
||||
1&1&1\\
|
||||
3&3&1
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 2-1+9-9-6+4\\
|
||||
&= -12
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is invertible
|
||||
|
||||
\end{multicols}
|
||||
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&0&1\\
|
||||
2&3&2\\
|
||||
-1&3&2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 6+0+6+3-6-0\\
|
||||
&= 9
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is invertible
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2&3\\
|
||||
-5&-10&-15\\
|
||||
6&12&18
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= -180-180-180+180+180+180\\
|
||||
&= 0
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is not invertible
|
||||
|
||||
\end{multicols}
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
For what value $a$ the matrix is not invertible?
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&1\\
|
||||
2&a&4\\
|
||||
-3&1&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
A matrix is not invertible if its determinant is zero.
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 2a-24+2+3a-4-8\\
|
||||
&= 5a-34\\
|
||||
5a-34 &= 0\\
|
||||
a &= \frac{34}{5} \approx 6.8
|
||||
\end{align*}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute and apply the Householder matrix which makes transforms the first column of
|
||||
to a multple of the first one-hot vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$ for
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
2&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
and for (hint: here subtracting is nicer)
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&-4&3\\
|
||||
3&1&1\\
|
||||
\sqrt{6}&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Verify that
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&0&0\\
|
||||
0&\cos(\alpha)&-\sin(\alpha)\\
|
||||
0&\sin(\alpha)&\cos(\alpha)
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Satisfies being an orthogonal matrix.
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 0 + (-\sin^2(\alpha)) + 0 - \cos^2(\alpha) -0 -0\\
|
||||
&=-(sin^2(\alpha) + cos^2(\alpha))\\
|
||||
&= -1
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
What is the cosine of the angle between
|
||||
|
||||
$$
|
||||
\left( 6,-6,-4,\sqrt{12} \right), \left( 6,4,2,\sqrt{25} \right)?
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle(\vec{a},\vec{b})) &= \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\|\vec{b}\|}\\
|
||||
&= \frac{6\cdot6 + (-6)\cdot4 + (-4)\cdot2 + \sqrt{12}\cdot\sqrt{25}}{\sqrt{6^2+(-6)^2+(-4)^2+\sqrt{12}^2}\sqrt{6^2+4^2+2^2+\sqrt{25}^2}}\\
|
||||
&= \frac{36-24-8+5\sqrt{12}}{\sqrt{100}\sqrt{81}}\\
|
||||
&= \frac{4}{90}\\
|
||||
\end{align*}
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Another $3\times3$ affine system
|
||||
|
||||
\begin{itemize}
|
||||
\item show the intermediate result when the first column is the one hot vector $\begin{bmatrix}1&0&0\end{bmatrix}$ for the first time
|
||||
\item show that the intermediate result when the matrix has row echelon form for the first time
|
||||
\item get the solution
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
2x-3y+2z &=-4\\
|
||||
7x+4.5y-1z &=16\\
|
||||
4x+3y+z &=2
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
&\left[ \begin{array}{rrr|r}
|
||||
2&-3&2&-4\\
|
||||
7&4.5&-1&16\\
|
||||
4&3&1&2
|
||||
\end{array} \right]\\
|
||||
\frac{1}{2} \rho_1 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
7&4.5&-1&16\\
|
||||
4&3&1&2
|
||||
\end{array} \right]\\
|
||||
(i) \rho_2 - 7\rho_1, \rho_3 - 4\rho_1 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&15&-8&30\\
|
||||
0&9&-3&10
|
||||
\end{array} \right]\\
|
||||
\frac{1}{15} \rho_2, -\frac{1}{3} \rho_3 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&1&-8/15&2\\
|
||||
0&-3&1&-\frac{10}{3}
|
||||
\end{array} \right]\\
|
||||
(ii) \rho_3 + 3\rho_2 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&1&-8/15&2\\
|
||||
0&0&1&-\frac{40}{9}
|
||||
\end{array} \right]\\
|
||||
\rho_2 + \frac{8}{15}\rho_3 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&1&0&-\frac{10}{27}\\
|
||||
0&0&1&-\frac{40}{9}
|
||||
\end{array} \right]\\
|
||||
\rho_1+\frac{3}{2}\rho_2 - \rho_3 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&0&0&-\frac{17}{9}\\
|
||||
0&1&0&-\frac{10}{27}\\
|
||||
0&0&1&-\frac{40}{9}
|
||||
\end{array} \right]\\
|
||||
\begin{bmatrix}
|
||||
x\\
|
||||
y\\
|
||||
z
|
||||
\end{bmatrix} &= \begin{bmatrix}
|
||||
-\frac{17}{9}\\
|
||||
-\frac{10}{27}\\
|
||||
-\frac{40}{9}
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{multicols}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\end{document}
|
||||
Binary file not shown.
|
|
@ -0,0 +1,915 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
\usepackage{multicol}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#10\ Solutions}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute $A^\top A$ for
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
2&-3\\
|
||||
-6&-9
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
0&1&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}
|
||||
2&-3\\
|
||||
-6&-9
|
||||
\end{bmatrix}\\
|
||||
A^\top &= \begin{bmatrix}
|
||||
2&-6\\
|
||||
-3&-9
|
||||
\end{bmatrix}\\
|
||||
A^\top A &= \begin{bmatrix}
|
||||
2&-6\\
|
||||
-3&-9
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2&-3\\
|
||||
-6&-9
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
40&48\\
|
||||
48&100
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix}
|
||||
0&1&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}\\
|
||||
A^\top &= \begin{bmatrix}
|
||||
0&2&4\\
|
||||
1&0&3\\
|
||||
-1&2&1
|
||||
\end{bmatrix}\\
|
||||
A^\top A &= \begin{bmatrix}
|
||||
0&2&4\\
|
||||
1&0&3\\
|
||||
-1&2&1
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
0&1&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
20&12&8\\
|
||||
12&10&2\\
|
||||
8&2&6
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inverse of
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 = \begin{bmatrix}
|
||||
-2&-3\\
|
||||
-6&-4
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_1 = \begin{bmatrix}
|
||||
3&1\\
|
||||
-2&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Use this inverses to solve
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
$$
|
||||
\begin{equation*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Note: it is not common to solve $Ax=b$ using matrix inversion.\\
|
||||
|
||||
Reasons:
|
||||
\begin{itemize}
|
||||
\item $Ax=b$ can be solvable when $A$ is not invertible.
|
||||
\item It is often slower / more costly
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A^{-1} &\implies \left[ \begin{array}{rr|rr}
|
||||
-2&-3&1&0\\
|
||||
-6&-4&0&1
|
||||
\end{array}\right]\\
|
||||
-\frac{1}{2}\rho_1, -\frac{1}{4}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{3}{2}&-\frac{1}{2}&0\\
|
||||
\frac{3}{2}&1&0&-\frac{1}{4}
|
||||
\end{array}\right]\\
|
||||
\frac{3}{2}\rho_1 - \rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{3}{2}&-\frac{1}{2}&0\\
|
||||
0&\frac{5}{4}&-\frac{3}{4}&\frac{1}{4}
|
||||
\end{array}\right]\\
|
||||
\frac{4}{5} \rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{3}{2}&-\frac{1}{2}&0\\
|
||||
0&1&-\frac{3}{5}&\frac{1}{5}
|
||||
\end{array}\right]\\
|
||||
\rho_1 - \frac{3}{2}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&0&\frac{2}{5}&-\frac{3}{10}\\
|
||||
0&1&-\frac{3}{5}&\frac{1}{5}
|
||||
\end{array}\right]\\\\
|
||||
A^{-1} &= \begin{bmatrix}
|
||||
\frac{2}{5}&-\frac{3}{10}\\
|
||||
\frac{3}{5}&\frac{1}{10}
|
||||
\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0 x = \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
A_0^{-1} A_0 x &= A_0^{-1} \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
x &= \begin{bmatrix}
|
||||
\frac{2}{5}\\
|
||||
\frac{3}{5}
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
0.9\\
|
||||
-1.6
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\pagebreak
|
||||
|
||||
|
||||
\part
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A^{-1} &\implies \left[ \begin{array}{rr|rr}
|
||||
3&1&1&0\\
|
||||
-2&2&0&1
|
||||
\end{array}\right]\\
|
||||
\rho_2 + \rho_1 &\implies \left[ \begin{array}{rr|rr}
|
||||
3&1&1&0\\
|
||||
1&3&1&1
|
||||
\end{array}\right]\\
|
||||
\frac{1}{3}\rho_1 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{1}{3}&\frac{1}{3}&0\\
|
||||
1&3&1&1
|
||||
\end{array}\right]\\
|
||||
\rho_1-\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{1}{3}&\frac{1}{3}&0\\
|
||||
0&1&\frac{1}{4}&\frac{3}{8}
|
||||
\end{array}\right]\\
|
||||
-\frac{3}{8}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&\frac{1}{3}&\frac{1}{3}&0\\
|
||||
0&1&\frac{1}{4}&\frac{3}{8}
|
||||
\end{array}\right]\\
|
||||
\rho_1 - \frac{1}{3}\rho_2 &\implies \left[ \begin{array}{rr|rr}
|
||||
1&0&\frac{1}{4}&-\frac{1}{8}\\
|
||||
0&1&\frac{1}{4}&\frac{3}{8}
|
||||
\end{array}\right]\\
|
||||
A^{-1} &= \begin{bmatrix}
|
||||
\frac{1}{4}&-\frac{1}{8}\\
|
||||
\frac{1}{4}&\frac{3}{8}
|
||||
\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A_0 x &= \begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}\\
|
||||
A_0^{-1} A_0 x &= A_0^{-1} \begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}\\
|
||||
x &= \begin{bmatrix}
|
||||
\frac{1}{4}&-\frac{1}{8}\\
|
||||
\frac{1}{4}&\frac{3}{8}
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
\frac{9}{8}\\
|
||||
-\frac{11}{8}
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\answer
|
||||
|
||||
\begin{multicols}{2}
|
||||
\textbf{A:}\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A_0) &= (-2)(-4)-(-3)(-6) = -10\\
|
||||
A_0^{-1} &= \frac{1}{\det(A_0)} \begin{bmatrix}
|
||||
-4&3\\
|
||||
6&-2
|
||||
\end{bmatrix}\\
|
||||
x &= A_0^{-1} \begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
&= -\frac{1}{10} \begin{bmatrix}
|
||||
-4&3\\
|
||||
6&-2
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3\\
|
||||
1
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
0.9\\
|
||||
-1.6
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
\textbf{B:}\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det{A_1} &= 3(2) - (-2)(1) = 8\\
|
||||
A_1^{-1} &= \frac{1}{8} \begin{bmatrix}
|
||||
2&-1\\
|
||||
2&3
|
||||
\end{bmatrix}\\
|
||||
x &= \frac{1}{8} \begin{bmatrix}
|
||||
2&-1\\
|
||||
2&3
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2\\
|
||||
-5
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
\frac{9}{8}\\
|
||||
-\frac{11}{8}
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\textbf{FORMULA:}\\
|
||||
|
||||
$$
|
||||
A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}
|
||||
d&-b\\-c&a
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the determinant of
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
2&-4&3\\
|
||||
1&1&1\\
|
||||
3&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&0&1\\
|
||||
2&3&2\\
|
||||
-1&3&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&3\\
|
||||
-5&-10&-15\\
|
||||
6&12&18
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\begin{itemize}
|
||||
\item Are they Invertible?
|
||||
\item Which of them has full rank? Which of them has lower rank and which one?
|
||||
\end{itemize}
|
||||
|
||||
\textbf{Notes:} \\
|
||||
|
||||
\begin{itemize}
|
||||
\item Rank is the number of linearly independent rows or columns.
|
||||
\item A matrix is invertible if they have full rank.
|
||||
\end{itemize}
|
||||
|
||||
\answer
|
||||
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
4&3&1
|
||||
\end{bmatrix} \rightarrow \begin{bmatrix}
|
||||
1&2&-1\\
|
||||
0&1&-1\\
|
||||
0&0&0
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 0+16-6-(0+6+4)\\
|
||||
&= 0
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is not invertible. Rank is 2.
|
||||
|
||||
\columnbreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
2&-4&3\\
|
||||
1&1&1\\
|
||||
3&3&1
|
||||
\end{bmatrix} \rightarrow \begin{bmatrix}
|
||||
1&1&-1\\0&1&-\frac{5}{6}\\0&0&1
|
||||
\end{bmatrix}
|
||||
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 2-1+9-9-6+4\\
|
||||
&= -12
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is invertible. Rank is 3.
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&0&1\\
|
||||
2&3&2\\
|
||||
-1&3&2
|
||||
\end{bmatrix} \rightarrow \begin{bmatrix}
|
||||
1&0&1\\
|
||||
0&1&1\\
|
||||
0&0&1
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 6+0+6+3-6-0\\
|
||||
&= 9
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is invertible. Rank is 3.
|
||||
|
||||
\columnbreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2&3\\
|
||||
-5&-10&-15\\
|
||||
6&12&18
|
||||
\end{bmatrix} \rightarrow \begin{bmatrix}
|
||||
1&2&3\\
|
||||
0&0&0\\
|
||||
0&0&0
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= -180-180-180+180+180+180\\
|
||||
&= 0
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
This matrix is not invertible and has rank 1.
|
||||
|
||||
\end{multicols}
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
For what value $a$ the matrix is not invertible?
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&1\\
|
||||
2&a&4\\
|
||||
-3&1&2
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
A matrix is not invertible if its determinant is zero.
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\det(A) &= 2a-24+2+3a-4-8\\
|
||||
&= 5a-34\\
|
||||
5a-34 &= 0\\
|
||||
a &= \frac{34}{5} \approx 6.8
|
||||
\end{align*}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute and apply the Householder matrix which makes transforms the first column of
|
||||
to a multple of the first one-hot vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$ for
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&2&-1\\
|
||||
2&0&2\\
|
||||
2&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
and for (hint: here subtracting is nicer)
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&-4&3\\
|
||||
3&1&1\\
|
||||
\sqrt{6}&3&1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
\textbf{Formula:} \\
|
||||
$$
|
||||
\begin{align*}
|
||||
u &= x \pm \left\lVert x \right\rVert_2 z\\
|
||||
H_u &= I - \frac{2}{u \cdot u} uu^T\\
|
||||
\text{Properties of H: }\\
|
||||
H_ux &= \mp \left\lVert x \right\rVert_2 z \text{ H maps x to $\mp \left\lvert x \right\rvert z$ given $u = x \pm \lvert x \rvert z$}\\
|
||||
H_u^\top H_u &= I \text{ H is an orthogonal matrix}\\
|
||||
H_u &= H_u^\top \text{ H is a symmetric matrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\textbf{Example:}\\
|
||||
|
||||
Here $A = \begin{bmatrix}3&1\\4&4\end{bmatrix}$
|
||||
|
||||
$x = \begin{bmatrix}3\\4\end{bmatrix}$
|
||||
|
||||
$\right \lVert x \right \rVert_2 = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
u &= \begin{bmatrix}3\\4\end{bmatrix} \pm 5 \begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}-2\\4\end{bmatrix}\text{ (Choose Minus)}\\
|
||||
H &= \begin{bmatrix}1&0\\0&1\end{bmatrix}-\frac{2}{(-2)^2+4^2}\begin{bmatrix}-2\\4\end{bmatrix}\begin{bmatrix}-2&4\end{bmatrix}\\
|
||||
&= \begin{bmatrix}1&0\\0&1\end{bmatrix}-\frac{2}{20}\begin{bmatrix}4&-8\\-8&16\end{bmatrix}\\
|
||||
&= \begin{bmatrix}0.6&0.8\\0.8&-0.6\end{bmatrix}\\\\
|
||||
HA &= \begin{bmatrix}0.6&0.8\\0.8&-0.6\end{bmatrix}\begin{bmatrix}3&1\\4&4\end{bmatrix}\\
|
||||
&= \begin{bmatrix}5&3.8\\0&-1.6\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
u &= \begin{bmatrix}3\\4\end{bmatrix} \pm 5 \begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}8\\4\end{bmatrix}\text{ (Choose Positive Sign)}\\
|
||||
H &= \begin{bmatrix}1&0\\0&1\end{bmatrix}-\frac{2}{(8)^2+4^2}\begin{bmatrix}8\\4\end{bmatrix}\begin{bmatrix}8&4\end{bmatrix}\\
|
||||
&= \begin{bmatrix}1&0\\0&1\end{bmatrix}-\frac{2}{80}\begin{bmatrix}64&32\\32&16\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-0.6&-0.8\\-0.8&0.6\end{bmatrix}\\\\
|
||||
HA &= \begin{bmatrix}-0.6&-0.8\\-0.8&0.6\end{bmatrix}\begin{bmatrix}3&1\\4&4\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-5&-3.8\\0&-1.6\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
u &= \begin{bmatrix}1\\2\\2\end{bmatrix} \pm \lVert \begin{pmatrix}1,2,2\end{pmatrix} \rVert \begin{bmatrix}1\\0\\0\end{bmatrix} \\
|
||||
&= \begin{bmatrix}1\\2\\2\end{bmatrix} \pm \sqrt{1^2 + 2^2 + 2^2} \begin{bmatrix}1\\0\\0\end{bmatrix}\\
|
||||
&= \begin{bmatrix}1\\2\\2\end{bmatrix} - 3 \begin{bmatrix}1\\0\\0\end{bmatrix}\\
|
||||
&= \begin{bmatrix}1\\2\\2\end{bmatrix} - \begin{bmatrix}3\\0\\0\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-2\\2\\2\end{bmatrix}\\
|
||||
H &= I - \frac{2}{12}uu^\top = I - \frac{1}{6}\begin{bmatrix}-2\\2\\2\end{bmatrix} \begin{bmatrix}-2&2&2\end{bmatrix} \\
|
||||
&= I - \frac{1}{6}\begin{bmatrix}4&-4&-4\\-4&4&4\\-4&4&4\end{bmatrix}\\
|
||||
&= \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix} - \frac{1}{6}\begin{bmatrix}4&-4&-4\\-4&4&4\\-4&4&4\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\\frac{2}{3}&\frac{1}{3}&\frac{1}{3}\\ \frac{2}{3}&\frac{1}{3}&\frac{1}{3}\end{bmatrix}\\
|
||||
HA &= \begin{bmatrix}3&2.67&1.67\\0&-0.67&-0.67\\0&2.33&-1.67\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
u &= x - \lVert x \lVert z = \begin{bmatrix}1\\3\\\sqrt{6}\end{bmatrix} - 4\begin{bmatrix} 1\\0\\0 \end{bmatrix} = \begin{bmatrix} -3\\3\\\sqrt{6}\end{bmatrix}\\
|
||||
H &= I - \frac{2}{u\cdot u} uu^\top\\
|
||||
&= I - \frac{2}{24}\begin{bmatrix}-3\\3\\\sqrt{6}\end{bmatrix}\begin{bmatrix}-3&3&\sqrt{6}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}\frac{1}{4}&\frac{3}{4}&\frac{\sqrt{6}}{4}\\\frac{3}{4}&\frac{1}{4}&-\frac{\sqrt{6}}{4}\\\frac{\sqrt{6}}{4}&-\frac{\sqrt{6}}{4}&\frac{1}{2}\end{bmatrix}\\
|
||||
HA &= \begin{bmatrix} 4&*&*\\0&*&*\\0&*&*\end{bmatrix}\\\\
|
||||
u &= x + \lVert x \lVert z = \begin{bmatrix}1\\3\\\sqrt{6}\end{bmatrix} + 4\begin{bmatrix} 1\\0\\0 \end{bmatrix} = \begin{bmatrix} 5\\3\\\sqrt{6}\end{bmatrix}\\
|
||||
H &= I - \frac{2}{u\cdot u} uu^\top\\
|
||||
&= \begin{bmatrix}-\frac{1}{4}&-\frac{3}{4}&-\frac{\sqrt{6}}{4}\\-\frac{3}{4}&\frac{11}{20}&-\frac{3\sqrt{6}}{70}\\-\frac{\sqrt{6}}{4}&-\frac{3\sqrt{6}}{70}&\frac{7}{10}\end{bmatrix}\\
|
||||
HA &= \begin{bmatrix} -4&*&*\\0&*&*\\0&*&*\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
|
||||
\end{multicols}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Verify that
|
||||
|
||||
$$
|
||||
\begin{equation*}
|
||||
A = \begin{bmatrix}
|
||||
1&0&0\\
|
||||
0&\cos(\alpha)&-\sin(\alpha)\\
|
||||
0&\sin(\alpha)&\cos(\alpha)
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
$$
|
||||
|
||||
Satisfies being an orthogonal matrix.
|
||||
|
||||
Note: Cannot use determinant to verify.
|
||||
|
||||
\answer
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A^\top A &= \begin{bmatrix}
|
||||
1&0&0\\
|
||||
0&\cos(\alpha)&\sin(\alpha)\\
|
||||
0&-\sin(\alpha)&\cos(\alpha)
|
||||
\end{bmatrix} \begin{bmatrix}
|
||||
1&0&0\\
|
||||
0&\cos(\alpha)&-\sin(\alpha)\\
|
||||
0&\sin(\alpha)&\cos(\alpha)
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
1&0&0\\
|
||||
0&\cos^2(\alpha) + \sin^2(\alpha)&-\sin(\alpha)\cos(\alpha) + \sin(\alpha)\cos(\alpha)\\
|
||||
0&-\sin(\alpha)\cos(\alpha) + \sin(\alpha)\cos(\alpha)&\sin^2(\alpha) + \cos^2(\alpha)
|
||||
\end{bmatrix}\\
|
||||
&= \begin{bmatrix}
|
||||
1&0&0\\
|
||||
0&1&0\\
|
||||
0&0&1
|
||||
\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
Thus A is an orthogonal matrix.
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
What is the cosine of the angle between
|
||||
|
||||
$$
|
||||
\left( 6,-6,-4,\sqrt{12} \right), \left( 6,4,2,\sqrt{25} \right)?
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle(\vec{a},\vec{b})) &= \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\|\vec{b}\|}\\
|
||||
&= \frac{6\cdot6 + (-6)\cdot4 + (-4)\cdot2 + \sqrt{12}\cdot\sqrt{25}}{\sqrt{6^2+(-6)^2+(-4)^2+\sqrt{12}^2}\sqrt{6^2+4^2+2^2+\sqrt{25}^2}}\\
|
||||
&= \frac{36-24-8+5\sqrt{12}}{\sqrt{100}\sqrt{81}}\\
|
||||
&= \frac{4 + 5\sqrt{12}}{90}\\
|
||||
$\approx 0.24
|
||||
\end{align*}
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Another $3\times3$ affine system
|
||||
|
||||
\begin{itemize}
|
||||
\item show the intermediate result when the first column is the one hot vector $\begin{bmatrix}1&0&0\end{bmatrix}$ for the first time
|
||||
\item show that the intermediate result when the matrix has row echelon form for the first time
|
||||
\item get the solution
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
2x-3y+2z &=-4\\
|
||||
7x+4.5y-1z &=16\\
|
||||
4x+3y+z &=2
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
&\left[ \begin{array}{rrr|r}
|
||||
2&-3&2&-4\\
|
||||
7&4.5&-1&16\\
|
||||
4&3&1&2
|
||||
\end{array} \right]\\
|
||||
\frac{1}{2} \rho_1 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
7&4.5&-1&16\\
|
||||
4&3&1&2
|
||||
\end{array} \right]\\
|
||||
(i) \rho_2 - 7\rho_1, \rho_3 - 4\rho_1 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&15&-8&30\\
|
||||
0&9&-3&10
|
||||
\end{array} \right]\\
|
||||
\frac{1}{15} \rho_2, -\frac{1}{3} \rho_3 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&1&-8/15&2\\
|
||||
0&-3&1&-\frac{10}{3}
|
||||
\end{array} \right]\\
|
||||
(ii) \rho_3 + 3\rho_2 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&1&-8/15&2\\
|
||||
0&0&1&-\frac{40}{9}
|
||||
\end{array} \right]\\
|
||||
\rho_2 + \frac{8}{15}\rho_3 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&-3/2&1&-2\\
|
||||
0&1&0&-\frac{10}{27}\\
|
||||
0&0&1&-\frac{40}{9}
|
||||
\end{array} \right]\\
|
||||
\rho_1+\frac{3}{2}\rho_2 - \rho_3 &\implies \left[ \begin{array}{rrr|r}
|
||||
1&0&0&-\frac{17}{9}\\
|
||||
0&1&0&-\frac{10}{27}\\
|
||||
0&0&1&-\frac{40}{9}
|
||||
\end{array} \right]\\
|
||||
\begin{bmatrix}
|
||||
x\\
|
||||
y\\
|
||||
z
|
||||
\end{bmatrix} &= \begin{bmatrix}
|
||||
-\frac{17}{9}\\
|
||||
-\frac{10}{27}\\
|
||||
-\frac{40}{9}
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{multicols}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\end{document}
|
||||
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|
|
@ -0,0 +1,246 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
\usepackage{multicol}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#11}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item Compute the eigenvalues and eigenvectors for the matrices below.
|
||||
\item For one of these matrices compute the matrix $P$ such that $P^{-1}DP = A$, and verify that $PAP^{-1}$ is the diagonal matrix of the eigenvalues
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix} 4&\sqrt{15}\\\sqrt{15}&2 \end{bmatrix} \\
|
||||
A &= \begin{bmatrix} 2&-2\\-3&1 \end{bmatrix} \\
|
||||
A &= \begin{bmatrix} 2&-4\\4&-6 \end{bmatrix} \\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item Compute an eigenvector for the eigenvalue $x = 3$ for the below $(3,3)$-matrix. Note: nobody asks you to compute its characteristic polynomial or to get all of its eigenvalues (Prof did it)
|
||||
\item Validate that the found eigenvector $v$ is indeed the correct one, that is, that $Av = 3v$ holds.
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix} 1&0&1\\0&1&-2\\2&4&4 \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
$$
|
||||
A = \begin{bmatrix} 1&1\\0&1 \end{bmatrix}
|
||||
$$
|
||||
\begin{itemize}
|
||||
\item Show that this matrix has only the eigenvalue $1$, twice.
|
||||
\item Prove that there cannot exist any matrix P such that $P^{-1}DP = A$. Hint: You know how $D$ in $P^{-1}DP$ must look like
|
||||
\item Find an eigenvector
|
||||
\end{itemize}
|
||||
Bonus knowldge: Shear matrices have an eigenspace of dimensionality $d-1$. In the above case the set of all eigenvectors must be $cv, c \in \mathbb{R}$ for some vector $v$.\\
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Show that $$\begin{bmatrix} 2&-4\\\frac{13}{4}&-4 \end{bmatrix}$$
|
||||
\begin{itemize}
|
||||
\item has no real eigenvalue
|
||||
\item Bonus: what are its complex-valued eigenvalues?
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Bonus Matrix:
|
||||
\begin{itemize}
|
||||
\item Get its eigenvalues and eigenvectors
|
||||
\end{itemize}
|
||||
Note: This is a symmetric one, so you can epect 2 eigenvalues and orthogonal eigenspace
|
||||
$$
|
||||
A = \begin{bmatrix}-2&\sqrt{24}\\\sqrt{24}&8 \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item Use numpy to get its eigenvalues and eigenvectors
|
||||
\item Solve $Ax = (3,17,1/3)$ using numpy
|
||||
\item Not in exam:
|
||||
\end{itemize}
|
||||
Compute the characteristic polynomial for
|
||||
$$
|
||||
A = \begin{bmatrix}1&2&4\\2&-2&1\\3&1&3 \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
What is the cosine of the angle between
|
||||
$$(6,-6,-4,\sqrt{12}),(6,4,2,\sqrt{25})$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Another 3x3 affine system
|
||||
\begin{itemize}
|
||||
\item Show the intermediate result when the first column is the one hot vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$ for the first time
|
||||
\item Show the intermediate result when the matrix has row echelon form for the first time
|
||||
\item Get the solution
|
||||
\end{itemize}
|
||||
$$
|
||||
\begin{align*}
|
||||
2x-3y+2z &=-4\\
|
||||
7x+4.5y-1z &=16\\
|
||||
4x +3y+z &=2
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\end{document}
|
||||
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|
|
@ -0,0 +1,541 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
\usepackage{multicol}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#11}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\def\therefore{\boldsymbol{\text{ }
|
||||
\leavevmode
|
||||
\lower0.4ex\hbox{$\cdot$}
|
||||
\kern-.5em\raise0.7ex\hbox{$\cdot$}
|
||||
\kern-0.55em\lower0.4ex\hbox{$\cdot$}
|
||||
\thinspace\text{ }}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item Compute the eigenvalues and eigenvectors for the matrices below.
|
||||
\item For one of these matrices compute the matrix $P$ such that $P^{-1}DP = A$, and verify that $PAP^{-1}$ is the diagonal matrix of the eigenvalues
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A &= \begin{bmatrix} 4&\sqrt{15}\\\sqrt{15}&2 \end{bmatrix} \\
|
||||
A &= \begin{bmatrix} 2&-2\\-3&1 \end{bmatrix} \\
|
||||
A &= \begin{bmatrix} 2&-4\\4&-6 \end{bmatrix} \\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix} 4&\sqrt{15}\\\sqrt{15}&2 \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\det{\begin{bmatrix} 4-\lambda&\sqrt{15}\\\sqrt{15}&2-\lambda \end{bmatrix}} = 0
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
f(\lambda) &= \lambda^2 - 6\lambda -7 = 0\\
|
||||
&(\lambda - 7)(\lambda + 1) = 0\\
|
||||
\lambda &= 7, -1
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\begin{multicols}{2}
|
||||
$$
|
||||
\begin{align*}
|
||||
\lambda_1 &= -1,\\
|
||||
A-(-1)I &= \begin{bmatrix} 5&\sqrt{15}\\\sqrt{15}&3 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} \sqrt{15} & 3\\0&0\end{bmatrix} \\
|
||||
\sqrt{15}x_0+3x_1 &= 0\\
|
||||
x_0 &= -\frac{3}{\sqrt{15}}x_1\\
|
||||
\begin{bmatrix}x_0\\x_1\end{bmatrix} &= x_1\begin{bmatrix}-\frac{3}{\sqrt{15}}\\1\end{bmatrix}\\
|
||||
\therefore v_1 &= \begin{bmatrix}-3\\\sqrt{15}\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\lambda_1 &= 7,\\
|
||||
A-(7)I &= \begin{bmatrix} -3&\sqrt{15}\\\sqrt{15}&-5 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} \sqrt{15} & -5\\0&0\end{bmatrix} \\
|
||||
\sqrt{15}x_0-5x_1 &= 0\\
|
||||
x_0 &= \frac{5}{\sqrt{15}}x_1\\
|
||||
\begin{bmatrix}x_0\\x_1\end{bmatrix} &= x_1\begin{bmatrix}\frac{5}{\sqrt{15}}\\1\end{bmatrix}\\
|
||||
\therefore v_1 &= \begin{bmatrix}5\\\sqrt{15}\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{multicols}
|
||||
|
||||
Case: Symmetric Matix, 2 different eigenvalues, 2 eigenvectors\\
|
||||
Eigenspaces are orthogonal, Check: $v_0\cdotv_1 = -15 + \sqrt{15}\sqrt{15} = 0$\\
|
||||
|
||||
|
||||
\textbf{Eigendecomposition}\\
|
||||
|
||||
$\lambda_1=-1, v_1 = \begin{bmatrix}-3\\\sqrt{15}\end{bmatrix}, \qquad \lambda_2=7, v_2 = \begin{bmatrix}5\\\sqrt{15}\end{bmatrix}$\\
|
||||
|
||||
$p^{-1} = \begin{bmatrix}v_1&v_2\end{bmatrix} = \begin{bmatrix}-3&5\\\sqrt{15}&\sqrt{15}\end{bmatrix}$\\
|
||||
$D = \begin{bmatrix}-1&0\\0&7\end{bmatrix}$\\
|
||||
$p = (p^{-1})^{-1} = \frac{1}{-3\sqrt{15}-5\sqrt{15}} \begin{bmatrix}\sqrt{15}&-5\\-\sqrt{15}&-3\end{bmatrix} = \begin{bmatrix}-\frac{1}{8}&\frac{5}{8\sqrt{15}}\\\frac{1}{8}&\frac{3}{8\sqrt{15}}\end{bmatrix}$\\
|
||||
Verify:
|
||||
$$
|
||||
\begin{align*}
|
||||
P^{-1}DP = \begin{bmatrix}-3&5\\\sqrt{15}&\sqrt{15}\end{bmatrix}\begin{bmatrix}-1&0\\0&7\end{bmatrix}\begin{bmatrix}-\frac{1}{8}&\frac{5}{8\sqrt{15}}\\\frac{1}{8}&\frac{3}{8\sqrt{15}}\end{bmatrix} = \begin{bmatrix}4&\sqrt{15}\\\sqrt{15}&2\end{bmatrix} = A\\
|
||||
PAP^{-1} = \begin{bmatrix}-\frac{1}{8}&\frac{5}{8\sqrt{15}}\\\frac{1}{8}&\frac{3}{8\sqrt{15}}\end{bmatrix}\begin{bmatrix}4&\sqrt{15}\\\sqrt{15}&2\end{bmatrix}\begin{bmatrix}-\frac{1}{8}&\frac{5}{8\sqrt{15}}\\\frac{1}{8}&\frac{3}{8\sqrt{15}}\end{bmatrix} = \begin{bmatrix}-1&0\\0&7\end{bmatrix} = D
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix} 2&-2\\-3&1 \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\det{\begin{bmatrix} 2-\lambda&-2\\-3&1-\lambda \end{bmatrix}} = 0
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
f(\lambda) &= \lambda^2 - 3\lambda - 4 = 0\\
|
||||
&(\lambda - 4)(\lambda + 1) = 0\\
|
||||
\therefore \lambda &= 4, -1
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\lambda_1 &= -1,\\
|
||||
A-(-1)I &= \begin{bmatrix} 3&-2\\-3&2 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} 3 & -2\\0&0\end{bmatrix} \\
|
||||
3x_0-2x_1 &= 0\\
|
||||
x_0 &= \frac{2}{3}x_1\\
|
||||
\begin{bmatrix}x_0\\x_1\end{bmatrix} &= x_1\begin{bmatrix}\frac{2}{3}\\1\end{bmatrix}\\
|
||||
\therefore v_1 &= \begin{bmatrix}2\\3\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\lambda_2 &= 4,\\
|
||||
A-(4)I &= \begin{bmatrix} -2&-2\\-3&-3 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} -2 & -2\\0&0\end{bmatrix} \\
|
||||
-2x_0-2x_1 &= 0\\
|
||||
x_0 &= -x_1\\
|
||||
\begin{bmatrix}x_0\\x_1\end{bmatrix} &= x_1\begin{bmatrix}-1\\1\end{bmatrix}\\
|
||||
\therefore v_2 &= \begin{bmatrix}1\\-1\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{multicols}
|
||||
|
||||
Eigenspaces are not orthogonal, but matrix is not symetric\\
|
||||
So even though we get 2 eigenvalues and 2 eigenvectors, it is okay that the eigenvectors are not orthogonal to each other.\\
|
||||
|
||||
\textbf{Eigendecomposition}\\
|
||||
|
||||
$\lambda_1=-1, v_1 = \begin{bmatrix}2\\3\end{bmatrix}, \qquad \lambda_2=4, v_2 = \begin{bmatrix}1\\-1\end{bmatrix}$\\
|
||||
|
||||
$p^{-1} = \begin{bmatrix}v_1&v_2\end{bmatrix} = \begin{bmatrix}2&1\\3&-1\end{bmatrix}$\\
|
||||
|
||||
$D = \begin{bmatrix}-1&0\\0&4\end{bmatrix}$\\
|
||||
|
||||
$p = (p^{-1})^{-1} = \frac{1}{-2-3} \begin{bmatrix}-1&-1\\-3&2\end{bmatrix} = \begin{bmatrix}\frac{1}{5}&\frac{1}{5}\\\frac{3}{5}&-\frac{2}{5}\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
P^{-1}DP &= \begin{bmatrix}2&1\\3&-1\end{bmatrix}\begin{bmatrix}-1&0\\0&4\end{bmatrix}\begin{bmatrix}\frac{1}{5}&\frac{1}{5}\\\frac{3}{5}&-\frac{2}{5}\end{bmatrix} = \begin{bmatrix}2&-2\\-3&1\end{bmatrix} = A\\
|
||||
PAP^{-1} &= \begin{bmatrix}\frac{1}{5}&\frac{1}{5}\\\frac{3}{5}&-\frac{2}{5}\end{bmatrix}\begin{bmatrix}2&-2\\-3&1\end{bmatrix}\begin{bmatrix}2&1\\3&-1\end{bmatrix} = \begin{bmatrix}-1&0\\0&4\end{bmatrix} = D
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix} 2&-4\\4&-6 \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\det{\begin{bmatrix} 2-\lambda&-4\\4&-6-\lambda \end{bmatrix}} = 0
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
f(\lambda) &= \lambda^2 +4\lambda +4 = 0\\
|
||||
&(\lambda +2)(\lambda +2) = 0\\
|
||||
\lambda &= -2, -2
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\lambda_1 &= -2,\\
|
||||
A-(-2)I &= \begin{bmatrix} 4&-4\\4&-4 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} 4 & -4\\0&0\end{bmatrix} \\
|
||||
4x_0-4x_1 &= 0\\
|
||||
x_0 &= x_1\\
|
||||
\begin{bmatrix}x_0\\x_1\end{bmatrix} &= x_1\begin{bmatrix}1\\1\end{bmatrix}\\
|
||||
\therefore v_1 &= \begin{bmatrix}1\\1\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Eigenspace just one dim, but matrix is not symmetric\\
|
||||
So it is okay that we do not have 2 eigenvalues nor 2 eigenvectors. (We cant compute D since there is only 1 eigenvector)\\
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item Compute an eigenvector for the eigenvalue $x = 3$ for the below $(3,3)$-matrix. Note: nobody asks you to compute its characteristic polynomial or to get all of its eigenvalues (Prof did it)
|
||||
\item Validate that the found eigenvector $v$ is indeed the correct one, that is, that $Av = 3v$ holds.
|
||||
\end{itemize}
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix} 1&0&1\\0&1&-1\\2&4&4 \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A-\lambda I &= \begin{bmatrix} 1-\lambda&0&1\\0&1-\lambda&-1\\2&4&4-\lambda \end{bmatrix} \\
|
||||
f{\lambda} &= - \lambda^3 + +6\lambda^2 - 11\lambda + 6 = 0\\
|
||||
f(3) &= 3^3 - 6\cdot3^2 + 11\cdot3 + 6 = 0\\
|
||||
\end{align*}
|
||||
$$
|
||||
When $\lambda = 3&,\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A-3I &= \begin{bmatrix} -2&0&1\\0&-2&-1\\2&4&1 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} -2&0&1\\0&-2&-1\\0&0&0 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$1^{st}$ row: $-2x_0 + x_2 = 0 \Rightarrow x_0 = \frac{1}{2}x_2$\\
|
||||
$2^{nd}$ row: $-2x_1 - x_2 = 0 \Rightarrow x_1 = -\frac{1}{2}x_2$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix}x_0\\x_1\\x_2\end{bmatrix} &= x_2 \begin{bmatrix}\frac{1}{2}\\-\frac{1}{2}\\1\end{bmatrix}\\
|
||||
\therefore v &= \begin{bmatrix}1\\-1\\2\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Verify:
|
||||
$$
|
||||
\begin{bmatrix}1&0&1\\0&1&-1\\2&4&4 \end{bmatrix}\begin{bmatrix}1\\-1\\2\end{bmatrix} = 3\begin{bmatrix}1\\-1\\2\end{bmatrix}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
$$
|
||||
A = \begin{bmatrix} 1&1\\0&1 \end{bmatrix}
|
||||
$$
|
||||
\begin{itemize}
|
||||
\item Show that this matrix has only the eigenvalue $1$, twice.
|
||||
\item Prove that there cannot exist any matrix P such that $P^{-1}DP = A$. Hint: You know how $D$ in $P^{-1}DP$ must look like
|
||||
\item Find an eigenvector
|
||||
\end{itemize}
|
||||
Bonus knowldge: Shear matrices have an eigenspace of dimensionality $d-1$. In the above case the set of all eigenvectors must be $cv, c \in \mathbb{R}$ for some vector $v$.\\
|
||||
\solution
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix} 1&1\\0&1 \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
A - \lambda I = \begin{bmatrix} 1-\lambda&1\\0&1-\lambda \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
f(\lambda) = (\lambda -1)^2 = 0
|
||||
$$
|
||||
|
||||
So only $\lambda = 1$ is a solution. It has 2 eigenvalues equal to 1 (\textit{eigenvalue `1` has algebraic mutiplicity of 2})\\
|
||||
|
||||
Suppose P exists such that $P^{-1}DP = A$, then $D = \begin{bmatrix}1&0\\0&1\end{bmatrix}$\\
|
||||
$P^{-1}DP = P^{-1}IP = P^{-1}P = I$\\
|
||||
But A is not the identity matrix. So there cannot exist any P as assumed above (Proof by Contradiction)
|
||||
|
||||
$$
|
||||
A-I = \begin{bmatrix} o&1\\0&0 \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$0\cdot x_0 + 1 \cdot x_1 = 0 \Rightarrow x_1 = 0$$
|
||||
$$\begin{bmatrix}x_0\\x_1\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}$$
|
||||
$$\therefore \text{ eigenvector } v = \begin{bmatrix}1\\0\end{bmatrix}$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Show that $$\begin{bmatrix} 2&-4\\\frac{13}{4}&-4 \end{bmatrix}$$
|
||||
\begin{itemize}
|
||||
\item has no real eigenvalue
|
||||
\item Bonus: what are its complex-valued eigenvalues?
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
A-\lambda I = \begin{bmatrix} 2-\lambda&-4\\ \frac{13}{4}&-4-\lambda \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$f(\lambda) = (\lambda - 2)(\lambda + 4) = \lambda^2 +2\lambda +5 = 0$$
|
||||
$$\lambda = \frac{-(2)\pm \sqrt{(2)^2 - 4\cdot 5}}{2} = \frac{-2\pm 2\sqrt{5}}{2} = \frac{-2\pm\sqrt{-1}\sqrt{16}}{2} = \frac{-2\pm i(4)}{2}=-1\pm2i$$
|
||||
|
||||
There are no real eigenvalues.
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Bonus Matrix:
|
||||
\begin{itemize}
|
||||
\item Get its eigenvalues and eigenvectors
|
||||
\end{itemize}
|
||||
Note: This is a symmetric one, so you can expect 2 eigenvalues and orthogonal eigenspace
|
||||
$$
|
||||
A = \begin{bmatrix}-2&\sqrt{24}\\\sqrt{24}&8 \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}-2&\sqrt{24}\\\sqrt{24}&8 \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
A-\lambda I = \begin{bmatrix}-2-\lambda&\sqrt{24}\\\sqrt{24}&8-\lambda \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
f(\lambda) = (\lambda + 2)(\lambda - 8) = \lambda^2 - 6\lambda - 40 = (\lambda - 10)(\lambda + 4) = 0
|
||||
$$
|
||||
So $\lambda = 10$ or $\lambda = -4$
|
||||
|
||||
\begin{multicols}{2}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A - -4I &= \begin{bmatrix}2&\sqrt{24}\\\sqrt{24}&12 \end{bmatrix}\\
|
||||
&= \begin{bmatrix}\sqrt{24}&12\\0&0\end{bmatrix}\\
|
||||
\sqrt{24}x_0+12x_1 &= 0\\
|
||||
x_0 &= -\frac{12}{\sqrt{24}}x_1\\
|
||||
\begin{bmatrix}x_0\\x_1\end{bmatrix} &= \begin{bmatrix}-\frac{12}{\sqrt{24}}\\1\end{bmatrix}\\
|
||||
\therefore v_1 &= \begin{bmatrix}-12\\\sqrt{24}\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\columnbreak
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
A - 10I &= \begin{bmatrix}-12&\sqrt{24}\\\sqrt{24}&-2 \end{bmatrix}\\
|
||||
&= \begin{bmatrix}-12&\sqrt{24}\\0&0\end{bmatrix}\\
|
||||
-12x_0+\sqrt{24}x_1 &= 0\\
|
||||
x_0 &= \frac{\sqrt{24}}{12}x_1\\
|
||||
\begin{bmatrix}x_0\\x_1\end{bmatrix} &= \begin{bmatrix}\frac{\sqrt{24}}{12}\\1\end{bmatrix}\\
|
||||
\therefore v_2 &= \begin{bmatrix}\sqrt{24}\\12\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{multicols}
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item Use numpy to get its eigenvalues and eigenvectors
|
||||
\item Solve $Ax = (3,17,1/3)$ using numpy
|
||||
\item Not in exam:
|
||||
\end{itemize}
|
||||
Compute the characteristic polynomial for
|
||||
$$
|
||||
A = \begin{bmatrix}1&2&4\\2&-2&1\\3&1&3 \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
A-\lambda I = \begin{bmatrix}1-\lambda&2&4\\2&-2-\lambda&1\\3&1&3-\lambda \end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
f(\lambda) = -\lambda^3 + 2\lambda^2 +22\lambda+19
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
What is the cosine of the angle between
|
||||
$$(6,-6,-4,\sqrt{12}),(6,4,2,\sqrt{25})$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Another 3x3 affine system
|
||||
\begin{itemize}
|
||||
\item Show the intermediate result when the first column is the one hot vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$ for the first time
|
||||
\item Show the intermediate result when the matrix has row echelon form for the first time
|
||||
\item Get the solution
|
||||
\end{itemize}
|
||||
$$
|
||||
\begin{align*}
|
||||
2x-3y+2z &=-4\\
|
||||
7x+4.5y-1z &=16\\
|
||||
4x +3y+z &=2
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\end{document}
|
||||
Binary file not shown.
Binary file not shown.
|
|
@ -0,0 +1,614 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#7}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the euclidean vector norm for vectors
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1,0,2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
3,4
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-7,2,-4, \sqrt{12}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$\lVert [1,0,2] \rVert_2 = \sqrt{1^2 + 0^2 + 2^2} = \boxed{\sqrt{5}}$$
|
||||
|
||||
\part
|
||||
|
||||
$$\lVert [3,4] \rVert_2 = \sqrt{3^2 + 4^2} = \sqrt{25} = \boxed{5}$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\lVert [-7,2,-4, \sqrt{12}] \rVert_2 &= \sqrt{(-7)^2 + 2^2 + (-4)^2 + \sqrt{12}^2} \\
|
||||
&= \sqrt{49 + 4 + 16 + 12} \\
|
||||
&= \sqrt{81}\\
|
||||
&= \boxed{9}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the corresponding unit length vector for these:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
3,4
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-1,-2,3
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-7,2,-4, \sqrt{12}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\lVert v \rVert_2 = 1
|
||||
$$
|
||||
$$
|
||||
v \neq 0 \implies \frac{v}{\lVert v \rVert_2}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{[3,4]}{\lVert [3,4] \rVert_2} &= \frac{[3,4]}{5} \\
|
||||
&= \frac{1}{5} \begin{bmatrix} 3,4 \end{bmatrix} \\
|
||||
&= \boxed{\begin{bmatrix} \frac{3}{5}, \frac{4}{5} \end{bmatrix}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{[-1,-2,3]}{\lVert [-1,-2,3] \rVert_2} &= \frac{[-1,-2,3]}{\sqrt{1^2 + 4^2 + 9^2}} \\
|
||||
&= \frac{1}{\sqrt{14}} \begin{bmatrix} -1,-2,3 \end{bmatrix} \\
|
||||
&= \boxed{\begin{bmatrix} -\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \end{bmatrix}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{[-7,2,-4, \sqrt{12}]}{\lVert [-7,2,-4, \sqrt{12}] \rVert_2} &= \frac{[-7,2,-4, \sqrt{12}]}{\sqrt{(-7)^2 + 2^2 + (-4)^2 + \sqrt{12}^2}} \\
|
||||
&= \frac{1}{\sqrt{81}} \begin{bmatrix} -7,2,-4, \sqrt{12} \end{bmatrix} \\
|
||||
&= \frac{1}{9} \begin{bmatrix} -7,2,-4,\sqrt{12}\end{bmatrix}\\
|
||||
&= \boxed{\begin{bmatrix} -\frac{7}{9}, \frac{2}{9}, -\frac{4}{9}, \frac{\sqrt{12}}{9} \end{bmatrix}}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inner product between these vectors and their angle in degrees:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
3,-2,2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
1,2,2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
Compute the inner product between these vectors and their angle in degrees:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1,0,1
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
2,1,-2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
\frac{1}{2\sqrt{2}}
|
||||
, -\frac{\sqrt{3}}{2}
|
||||
, \frac{1}{2\sqrt{2}}}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$\frac{u}{\lVert u \rVert_2} \cdot \frac{v}{\lVert v \rVert_2} = \cos(\angle (u,v))$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\langle [3,-2,2], [1,2,2] \rangle &= 3 \cdot 1 + (-2) \cdot 2 + 2 \cdot 2 \\
|
||||
&= 3 + (-4) + 4 \\
|
||||
&= \boxed{3}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Angle:\\
|
||||
|
||||
Let $u$ be $\begin{bmatrix}3,-2,2\end{bmatrix}$ and $v$ be $\begin{bmatrix}1,2,2\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle (u,v)) &= \frac{\langle u,v \rangle}{\lVert u \rVert_2 \cdot \lVert v \rVert_2} \\
|
||||
&= \frac{\begin{bmatrix}3,-2,2\end{bmatrix} \cdot \begin{bmatrix}1,2,2\end{bmatrix}}{\lVert \begin{bmatrix}3,-2,2\end{bmatrix} \rVert_2 \cdot \lVert \begin{bmatrix}1,2,2\end{bmatrix} \rVert_2} \\
|
||||
&= \frac{3}{\sqrt{17}\cdot\sqrt{9}} \\
|
||||
&= \frac{3}{\sqrt{153}} \\\\
|
||||
\angle (u,v) &= \cos^{-1}(\frac{3}{\sqrt{153}}) \\
|
||||
&= \boxed{75.96^{\circ}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\langle [1,0,1], [2,1,-2] \rangle &= 1 \cdot 2 + 0 \cdot 1 + 1 \cdot (-2) \\
|
||||
&= 2 + 0 + (-2) \\
|
||||
&= \boxed{0}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Angle:\\
|
||||
|
||||
Let $u$ be $\begin{bmatrix}1,0,1\end{bmatrix}$ and $v$ be $\begin{bmatrix}2,1,-2\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle (u,v)) &= \frac{\langle u,v \rangle}{\lVert u \rVert_2 \cdot \lVert v \rVert_2} \\ &= \frac{\begin{bmatrix}1,0,1\end{bmatrix} \cdot \begin{bmatrix}2,1,-2\end{bmatrix}}{\lVert \begin{bmatrix}1,0,1\end{bmatrix} \rVert_2 \cdot \lVert \begin{bmatrix}2,1,-2\end{bmatrix} \rVert_2} \\
|
||||
&= \frac{0}{\sqrt{2}\cdot\sqrt{6}} \\
|
||||
&= 0 \\\\
|
||||
\angle (u,v) &= \cos^{-1}(0) \\
|
||||
&= \boxed{90^{\circ}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
[2,1,2], \left[\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}}\right] &= \left(2 \cdot \frac{1}{2\sqrt(2)}\right) + \left(1 \cdot -\frac{\sqrt{3}}{2}\right) + \left(2 \cdot \frac{1}{2\sqrt(2)}\right) \\
|
||||
&= \frac{1}{\sqrt{2}} - \frac{\sqrt{2}}{2} + \frac{1}{\sqrt{2}}\\
|
||||
&= \frac{\sqrt{2}}{2}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
Angle:\\
|
||||
|
||||
Let $u$ be $\begin{bmatrix}2,1,2\end{bmatrix}$ and $v$ be $\begin{bmatrix}\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}}\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle (u,v)) &= \frac{\langle u,v \rangle}{\lVert u \rVert_2 \cdot \lVert v \rVert_2} \\
|
||||
&= \frac{\begin{bmatrix}2,1,2\end{bmatrix} \cdot \begin{bmatrix}\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}}\end{bmatrix}}{\lVert \begin{bmatrix}2,1,2\end{bmatrix} \rVert_2 \cdot \lVert \begin{bmatrix}\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}}\end{bmatrix} \rVert_2} \\
|
||||
&= \frac{\frac{\sqrt{2}}{2}}{\sqrt{9}\cdot\sqrt{1}}\\
|
||||
&= \frac{\frac{\sqrt{2}}{2}}{\sqrt{9}}\\
|
||||
&= \frac{\sqrt{2}}{2\sqrt{9}}\\
|
||||
&= \cos^{-1}(\frac{\sqrt{2}}{2\sqrt{9}})\\
|
||||
&= \boxed{76.36^{\circ}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item What is the projection of $\begin{bmatrix}5,2 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,1 \end{bmatrix}$?
|
||||
\item What is the projection of $\begin{bmatrix}0,2,1 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,-1,-1 \end{bmatrix}$?
|
||||
\item Project $\begin{bmatrix} 5,2 \end{bmatrix}$ onto the subspace spanned by vectors $\begin{bmatrix} 2,3 \end{bmatrix}$, $\begin{bmatrix}1,1 \end{bmatrix}$
|
||||
\item What is the projection of $\begin{bmatrix} 1,-1,1 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,1,1 \end{bmatrix}$ onto the subspace spanned by vectors $\begin{bmatrix} 0,0,-1 \end{bmatrix}, \begin{bmatrix} 2,0,1 \end{bmatrix}$? Hint: this one is more tricky. Reason: $\begin{bmatrix} 0,-1,-1 \end{bmatrix} \cdot \begin{bmatrix}2,0,1\end{bmatrix} \neq 0$
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
$$ x_{\parallel} v = \frac{x\cdot v}{v\cdot v}v =\left(x\cdot \frac{v}{\lVert v \rVert_2}\right) \frac{v}{\lVert v \rVert_2} $$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{\begin{bmatrix} 5,2 \end{bmatrix} \cdot \begin{bmatrix} 1,1 \end{bmatrix}}{2} \begin{bmatrix} 1,1 \end{bmatrix} &= \frac{7}{2} \begin{bmatrix}1,1\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{\begin{bmatrix} 0,2,1 \end{bmatrix} \cdot \begin{bmatrix} 1,-1,-1 \end{bmatrix}}{3} \begin{bmatrix} 1,-1,-1 \end{bmatrix} &= \frac{-3}{3} \begin{bmatrix}1,-1,-1\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-1,1,1\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{\begin{bmatrix} 5,2 \end{bmatrix} \cdot \begin{bmatrix} 2,3 \end{bmatrix}}{13} \begin{bmatrix} 2,3 \end{bmatrix} + \frac{\begin{bmatrix} 5,2 \end{bmatrix} \cdot \begin{bmatrix} 1,1 \end{bmatrix}}{2} \begin{bmatrix} 1,1 \end{bmatrix} &= \frac{16}{13} \begin{bmatrix}2,3\end{bmatrix} + \frac{7}{2} \begin{bmatrix}1,1\end{bmatrix}\\
|
||||
&= \begin{bmatrix} \frac{32}{13}, \frac{48}{13} \end{bmatrix} + \begin{bmatrix} \frac{7}{2}, \frac{7}{2} \end{bmatrix}\\
|
||||
&= \begin{bmatrix} \frac{32+7}{13}, \frac{48+7}{13} \end{bmatrix}\\
|
||||
&= \begin{bmatrix} \frac{39}{13}, \frac{55}{13} \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute these matrix multiplications:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
2 & 1\\3&-2
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
-1 &0\\-4&-2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
-3\\2\\1
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2&4&-2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
Question: Do you need more of them to practice? If so, you can do at home:\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2\\2&4
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3&0&1\\0&1&2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
0.5&2.5\\-3.5&1.5
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
6\\4
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
2 & 1\\3&-2
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
-1 &0\\-4&-2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
2\cdot(-1)+1\cdot(-4) & 2\cdot0+1\cdot(-2)\\3\cdot(-1)+(-2)\cdot(-4) & 3\cdot0+(-2)\cdot(-2)
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
-6 & -2\\5 & 4
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
-3\\2\\1
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2&4&-2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
-3\cdot2 & -3\cdot4 & -3\cdot(-2)\\2\cdot2 & 2\cdot4 & 2\cdot(-2)\\1\cdot2 & 1\cdot4 & 1\cdot(-2)
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
-6 & -12 & 6\\4 & 8 & -4\\2 & 4 & -2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2\\2&4
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3&0&1\\0&1&2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1\cdot3+2\cdot0 & 1\cdot0+2\cdot1 & 1\cdot1+2\cdot2\\2\cdot3+4\cdot0 & 2\cdot0+4\cdot1 & 2\cdot1+4\cdot2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
3 & 2 & 4\\6 & 4 & 10
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
0.5&2.5\\-3.5&1.5
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
6\\4
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
0.5\cdot6+2.5\cdot4 \\ 0.5\cdot4+2.5\cdot6
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
13 \\ -15
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
\begin{itemize}
|
||||
\item Project $\begin{bmatrix} 5,2 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} 2,-3 \end{bmatrix}$
|
||||
\item Project $\begin{bmatrix} 1,-1,3 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} -3,1,1\end{bmatrix}$
|
||||
\item Project $\begin{bmatrix} 1,-1,3,1 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} -2,2,0,0\end{bmatrix}$, $\begin{bmatrix} 0,0,\sqrt{2}, \sqrt{2}\end{bmatrix}$
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 5,2 \end{bmatrix} \cdot \begin{bmatrix} 2,-3 \end{bmatrix} \cdot \frac{\begin{bmatrix} 2,-3 \end{bmatrix}}{\lVert \begin{bmatrix} 2,-3 \end{bmatrix} \rVert^2} &= \frac{2\cdot5+(-3)\cdot2}{\lVert \begin{bmatrix} 2,-3 \end{bmatrix} \rVert^2} \\
|
||||
&= \frac{13}{13} \\
|
||||
&= 1
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 1,-1,3 \end{bmatrix} \cdot \begin{bmatrix} -3,1,1\end{bmatrix} \cdot \frac{\begin{bmatrix} -3,1,1\end{bmatrix}}{\lVert \begin{bmatrix} -3,1,1\end{bmatrix} \rVert^2} &= \frac{(-3)\cdot1+1\cdot(-1)+1\cdot3}{\lVert \begin{bmatrix} -3,1,1\end{bmatrix} \rVert^2} \\
|
||||
&= \frac{1}{\sqrt{13}} \\
|
||||
&= \frac{1}{\sqrt{13}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 1,-1,3,1 \end{bmatrix} \cdot \begin{bmatrix} -2,2,0,0\end{bmatrix} \cdot \frac{\begin{bmatrix} -2,2,0,0\end{bmatrix}}{\lVert \begin{bmatrix} -2,2,0,0\end{bmatrix} \rVert^2} &= \frac{(-2)\cdot1+2\cdot(-1)+0\cdot3+0\cdot1}{\lVert \begin{bmatrix} -2,2,0,0\end{bmatrix} \rVert^2} \\
|
||||
&= \frac{1}{\sqrt{8}} \\
|
||||
&= \frac{1}{\sqrt{8}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 1,-1,3,1 \end{bmatrix} \cdot \begin{bmatrix} 0,0,\sqrt{2}, \sqrt{2}\end{bmatrix} \cdot \frac{\begin{bmatrix} 0,0,\sqrt{2}, \sqrt{2}\end{bmatrix}}{\lVert \begin{bmatrix} 0,0,\sqrt{2}, \sqrt{2}\end{bmatrix} \rVert^2} &= \frac{0\cdot1+0\cdot(-1)+\sqrt{2}\cdot3+\sqrt{2}\cdot1}{\lVert \begin{bmatrix} 0,0,\sqrt{2}, \sqrt{2}\end{bmatrix} \rVert^2} \\
|
||||
&= \frac{3+\sqrt{2}}{\sqrt{2}} \\
|
||||
&= \frac{3+\sqrt{2}}{\sqrt{2}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Run Gram-Schmid-orthogonalization on the vectors
|
||||
$$\begin{bmatrix} 12,12,6 \end{bmatrix}, \begin{bmatrix} 2,-2,4 \end{bmatrix}, \begin{bmatrix} -2,-2,1 \end{bmatrix}$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 12,12,6 \end{bmatrix} &= \begin{bmatrix} 12,12,6 \end{bmatrix} \\
|
||||
\begin{bmatrix} 2,-2,4 \end{bmatrix} &= \begin{bmatrix} 2,-2,4 \end{bmatrix} - \frac{2\cdot12+(-2)\cdot12+4\cdot6}{\lVert \begin{bmatrix} 12,12,6 \end{bmatrix} \rVert^2} \cdot \begin{bmatrix} 12,12,6 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} 2,-2,4 \end{bmatrix} - \frac{144}{144} \cdot \begin{bmatrix} 12,12,6 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} 2,-2,4 \end{bmatrix} - \begin{bmatrix} 12,12,6 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} -10,-14,-2 \end{bmatrix} \\
|
||||
\begin{bmatrix} -2,-2,1 \end{bmatrix} &= \begin{bmatrix} -2,-2,1 \end{bmatrix} - \frac{(-2)\cdot12+(-2)\cdot12+1\cdot6}{\lVert \begin{bmatrix} 12,12,6 \end{bmatrix} \rVert^2} \cdot \begin{bmatrix} 12,12,6 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} -2,-2,1 \end{bmatrix} - \frac{0}{144} \cdot \begin{bmatrix} 12,12,6 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} -2,-2,1 \end{bmatrix} - \begin{bmatrix} 0,0,0 \end{bmatrix} \\
|
||||
&= \begin{bmatrix} -2,-2,1 \end{bmatrix}
|
||||
\end{align*}
|
||||
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
{\large \textbf{Understanding Distances coming from $\ell_p$-norms}}\\
|
||||
|
||||
Coding: plot in python or similar the set of points $x \in \mathbb{R}^2$ usuch that $\lVert x \rVert_p = 1$ for
|
||||
\begin{itemize}
|
||||
\item $p = 0.2$
|
||||
\item $p = 0.5$
|
||||
\item $p = 1$
|
||||
\item $p = 1.5$
|
||||
\item $p = 2$
|
||||
\item $p = 4$
|
||||
\item $p = 8$
|
||||
\item $p = 16$
|
||||
\end{itemize}
|
||||
Hint: in 2 dimensions for $p=2$ the solution is given by
|
||||
$$x(t) = \left(\cos(t), \sin(t)\right)$$
|
||||
due to $\cos^2(t) + \sin^2(t) = 1$.
|
||||
|
||||
you can use the same idea with different powers. You can start by considering $(\cos^r(t), \sin^r(t))$. One thing to node: $\cos(t)^r + \sin(t)^r$ is not always defined for negative values and certain $r$.\\
|
||||
|
||||
For $p \neq 2$, you can consider this, thich deals with the signs:
|
||||
$$x(t) = (sign(\cos(t))|\cos(t)|^r, sign(\sin(t))|\sin(t)|^r)$$
|
||||
for the right choice of r. Find out which $r$ is suitable for a general $p > 0$ such that $\lVert x \rVert_p = 1$. Then plot it in python.\\
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 1,-1,3,1 \end{bmatrix} \cdot \begin{bmatrix} 1,1,1,1\end{bmatrix} \cdot \frac{\begin{bmatrix} 1,1,1,1\end{bmatrix}}{\lVert \begin{bmatrix} 1,1,1,1\end{bmatrix} \rVert^2} &= \frac{1\cdot1+(-1)\cdot1+3\cdot1+1\cdot1}{\lVert \begin{bmatrix} 1,1,1,1\end{bmatrix} \rVert^2} \\
|
||||
&= \frac{4}{2} \\
|
||||
&= 2
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,10 @@
|
|||
\relax
|
||||
\@writefile{toc}{\contentsline {section}{Question 1}{2}{}\protected@file@percent }
|
||||
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|
||||
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|
||||
\gdef \@abspage@last{11}
|
||||
|
|
@ -0,0 +1,708 @@
|
|||
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||||
entering extended mode
|
||||
restricted \write18 enabled.
|
||||
file:line:error style messages enabled.
|
||||
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|
||||
**INF1004_W8_Tutorial_2200624_Blank.tex
|
||||
(./INF1004_W8_Tutorial_2200624_Blank.tex
|
||||
LaTeX2e <2021-11-15> patch level 1
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Document Class: article 2021/10/04 v1.4n Standard LaTeX document class
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|
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File: size10.clo 2021/10/04 v1.4n Standard LaTeX file (size option)
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||||
)
|
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\c@part=\count185
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\c@section=\count186
|
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\c@subsection=\count187
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\c@subsubsection=\count188
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\c@paragraph=\count189
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\c@subparagraph=\count190
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\c@figure=\count191
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\c@table=\count192
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\abovecaptionskip=\skip47
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\belowcaptionskip=\skip48
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\bibindent=\dimen138
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) (/usr/share/texmf-dist/tex/latex/fancyhdr/fancyhdr.sty
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Package: fancyhdr 2021/01/28 v4.0.1 Extensive control of page headers and footers
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\f@nch@headwidth=\skip49
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\f@nch@O@olf=\skip56
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\f@nch@O@orf=\skip57
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) (/usr/share/texmf-dist/tex/latex/fancyhdr/extramarks.sty
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Package: extramarks 2021/01/28 v4.0.1 Extra marks for LaTeX
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\@temptokenb=\toks16
|
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) (/usr/share/texmf-dist/tex/latex/amsmath/amsmath.sty
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||||
Package: amsmath 2021/10/15 v2.17l AMS math features
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13 words of extra memory for PDF output out of 10000 (max. 10000000)
|
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\documentclass[a4paper]{article}
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||||
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||||
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||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
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||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
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||||
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|
||||
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||||
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||||
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|
||||
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|
||||
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||||
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||||
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||||
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|
||||
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|
||||
\rhead{\firstxmark}
|
||||
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|
||||
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|
||||
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||||
\renewcommand\headrulewidth{0.4pt}
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||||
\renewcommand\footrulewidth{0.4pt}
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||||
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||||
\setlength\parindent{0pt}
|
||||
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||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
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|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
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||||
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||||
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|
||||
}{
|
||||
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|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#7}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the euclidean vector norm for vectors
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1,0,2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
3,4
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-7,2,-4, \sqrt{12}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the corresponding unit length vector for these:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
3,4
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-1,-2,3
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-7,2,-4, \sqrt{12}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inner product between these vectors and their angle in degrees:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
3,-2,2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
1,2,2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
Compute the inner product between these vectors and their angle in degrees:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1,0,1
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
2,1,-2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
\frac{1}{2\sqrt{2}}
|
||||
, -\frac{\sqrt{3}}{2}
|
||||
, \frac{1}{2\sqrt{2}}}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
$$\frac{u}{\lVert u \rVert_2} \cdot \frac{v}{\lVert v \rVert_2} = \cos(\angle (u,v))$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\langle [3,-2,2], [1,2,2] \rangle &= 3 \cdot 1 + (-2) \cdot 2 + 2 \cdot 2 \\
|
||||
&= 3 + (-4) + 4 \\
|
||||
&= \boxed{3}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Angle:\\
|
||||
|
||||
Let $u$ be $\begin{bmatrix}3,-2,2\end{bmatrix}$ and $v$ be $\begin{bmatrix}1,2,2\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle (u,v)) &= \frac{\langle u,v \rangle}{\lVert u \rVert_2 \cdot \lVert v \rVert_2} \\
|
||||
&= \frac{\begin{bmatrix}3,-2,2\end{bmatrix} \cdot \begin{bmatrix}1,2,2\end{bmatrix}}{\lVert \begin{bmatrix}3,-2,2\end{bmatrix} \rVert_2 \cdot \lVert \begin{bmatrix}1,2,2\end{bmatrix} \rVert_2} \\
|
||||
&= \frac{3}{\sqrt{17}\cdot\sqrt{9}} \\
|
||||
&= \frac{3}{\sqrt{153}} \\
|
||||
&= \frac{1}{\sqrt{17}}\\
|
||||
\angle (u,v) &= \cos^{-1}(\frac{3}{\sqrt{153}}) \\
|
||||
&= \boxed{75.96^{\circ}} \text{ \textbf{OR}}\\
|
||||
&= 360^{\circ} - 75.96^{\circ}\\
|
||||
&= \boxed{284.04^{\circ}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\langle [1,0,1], [2,1,-2] \rangle &= 1 \cdot 2 + 0 \cdot 1 + 1 \cdot (-2) \\
|
||||
&= 2 + 0 + (-2) \\
|
||||
&= \boxed{0}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Angle:\\
|
||||
|
||||
Let $u$ be $\begin{bmatrix}1,0,1\end{bmatrix}$ and $v$ be $\begin{bmatrix}2,1,-2\end{bmatrix}$ and w be $\begin{bmatrix}\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}}\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle (u,v)) &= \frac{\langle u,v \rangle}{\lVert u \rVert_2 \cdot \lVert v \rVert_2} \\ &= \frac{\begin{bmatrix}1,0,1\end{bmatrix} \cdot \begin{bmatrix}2,1,-2\end{bmatrix}}{\lVert \begin{bmatrix}1,0,1\end{bmatrix} \rVert_2 \cdot \lVert \begin{bmatrix}2,1,-2\end{bmatrix} \rVert_2} \\
|
||||
&= \frac{0}{\sqrt{2}\cdot\sqrt{6}} \\
|
||||
&= 0 \\\\
|
||||
\angle (u,v) &= \cos^{-1}(0) \\
|
||||
&= \boxed{90^{\circ}} \text{ \textbf{OR}}\\
|
||||
&= 360^{\circ} - 90^{\circ}\\
|
||||
&= \boxed{270^{\circ}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\left \langle [1,0,1] \cdot \left[ \frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}} \right] \right \rangle = \frac{1}{\sqrt{2}}
|
||||
$$\\
|
||||
|
||||
$$
|
||||
\left\lVert \left [\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}} \right] \right \rVert = \sqrt{\frac{1}{4\cdot 2} + \frac{3}{4} + \frac{1}{4 \cdot 2}} = 1
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
\cos \angle(u,w) &= \frac{1}{\sqrt{2}} \cdot\frac{1}{\sqrt{2}}\cdot \frac{1}{1} \\
|
||||
&= \frac{1}{2}\\
|
||||
\angle(u,w) &= \cos^{-1}(\frac{1}{2}) \\
|
||||
&= 60^{\circ} \text{ \textbf{OR}}\\
|
||||
\angle(u,w) &= 360^{\circ} - 60^{\circ} = 300^{\circ}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
[2,1,-2] \cdot \left[\frac{1}{2\sqrt{2}},-\frac{\sqrt{3}}{2},\frac{1}{2\sqrt{2}}\right] &= \frac{1}{2\sqrt{2}} \cdot 2 + \frac{\sqrt{3}}{2} \cdot 1 + \frac{1}{2\sqrt{2}} \cdot (-2) \\
|
||||
&= - \frac{\sqrt{3}}{2}\\\\
|
||||
\cos \angle(v,w) &= -\frac{\sqrt{3}}{2} \cdot \frac{1}{3} \cdot \frac{1}{1} \\
|
||||
&= -\frac{1}{2\sqrt{3}}\\
|
||||
\angle(v,w) &= \cos^{-1}(-\frac{1}{2\sqrt{3}}) \\
|
||||
&= 106.74^{\circ} \text{ \textbf{OR}}\\
|
||||
\angle(v,w) &= 360^{\circ} - 106.74^{\circ} = 253.26^{\circ}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item What is the projection of $\begin{bmatrix}5,2 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,1 \end{bmatrix}$?
|
||||
\item What is the projection of $\begin{bmatrix}0,2,1 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,-1,-1 \end{bmatrix}$?
|
||||
\item Project $\begin{bmatrix} 5,2 \end{bmatrix}$ onto the subspace spanned by vectors $\begin{bmatrix} 2,3 \end{bmatrix}$, $\begin{bmatrix}1,1 \end{bmatrix}$
|
||||
\item What is the projection of $\begin{bmatrix} 1,-1,1 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,1,1 \end{bmatrix}$ onto the subspace spanned by vectors $\begin{bmatrix} 0,0,-1 \end{bmatrix}, \begin{bmatrix} 2,0,1 \end{bmatrix}$? Hint: this one is more tricky. Reason: $\begin{bmatrix} 0,-1,-1 \end{bmatrix} \cdot \begin{bmatrix}2,0,1\end{bmatrix} \neq 0$
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute these matrix multiplications:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
2 & 1\\3&-2
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
-1 &0\\-4&-2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
-3\\2\\1
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2&4&-2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
Question: Do you need more of them to practice? If so, you can do at home:\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2\\2&4
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3&0&1\\0&1&2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
0.5&2.5\\-3.5&1.5
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
6\\4
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
\begin{itemize}
|
||||
\item Project $\begin{bmatrix} 5,2 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} 2,-3 \end{bmatrix}$
|
||||
\item Project $\begin{bmatrix} 1,-1,3 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} -3,1,1\end{bmatrix}$
|
||||
\item Project $\begin{bmatrix} 1,-1,3,1 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} -2,2,0,0\end{bmatrix}$, $\begin{bmatrix} 0,0,\sqrt{2}, \sqrt{2}\end{bmatrix}$
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Run Gram-Schmid-orthogonalization on the vectors
|
||||
$$\begin{bmatrix} 12,12,6 \end{bmatrix}, \begin{bmatrix} 2,-2,4 \end{bmatrix}, \begin{bmatrix} -2,-2,1 \end{bmatrix}$$
|
||||
|
||||
\solution
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
{\large \textbf{Understanding Distances coming from $\ell_p$-norms}}\\
|
||||
|
||||
Coding: plot in python or similar the set of points $x \in \mathbb{R}^2$ usuch that $\lVert x \rVert_p = 1$ for
|
||||
\begin{itemize}
|
||||
\item $p = 0.2$
|
||||
\item $p = 0.5$
|
||||
\item $p = 1$
|
||||
\item $p = 1.5$
|
||||
\item $p = 2$
|
||||
\item $p = 4$
|
||||
\item $p = 8$
|
||||
\item $p = 16$
|
||||
\end{itemize}
|
||||
Hint: in 2 dimensions for $p=2$ the solution is given by
|
||||
$$x(t) = \left(\cos(t), \sin(t)\right)$$
|
||||
due to $\cos^2(t) + \sin^2(t) = 1$.
|
||||
|
||||
you can use the same idea with different powers. You can start by considering $(\cos^r(t), \sin^r(t))$. One thing to node: $\cos(t)^r + \sin(t)^r$ is not always defined for negative values and certain $r$.\\
|
||||
|
||||
For $p \neq 2$, you can consider this, thich deals with the signs:
|
||||
$$x(t) = (sign(\cos(t))|\cos(t)|^r, sign(\sin(t))|\sin(t)|^r)$$
|
||||
for the right choice of r. Find out which $r$ is suitable for a general $p > 0$ such that $\lVert x \rVert_p = 1$. Then plot it in python.\\
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\end{document}
|
||||
Binary file not shown.
|
|
@ -0,0 +1,714 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#7 Solution}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the euclidean vector norm for vectors
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1,0,2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
3,4
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-7,2,-4, \sqrt{12}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$\lVert [1,0,2] \rVert_2 = \sqrt{1^2 + 0^2 + 2^2} = \boxed{\sqrt{5}}$$
|
||||
|
||||
\part
|
||||
|
||||
$$\lVert [3,4] \rVert_2 = \sqrt{3^2 + 4^2} = \sqrt{25} = \boxed{5}$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\lVert [-7,2,-4, \sqrt{12}] \rVert_2 &= \sqrt{(-7)^2 + 2^2 + (-4)^2 + \sqrt{12}^2} \\
|
||||
&= \sqrt{49 + 4 + 16 + 12} \\
|
||||
&= \sqrt{81}\\
|
||||
&= \boxed{9}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the corresponding unit length vector for these:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
3,4
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-1,-2,3
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
-7,2,-4, \sqrt{12}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\lVert v \rVert_2 = 1
|
||||
$$
|
||||
$$
|
||||
v \neq 0 \implies \frac{v}{\lVert v \rVert_2}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{[3,4]}{\lVert [3,4] \rVert_2} &= \frac{[3,4]}{5} \\
|
||||
&= \frac{1}{5} \begin{bmatrix} 3,4 \end{bmatrix} \\
|
||||
&= \boxed{\begin{bmatrix} \frac{3}{5}, \frac{4}{5} \end{bmatrix}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{[-1,-2,3]}{\lVert [-1,-2,3] \rVert_2} &= \frac{[-1,-2,3]}{\sqrt{1^2 + 4^2 + 9^2}} \\
|
||||
&= \frac{1}{\sqrt{14}} \begin{bmatrix} -1,-2,3 \end{bmatrix} \\
|
||||
&= \boxed{\begin{bmatrix} -\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \end{bmatrix}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{[-7,2,-4, \sqrt{12}]}{\lVert [-7,2,-4, \sqrt{12}] \rVert_2} &= \frac{[-7,2,-4, \sqrt{12}]}{\sqrt{(-7)^2 + 2^2 + (-4)^2 + \sqrt{12}^2}} \\
|
||||
&= \frac{1}{\sqrt{81}} \begin{bmatrix} -7,2,-4, \sqrt{12} \end{bmatrix} \\
|
||||
&= \frac{1}{9} \begin{bmatrix} -7,2,-4,\sqrt{12}\end{bmatrix}\\
|
||||
&= \boxed{\begin{bmatrix} -\frac{7}{9}, \frac{2}{9}, -\frac{4}{9}, \frac{\sqrt{12}}{9} \end{bmatrix}}
|
||||
\end{align*}
|
||||
$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute the inner product between these vectors and their angle in degrees:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
3,-2,2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
1,2,2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
Compute the inner product between these vectors and their angle in degrees:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1,0,1
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
2,1,-2
|
||||
\end{bmatrix}
|
||||
,
|
||||
\begin{bmatrix}
|
||||
\frac{1}{2\sqrt{2}}
|
||||
, -\frac{\sqrt{3}}{2}
|
||||
, \frac{1}{2\sqrt{2}}}
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
$$\frac{u}{\lVert u \rVert_2} \cdot \frac{v}{\lVert v \rVert_2} = \cos(\angle (u,v))$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\langle [3,-2,2], [1,2,2] \rangle &= 3 \cdot 1 + (-2) \cdot 2 + 2 \cdot 2 \\
|
||||
&= 3 + (-4) + 4 \\
|
||||
&= \boxed{3}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Angle:\\
|
||||
|
||||
Let $u$ be $\begin{bmatrix}3,-2,2\end{bmatrix}$ and $v$ be $\begin{bmatrix}1,2,2\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle (u,v)) &= \frac{\langle u,v \rangle}{\lVert u \rVert_2 \cdot \lVert v \rVert_2} \\
|
||||
&= \frac{\begin{bmatrix}3,-2,2\end{bmatrix} \cdot \begin{bmatrix}1,2,2\end{bmatrix}}{\lVert \begin{bmatrix}3,-2,2\end{bmatrix} \rVert_2 \cdot \lVert \begin{bmatrix}1,2,2\end{bmatrix} \rVert_2} \\
|
||||
&= \frac{3}{\sqrt{17}\cdot\sqrt{9}} \\
|
||||
&= \frac{3}{\sqrt{153}} \\
|
||||
&= \frac{1}{\sqrt{17}}\\
|
||||
\angle (u,v) &= \cos^{-1}(\frac{3}{\sqrt{153}}) \\
|
||||
&= \boxed{75.96^{\circ}} \text{ \textbf{OR}}\\
|
||||
&= 360^{\circ} - 75.96^{\circ}\\
|
||||
&= \boxed{284.04^{\circ}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\langle [1,0,1], [2,1,-2] \rangle &= 1 \cdot 2 + 0 \cdot 1 + 1 \cdot (-2) \\
|
||||
&= 2 + 0 + (-2) \\
|
||||
&= \boxed{0}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Angle:\\
|
||||
|
||||
Let $u$ be $\begin{bmatrix}1,0,1\end{bmatrix}$ and $v$ be $\begin{bmatrix}2,1,-2\end{bmatrix}$ and w be $\begin{bmatrix}\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}}\end{bmatrix}$\\
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\cos(\angle (u,v)) &= \frac{\langle u,v \rangle}{\lVert u \rVert_2 \cdot \lVert v \rVert_2} \\ &= \frac{\begin{bmatrix}1,0,1\end{bmatrix} \cdot \begin{bmatrix}2,1,-2\end{bmatrix}}{\lVert \begin{bmatrix}1,0,1\end{bmatrix} \rVert_2 \cdot \lVert \begin{bmatrix}2,1,-2\end{bmatrix} \rVert_2} \\
|
||||
&= \frac{0}{\sqrt{2}\cdot\sqrt{6}} \\
|
||||
&= 0 \\\\
|
||||
\angle (u,v) &= \cos^{-1}(0) \\
|
||||
&= \boxed{90^{\circ}} \text{ \textbf{OR}}\\
|
||||
&= 360^{\circ} - 90^{\circ}\\
|
||||
&= \boxed{270^{\circ}}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\left \langle [1,0,1] \cdot \left[ \frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}} \right] \right \rangle = \frac{1}{\sqrt{2}}
|
||||
$$\\
|
||||
|
||||
$$
|
||||
\left\lVert \left [\frac{1}{2\sqrt{2}}, -\frac{\sqrt{3}}{2}, \frac{1}{2\sqrt{2}} \right] \right \rVert = \sqrt{\frac{1}{4\cdot 2} + \frac{3}{4} + \frac{1}{4 \cdot 2}} = 1
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
\cos \angle(u,w) &= \frac{1}{\sqrt{2}} \cdot\frac{1}{\sqrt{2}}\cdot \frac{1}{1} \\
|
||||
&= \frac{1}{2}\\
|
||||
\angle(u,w) &= \cos^{-1}(\frac{1}{2}) \\
|
||||
&= 60^{\circ} \text{ \textbf{OR}}\\
|
||||
\angle(u,w) &= 360^{\circ} - 60^{\circ} = 300^{\circ}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
[2,1,-2] \cdot \left[\frac{1}{2\sqrt{2}},-\frac{\sqrt{3}}{2},\frac{1}{2\sqrt{2}}\right] &= \frac{1}{2\sqrt{2}} \cdot 2 + \frac{\sqrt{3}}{2} \cdot 1 + \frac{1}{2\sqrt{2}} \cdot (-2) \\
|
||||
&= - \frac{\sqrt{3}}{2}\\\\
|
||||
\cos \angle(v,w) &= -\frac{\sqrt{3}}{2} \cdot \frac{1}{3} \cdot \frac{1}{1} \\
|
||||
&= -\frac{1}{2\sqrt{3}}\\
|
||||
\angle(v,w) &= \cos^{-1}(-\frac{1}{2\sqrt{3}}) \\
|
||||
&= 106.74^{\circ} \text{ \textbf{OR}}\\
|
||||
\angle(v,w) &= 360^{\circ} - 106.74^{\circ} = 253.26^{\circ}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
\begin{itemize}
|
||||
\item What is the projection of $\begin{bmatrix}5,2 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,1 \end{bmatrix}$?
|
||||
\item What is the projection of $\begin{bmatrix}0,2,1 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,-1,-1 \end{bmatrix}$?
|
||||
\item Project $\begin{bmatrix} 5,2 \end{bmatrix}$ onto the subspace spanned by vectors $\begin{bmatrix} 2,3 \end{bmatrix}$, $\begin{bmatrix}1,1 \end{bmatrix}$
|
||||
\item What is the projection of $\begin{bmatrix} 1,-1,1 \end{bmatrix}$ onto the subspace spanned by vector $\begin{bmatrix} 1,1,1 \end{bmatrix}$ onto the subspace spanned by vectors $\begin{bmatrix} 0,0,-1 \end{bmatrix}, \begin{bmatrix} 2,0,1 \end{bmatrix}$? Hint: this one is more tricky. Reason: $\begin{bmatrix} 0,-1,-1 \end{bmatrix} \cdot \begin{bmatrix}2,0,1\end{bmatrix} \neq 0$
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
$$ x_{\parallel} v = \frac{x\cdot v}{v\cdot v}v =\left(x\cdot \frac{v}{\lVert v \rVert_2}\right) \frac{v}{\lVert v \rVert_2} $$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{\begin{bmatrix} 5,2 \end{bmatrix} \cdot \begin{bmatrix} 1,1 \end{bmatrix}}{2} \begin{bmatrix} 1,1 \end{bmatrix} &= \frac{7}{2} \begin{bmatrix}1,1\end{bmatrix}\\
|
||||
&= \left [3.5, 3.5 \right]
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{\begin{bmatrix} 0,2,1 \end{bmatrix} \cdot \begin{bmatrix} 1,-1,-1 \end{bmatrix}}{3} \begin{bmatrix} 1,-1,-1 \end{bmatrix} &= \frac{-3}{3} \begin{bmatrix}1,-1,-1\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-1,1,1\end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
Use Gran Schmidts to get 2 orthogonal Vectors $\tilde{v}^{\{0\}}$ and $\tilde{v}^{\{1\}}$ \\
|
||||
Project $\begin{bmatrix}5,2\end{bmatrix}$ onto the subspace spanned by $\tilde{v}^{\{0\}}$ and $\tilde{v}^{\{1\}}$ and sum up projections
|
||||
$$
|
||||
\begin{align*}
|
||||
\tilde{v}^{\{0\}} &= \begin{bmatrix}1,1\end{bmatrix}\\
|
||||
\tilde{v}^{\{1\}} &= \begin{bmatrix}2,3\end{bmatrix}\\
|
||||
\tilde{v}^{\{1\}} &= \begin{bmatrix}2,3\end{bmatrix} - \frac{\begin{bmatrix}2,3\end{bmatrix} \cdot \begin{bmatrix}1,1\end{bmatrix}}{\begin{bmatrix}1,1\end{bmatrix} \cdot \begin{bmatrix}1,1\end{bmatrix}} \begin{bmatrix}1,1\end{bmatrix}\\
|
||||
&= \begin{bmatrix}2,3\end{bmatrix} - \frac{5}{2} \begin{bmatrix}1,1\end{bmatrix}\\
|
||||
&= \begin{bmatrix}-\frac{1}{2},\frac{1}{2}\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\textbf{Projection:}\\
|
||||
$$
|
||||
\begin{align*}
|
||||
\frac{\begin{bmatrix}5,2\end{bmatrix} \cdot \begin{bmatrix}1,1\end{bmatrix}}{2} \begin{bmatrix}1,1\end{bmatrix} + \frac{\begin{bmatrix}5,2\end{bmatrix} \cdot \begin{bmatrix}-\frac{1}{2},\frac{1}{2}\end{bmatrix}}{\begin{bmatrix}-\frac{1}{2}, -\frac{1}{2}\end{bmatrix}} \begin{bmatrix}-\frac{1}{2}, \frac{1}{2}\end{bmatrix} &= \frac{7}{2} \begin{bmatrix}1,1\end{bmatrix} + \frac{-\frac{3}{2}}{\frac{1}{2}} \begin{bmatrix}-\frac{1}{2}, \frac{1}{2}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}3.5, 3.5\end{bmatrix} + \begin{bmatrix}-\frac{3}{2}, \frac{3}{2}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}3.5 -\frac{3}{2}, 3.5 + \frac{3}{2}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}5,2\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
It must be $\begin{bmatrix}5,2\end{bmatrix}$.\\
|
||||
Reason: $\begin{bmatrix}2,3\end{bmatrix} \cdot \begin{bmatrix}1,1\end{bmatrix}$ are 2
|
||||
independent vectors $\mathbb{R}^2$ span the whole vector space.\\
|
||||
|
||||
2 independent vectors a,b, therefore dimension of subspace spanned by a and b is $\mathbb{R}^2$\\
|
||||
$\begin{bmatrix}5,2\end{bmatrix}$ projected onto $\mathbb{R}^2$ is $\begin{bmatrix}5,2\end{bmatrix}$
|
||||
\\\\
|
||||
\part
|
||||
\\\\
|
||||
let
|
||||
$
|
||||
\begin{cases}
|
||||
v^{\{0\}} = \begin{bmatrix}0,0,-1\end{bmatrix}\\\\
|
||||
v^{\{1\}} = \begin{bmatrix}2,0,1\end{bmatrix}
|
||||
\end{cases}
|
||||
$
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
v^{\{1\}} &= \begin{bmatrix}2,0,1\end{bmatrix} - \frac{\begin{bmatrix}2,0,1\end{bmatrix} \cdot \begin{bmatrix}0,0,-1\end{bmatrix}}{\begin{bmatrix}0,0,-1\end{bmatrix} \cdot \begin{bmatrix}0,0,-1\end{bmatrix}} \begin{bmatrix}0,0,-1\end{bmatrix}\\
|
||||
&= \begin{bmatrix}2,0,1\end{bmatrix} - \frac{1}{1} \begin{bmatrix}0,0,-1\end{bmatrix}\\
|
||||
&= \begin{bmatrix}2,0,1\end{bmatrix} - \begin{bmatrix}0,0,-1\end{bmatrix}\\
|
||||
&= \begin{bmatrix}2,0,0\end{bmatrix}\\
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
Now project $\begin{bmatrix}1,-1,1\end{bmatrix}$ onto the subspace spanned by $v^{\{0\}}$ and $v^{\{1\}}$\\
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
x_{\parallel v} &= \frac{\begin{bmatrix}1,-1,1\end{bmatrix} \cdot \begin{bmatrix}0,0,-1\end{bmatrix}}{\begin{bmatrix}0,0,-1\end{bmatrix} \cdot \begin{bmatrix}0,0,-1\end{bmatrix}} \begin{bmatrix}0,0,-1\end{bmatrix} + \frac{\begin{bmatrix}1,-1,1\end{bmatrix} \cdot \begin{bmatrix}2,0,0\end{bmatrix}}{\begin{bmatrix}2,0,0\end{bmatrix} \cdot \begin{bmatrix}2,0,0\end{bmatrix}} \begin{bmatrix}2,0,0\end{bmatrix}\\
|
||||
&= \frac{-1}{1} \begin{bmatrix}0,0,-1\end{bmatrix} + \frac{2}{4} \begin{bmatrix}2,0,0\end{bmatrix}\\
|
||||
&= \begin{bmatrix}1,0,1\end{bmatrix}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Compute these matrix multiplications:
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
2 & 1\\3&-2
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
-1 &0\\-4&-2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
-3\\2\\1
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2&4&-2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
Question: Do you need more of them to practice? If so, you can do at home:\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2\\2&4
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3&0&1\\0&1&2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
0.5&2.5\\-3.5&1.5
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
6\\4
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
2 & 1\\3&-2
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
-1 &0\\-4&-2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
2\cdot(-1)+1\cdot(-4) & 2\cdot0+1\cdot(-2)\\3\cdot(-1)+(-2)\cdot(-4) & 3\cdot0+(-2)\cdot(-2)
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
-6 & -2\\5 & 4
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
-3\\2\\1
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
2&4&-2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
-3\cdot2 & -3\cdot4 & -3\cdot(-2)\\2\cdot2 & 2\cdot4 & 2\cdot(-2)\\1\cdot2 & 1\cdot4 & 1\cdot(-2)
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
-6 & -12 & 6\\4 & 8 & -4\\2 & 4 & -2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
1&2\\2&4
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
3&0&1\\0&1&2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1\cdot3+2\cdot0 & 1\cdot0+2\cdot1 & 1\cdot1+2\cdot2\\2\cdot3+4\cdot0 & 2\cdot0+4\cdot1 & 2\cdot1+4\cdot2
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
3 & 2 & 5\\6 & 4 & 10
|
||||
\end{bmatrix}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
0.5&2.5\\-3.5&1.5
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
6\\4
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
0.5\cdot5+2.5\cdot4 \\ 0.5\cdot4+2.5\cdot6
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
13 \\ -15
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
\begin{itemize}
|
||||
\item Project $\begin{bmatrix} 5,2 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} 2,-3 \end{bmatrix}$
|
||||
\item Project $\begin{bmatrix} 1,-1,3 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} -3,1,1\end{bmatrix}$
|
||||
\item Project $\begin{bmatrix} 1,-1,3,1 \end{bmatrix}$ onto the orthogonal space of vector $\begin{bmatrix} -2,2,0,0\end{bmatrix}$, $\begin{bmatrix} 0,0,\sqrt{2}, \sqrt{2}\end{bmatrix}$
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x_{\perp v} &= \begin{bmatrix} 5,2 \end{bmatrix} - \frac{\begin{bmatrix} 5,2 \end{bmatrix}\cdot\begin{bmatrix} 2,-3 \end{bmatrix}}{\begin{bmatrix} 2,-3 \end{bmatrix}\cdot\begin{bmatrix} 2,-3 \end{bmatrix}}\begin{bmatrix} 2,-3 \end{bmatrix}\\
|
||||
&= \begin{bmatrix} 5,2 \end{bmatrix} - \frac{5\cdot2+2\cdot(-3)}{2\cdot2+(-3)\cdot(-3)}\begin{bmatrix} 2,-3 \end{bmatrix}\\
|
||||
&= \begin{bmatrix} 5,2 \end{bmatrix} - \frac{4}{13}\begin{bmatrix} 2,-3 \end{bmatrix}\\
|
||||
&= \begin{bmatrix} 5-\frac{8}{13},2+\frac{12}{13} \end{bmatrix}\\
|
||||
&= \begin{bmatrix} \frac{57}{13},\frac{38}{13} \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
\\
|
||||
Verify:\\
|
||||
$$
|
||||
\begin{bmatrix}5-\frac{8}{13},2+\frac{12}{13}\end{bmatrix}\cdot\begin{bmatrix}2,-3\end{bmatrix} = 0
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x_{\perp v} &= \begin{bmatrix} 1,-1,3 \end{bmatrix} - \frac{\begin{bmatrix} 1,-1,3 \end{bmatrix}\cdot\begin{bmatrix} -3,1,1\end{bmatrix}}{\begin{bmatrix} -3,1,1\end{bmatrix}\cdot\begin{bmatrix} -3,1,1\end{bmatrix}}\begin{bmatrix} -3,1,1\end{bmatrix}\\
|
||||
&= \begin{bmatrix} 1,-1,3 \end{bmatrix} - \frac{1\cdot(-3)+(-1)\cdot1+3\cdot1}{(-3)\cdot(-3)+1\cdot1+1\cdot1}\begin{bmatrix} -3,1,1\end{bmatrix}\\
|
||||
&= \begin{bmatrix} 1,-1,3 \end{bmatrix} - \frac{-1}{11}\begin{bmatrix} -3,1,1\end{bmatrix}\\
|
||||
&= \begin{bmatrix} 1-\frac{3}{11},-1+\frac{1}{11},3+\frac{1}{11} \end{bmatrix}\\
|
||||
&= \left[ \frac{8}{11},-\frac{10}{11},\frac{34}{11} \right ]
|
||||
\end{align*}
|
||||
$$
|
||||
\\
|
||||
Verify:\\
|
||||
$$
|
||||
\begin{bmatrix}1-\frac{3}{11},-1+\frac{1}{11},3+\frac{1}{11}\end{bmatrix}\cdot\begin{bmatrix}-3,1,1\end{bmatrix} = 0
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{bmatrix}-2,2,0,0\end{bmatrix}\cdot\begin{bmatrix}0,0,\sqrt{2},\sqrt{2}\end{bmatrix} = 0 \text{, thus:}\\
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x_{\perp v} &= \begin{bmatrix}1,-1,3,1\end{bmatrix}-\frac{\begin{bmatrix}1,-1,3,1\end{bmatrix}\cdot\begin{bmatrix}-2,2,0,0\end{bmatrix}}{\begin{bmatrix}-2,2,0,0\end{bmatrix}\cdot\begin{bmatrix}-2,2,0,0\end{bmatrix}}\begin{bmatrix}-2,2,0,0\end{bmatrix} - \frac{\begin{bmatrix}1,-1,3,1\end{bmatrix}\cdot\begin{bmatrix}0,0,\sqrt{2},\sqrt{2}\end{bmatrix}}{\begin{bmatrix}0,0,\sqrt{2},\sqrt{2}\end{bmatrix}\cdot\begin{bmatrix}0,0,\sqrt{2},\sqrt{2}\end{bmatrix}}\begin{bmatrix}0,0,\sqrt{2},\sqrt{2}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}1,-1,3,1 \end{bmatrix} - \frac{-4}{8}\begin{bmatrix}-2,2,0,0\end{bmatrix} - \frac{0}{2}\begin{bmatrix}0,0,\sqrt{2},\sqrt{2}\end{bmatrix}\\
|
||||
&= \begin{bmatrix}1,-1,3,1 \end{bmatrix} - \begin{bmatrix}-1,1,0,0\end{bmatrix} - \begin{bmatrix}0,0,2,2\end{bmatrix}\\
|
||||
&= \begin{bmatrix}0,0,1,-1 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
Verify:\\
|
||||
Verify that $\begin{bmatrix}0,0,1,-1\end{bmatrix}\cdot\begin{bmatrix}-2,2,0,0\end{bmatrix} = 0$ and $\begin{bmatrix}0,0,1,-1\end{bmatrix}\cdot\begin{bmatrix}0,0,\sqrt{2},\sqrt{2}\end{bmatrix} = 0$.
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Run Gram-Schmid-orthogonalization on the vectors
|
||||
$$\begin{bmatrix} 12,12,6 \end{bmatrix}, \begin{bmatrix} 2,-2,4 \end{bmatrix}, \begin{bmatrix} -2,-2,1 \end{bmatrix}$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
\tilde{v}^{\{0\}} &= \begin{bmatrix} 12,12,6 \end{bmatrix}\\\\
|
||||
\tilde{v}^{\{1\}} &= \begin{bmatrix}2,-2,4\end{bmatrix}-\frac{\begin{bmatrix}2,-2,4\end{bmatrix}\cdot\begin{bmatrix}12,12,6\end{bmatrix}}{\begin{bmatrix}12,12,6\end{bmatrix}\cdot\begin{bmatrix}12,12,6\end{bmatrix}}\begin{bmatrix}12,12,6\end{bmatrix}\\
|
||||
&= \begin{bmatrix}2,-2,4\end{bmatrix}-\frac{2\cdot12+(-2)\cdot12+4\cdot6}{12\cdot12+12\cdot12+6\cdot6}\begin{bmatrix}12,12,6\end{bmatrix}\\
|
||||
&= \begin{bmatrix}2,-2,4\end{bmatrix}-\frac{24}{288+36}\begin{bmatrix}12,12,6\end{bmatrix}\\
|
||||
&= \begin{bmatrix}2,-2,4\end{bmatrix}-\left[ \frac{12 \cdot 24}{9 \cdot 36},\frac{12 \cdot 24}{9 \cdot 36},\frac{6 \cdot 24}{9 \cdot 36}\right]\\
|
||||
&= \begin{bmatrix}2,-2,4\end{bmatrix}-\left[ \frac{288}{324},\frac{288}{324},\frac{144}{324}\right]\\
|
||||
&= \left [2-\frac{8}{9},-2-\frac{8}{9},4-\frac{4}{9}\right]\\\\
|
||||
\tilde{v}^{\{2\}} &= \begin{bmatrix}-2,-2,1\end{bmatrix}\\
|
||||
&-\frac{\begin{bmatrix}-2,-2,1\end{bmatrix}\cdot\begin{bmatrix}12,12,6\end{bmatrix}}{\begin{bmatrix}12,12,6\end{bmatrix}\cdot\begin{bmatrix}12,12,6\end{bmatrix}}\begin{bmatrix}12,12,6\end{bmatrix}\\
|
||||
&-\frac{\begin{bmatrix}-2,-2,1\end{bmatrix}\cdot\left [2-\frac{8}{9},-2-\frac{8}{9},4-\frac{4}{9}\right]}{\left [2-\frac{8}{9},-2-\frac{8}{9},4-\frac{4}{9}\right]\cdot\left [2-\frac{8}{9},-2-\frac{8}{9},4-\frac{4}{9}\right]}\left [2-\frac{8}{9},-2-\frac{8}{9},4-\frac{4}{9}\right]\\
|
||||
&= \begin{bmatrix}-2,-2,1\end{bmatrix}-\frac{(-2)\cdot12+(-2)\cdot12+1\cdot6}{12\cdot12+12\cdot12+6\cdot6}\begin{bmatrix}12,12,6\end{bmatrix}\\
|
||||
&-\frac{(-2)\cdot2+(-2)\cdot(-2)+1\cdot4}{2\cdot2+(-2)\cdot(-2)+4\cdot4}\left [2-\frac{8}{9},-2-\frac{8}{9},4-\frac{4}{9}\right]\\
|
||||
&= \begin{bmatrix}-2,-2,1\end{bmatrix}-\left[ \frac{14}{9},\frac{14}{9},\frac{7}{9}\right] - \frac{\frac{36+28}{9}}{24+\frac{128}{28}-\frac{32\cdot 9}{81}} \left[2 - \frac{8}{9},-2 - \frac{8}{9},4 - \frac{4}{9}\right]\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
{\large \textbf{Understanding Distances coming from $\ell_p$-norms}}\\
|
||||
|
||||
Coding: plot in python or similar the set of points $x \in \mathbb{R}^2$ usuch that $\lVert x \rVert_p = 1$ for
|
||||
\begin{itemize}
|
||||
\item $p = 0.2$
|
||||
\item $p = 0.5$
|
||||
\item $p = 1$
|
||||
\item $p = 1.5$
|
||||
\item $p = 2$
|
||||
\item $p = 4$
|
||||
\item $p = 8$
|
||||
\item $p = 16$
|
||||
\end{itemize}
|
||||
Hint: in 2 dimensions for $p=2$ the solution is given by
|
||||
$$x(t) = \left(\cos(t), \sin(t)\right)$$
|
||||
due to $\cos^2(t) + \sin^2(t) = 1$.
|
||||
|
||||
you can use the same idea with different powers. You can start by considering $(\cos^r(t), \sin^r(t))$. One thing to node: $\cos(t)^r + \sin(t)^r$ is not always defined for negative values and certain $r$.\\
|
||||
|
||||
For $p \neq 2$, you can consider this, thich deals with the signs:
|
||||
$$x(t) = (sign(\cos(t))|\cos(t)|^r, sign(\sin(t))|\sin(t)|^r)$$
|
||||
for the right choice of r. Find out which $r$ is suitable for a general $p > 0$ such that $\lVert x \rVert_p = 1$. Then plot it in python.\\
|
||||
|
||||
\solution
|
||||
\begin{tikzpicture}[>=Stealth]
|
||||
\foreach \i in {0,...,3}{
|
||||
\begin{scope}[xshift=\i*4.5cm]
|
||||
\draw [<->] (-1.2,0)--(2.5,0);
|
||||
\draw [<->] (0,-1.2)--(0,1.7);
|
||||
\draw[shorten <=-1cm, shorten >=-3mm] (0,1)--(2,0) node [midway, above] {$A$};
|
||||
\end{scope}
|
||||
}
|
||||
\begin{scope}[draw=blue, densely dashed]
|
||||
\draw [] (-1,0)--(0,1)--(1,0)--(0,-1)--cycle;
|
||||
\draw [](4.5,0) circle (0.88cm);
|
||||
\draw [xshift=9cm] (-.66,-.66) rectangle (.66,.66);
|
||||
\begin{scope}[xshift=13.5cm]
|
||||
\draw [domain=0:90,samples=100,smooth,variable=\t] plot({-1*cos(\t)^(3)},{1*sin(\t)^(3)});
|
||||
\draw [domain=0:90,samples=100,smooth,variable=\t] plot({-1*cos(\t)^(3)},{-1*sin(\t)^(3)});
|
||||
\draw [domain=0:90,samples=100,smooth,variable=\t] plot({1*cos(\t)^(3)},{-1*sin(\t)^(3)});
|
||||
\draw [domain=0:90,samples=100,smooth,variable=\t] plot({1*cos(\t)^(3)},{1*sin(\t)^(3)});
|
||||
\end{scope}
|
||||
\foreach \i [count=\j from 0] in {(0,1),(.39,.79),(.66,.66),(0,1)} \scoped [xshift=\j*4.5cm] { \draw [{Circle[width=3pt, length=3pt, fill=black, black]}-{Circle[width=3pt, length=3pt, fill=black, black]}, shorten <=-1.5pt, shorten >=-1.5pt] (0,0) node [below left] {$x$} -- \i node [above right] {$\hat x$} ; };
|
||||
\end{scope}
|
||||
\foreach \i [count=\j from 0] in {1,2,\infty,\frac{1}{2}} \scoped [xshift=\j*4.5cm] { \node [anchor=mid west] at (0,-1.5) {$p=\i$}; };
|
||||
\end{tikzpicture}
|
||||
\end{homeworkProblem}
|
||||
\end{document}
|
||||
Binary file not shown.
Binary file not shown.
|
|
@ -0,0 +1,375 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#8}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
\part
|
||||
|
||||
Do this one together:
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
x+2y-z&=3\\
|
||||
2x-3y+2z&=5\\
|
||||
-3x+y+5z&=13\\
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
Solve the following affine equation systems
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
3x+4y&=1\\
|
||||
2x+3y&=12
|
||||
\end{aligned}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{aligned}
|
||||
3x-2y&=4\\
|
||||
-6x+4y&=7
|
||||
\end{aligned}
|
||||
$$
|
||||
\\
|
||||
$$
|
||||
\begin{aligned}
|
||||
2x+y+z-6&=0\\
|
||||
4y+z+x&=5\\
|
||||
2x+z+3y&=7
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Modelling Problem: Solve using Gauss-Jordan elimination
|
||||
|
||||
\begin{framed}
|
||||
\begin{itemize}
|
||||
\item are there linear relationships?
|
||||
\item If so, understanding and deriving constants
|
||||
\end{itemize}
|
||||
\end{framed}
|
||||
In 2010, the average salary for all accountants together in the two cities San Diego, California, and Salt Lake City, Utah, was \$45091.50.\\
|
||||
The average salary in San Diego alone, however, was \$5231 greater than the average salary in Salt Lake
|
||||
City alone. What is the average salary of an accountant in each city, assuming that there are the same
|
||||
number of accountants in each city?
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Modelling Problem: Solve using Gauss-Jordan elimination. A chemist has prepared two acid solutions, one
|
||||
of which is 2\% by volume, the other 7\% by volume. How many cubic centimetres of each should the chemist
|
||||
mix together to obtain 40cm3 of a 3.2\% acid solution?\\
|
||||
Hint: If we multiply acidity per volume with a certain volume, we get a total amount of acid in this volume. If
|
||||
we sum 2 total amounts, we get another total amount of acid - which is the total amount for the union of the
|
||||
two volumes.\\
|
||||
In order to get back to an acidity per volume, we have to divide by the volume.
|
||||
\\
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
In a hack and slay game, you need bags of three items which you can use to increase your attack, defence
|
||||
and dexterity points. The counts of each item are x, y and z respectively.
|
||||
The contributions of each item are shown below:
|
||||
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
Item & Attack & Defence & Dexterity\\
|
||||
\hline
|
||||
Aunties Old Table Cloth (x) & -20 & 40 & 10\\
|
||||
Rusty old looking dagger (y) & 50&10&-10\\
|
||||
Geylang Gift Shop Crystal (z) &10&10&60\\
|
||||
\hline
|
||||
\end{tabular}\\\\
|
||||
In order to clear a final boss, you need to have 320 attack and 280 defense stats. Note that the stats scale
|
||||
linearly on the item equipped. Also you have 16 slots, which allow you to equip 16 items in total.\\
|
||||
PS: Isnt it weird how a small monster can drop a huge item on death ?
|
||||
|
||||
\begin{enumerate}
|
||||
\item Derive an affine equation. - Let us check in to your progress after 5 mins
|
||||
\item and solve it
|
||||
\end{enumerate}
|
||||
\\
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Solve the following affine equation systems. Follow these steps:
|
||||
|
||||
\begin{enumerate}
|
||||
\item Write down the augmented matrix $[A|b]$ of the equation system above
|
||||
\item Compute the reduced row echelon form.
|
||||
\begin{itemize}
|
||||
\item Show as an intermediate step the augmented matrix when for the first time the zero-th column A[:, 0] became a one-hot vector after performing transformations .
|
||||
\item Show as an intermediate step the augmented matrix when for the first time the augmented matrix is in row echelon form.
|
||||
\item Show as final answer the augmented matrix in reduced row echelon form.
|
||||
\end{itemize}
|
||||
\item Provide one solution which solves the equation system.
|
||||
\item Write the set of all solutions as a single vector like this, if there is only one solution
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
x\\y\\z
|
||||
\end{bmatrix} =
|
||||
\begin{bmatrix}
|
||||
u_0\\u_1\\u_2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
or an affine equation, if there is more than one solution, like this
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
x\\y\\z
|
||||
\end{bmatrix} =
|
||||
\begin{bmatrix}
|
||||
u_0\\u_1\\u_2
|
||||
\end{bmatrix} +
|
||||
s
|
||||
\begin{bmatrix}
|
||||
v_0\\v_1\\v_2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
or like this
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
x\\y\\z
|
||||
\end{bmatrix} =
|
||||
\begin{bmatrix}
|
||||
u_0\\u_1\\u_2
|
||||
\end{bmatrix} +
|
||||
s
|
||||
\begin{bmatrix}
|
||||
v_0\\v_1\\v_2
|
||||
\end{bmatrix} +
|
||||
t
|
||||
\begin{bmatrix}
|
||||
w_0\\w_1\\w_2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
or state that there is no solution, if it has no solution
|
||||
$$
|
||||
\begin{align*}
|
||||
x +y+z&=1\\
|
||||
2x-y+z&=-1\\
|
||||
x+3y-z&=7
|
||||
\end{align*}
|
||||
$$
|
||||
$$
|
||||
\begin{align*}
|
||||
3x-4y &=8\\
|
||||
x+y+z&=2\\
|
||||
2x-5y-z &=6
|
||||
\end{align*}
|
||||
$$
|
||||
\end{enumerate}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Solve this affine equation systems.
|
||||
|
||||
Again, write the set of all solutions as a single vector like this, if there is only on solution
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
x\\y\\z
|
||||
\end{bmatrix} =
|
||||
\begin{bmatrix}
|
||||
u_0\\u_1\\u_2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
or an affine equation, if there is more than one solution, like this
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
x\\y\\z
|
||||
\end{bmatrix} =
|
||||
\begin{bmatrix}
|
||||
u_0\\u_1\\u_2
|
||||
\end{bmatrix} +
|
||||
s
|
||||
\begin{bmatrix}
|
||||
v_0\\v_1\\v_2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
or like this
|
||||
|
||||
$$
|
||||
\begin{bmatrix}
|
||||
x\\y\\z
|
||||
\end{bmatrix} =
|
||||
\begin{bmatrix}
|
||||
u_0\\u_1\\u_2
|
||||
\end{bmatrix} +
|
||||
s
|
||||
\begin{bmatrix}
|
||||
v_0\\v_1\\v_2
|
||||
\end{bmatrix} +
|
||||
t
|
||||
\begin{bmatrix}
|
||||
w_0\\w_1\\w_2
|
||||
\end{bmatrix}
|
||||
$$
|
||||
or state that there is no solution, if it has no solution\\
|
||||
|
||||
\begin{align*}
|
||||
3x_0 - 9x_1 -6x_2 + 2x_3 &=5\\
|
||||
-2x_0 + 3x_1 + 4x_2 - 2x_3 &=-2\\\\
|
||||
2x_0-6x_1-6x_2+3x_3&=5\\
|
||||
-x_0-2x_1+3x_2-2x_3&=-2\\
|
||||
2x_0+4x_1-6x_2+4x_3&=7
|
||||
\end{align*}
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Define 2 or 3 equations in 3 variables with bias terms of your own choosing and solve it!\\
|
||||
Verify your obtained solution x by checking that it satisfies $Ax = b$.\\
|
||||
Do you need a Prof to write such things down?
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
\end{document}
|
||||
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|
|
@ -0,0 +1,265 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#9}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Draw these affine spaces
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 2\\-1 \end{bmatrix} \cdot x + 1.5 &= 0\\
|
||||
\begin{bmatrix} -5\\-1 \end{bmatrix} \cdot x + 6 &= 0\\
|
||||
\begin{bmatrix} -5\\-1 \end{bmatrix} \cdot x - 6 &= 0\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
Find two non-parallel vectors $x$ solving
|
||||
$$
|
||||
\begin{align*}
|
||||
&w \cdot x= 3\\
|
||||
w &= \begin{bmatrix} 1\\-2\\4 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
Show that the line given by
|
||||
$$
|
||||
\begin{align*}
|
||||
f(t) &= \begin{bmatrix} 2\\3\\1 \end{bmatrix} + t \begin{bmatrix} -1\\4\\2 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
does not intersect the plane given by
|
||||
$$
|
||||
2x+z=9
|
||||
$$
|
||||
Note:
|
||||
$$
|
||||
f(t) = \begin{bmatrix} x(t)\\y(t)\\z(t) \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
Show that the line given by
|
||||
$$
|
||||
\begin{align*}
|
||||
f(t) &= \begin{bmatrix} 1\\-3\\2 \end{bmatrix} + t \begin{bmatrix} 2\\3\\-5 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
does not intersect the plane given by
|
||||
$$
|
||||
3x-2y+2z = 18
|
||||
$$
|
||||
Note:
|
||||
$$
|
||||
f(t) = \begin{bmatrix} x(t)\\y(t)\\z(t) \end{bmatrix}
|
||||
$$
|
||||
|
||||
\solution
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Check whether the plane given by
|
||||
$$
|
||||
f(s,t) = \begin{bmatrix} 0\\1\\-3 \end{bmatrix} + t \begin{bmatrix} -2\\0\\1 \end{bmatrix} + s \begin{bmatrix} 1\\2\\-1 \end{bmatrix}
|
||||
$$
|
||||
has an intersection with the line given by
|
||||
$$
|
||||
\begin{align*}
|
||||
x+2y-z &= 3\\
|
||||
2x-y+z &= 6
|
||||
\end{align*}
|
||||
$$
|
||||
Note:
|
||||
$$
|
||||
f(s,t) = \begin{bmatrix} x(s,t)\\y(s,t)\\z(s,t) \end{bmatrix}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Convert the plane equation into the form $Ax=b$ for
|
||||
$$
|
||||
\begin{bmatrix}x_0\\y_0\\z_0 \end{bmatrix} = \begin{bmatrix} 5\\-2\\-3 \end{bmatrix}
|
||||
+ s \begin{bmatrix} 1\\2\\3 \end{bmatrix} + t \begin{bmatrix} 2\\-1\\2 \end{bmatrix}
|
||||
$$\\
|
||||
Steps:
|
||||
\begin{itemize}
|
||||
\item What is the dimensionality of the whole vector space in which these equations are defined?
|
||||
\item What is the dimensionality of the affine space spanned by the plane equation?
|
||||
\item How does the matrix B look like for which we seek solutions $x$ such that $Bx = 0$ ?
|
||||
\item Conclude based on the dimensionality of the whole vector space and the dimensionality of the plane, what is the dimensionality of solutions $x$ which we are searching for ?
|
||||
\item Find a basis for these solutions. Turn it into a matrix $A$
|
||||
\item Get the correct bias vector $b$ based the $A$ which you found
|
||||
\end{itemize}
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Convert the plane equation into the form $Ax=b$ for
|
||||
$$
|
||||
\begin{bmatrix}x_0\\y_0\\z_0 \end{bmatrix} = \begin{bmatrix} -1\\0\\1 \end{bmatrix}
|
||||
+ s \begin{bmatrix} -3\\1\\6 \end{bmatrix} + t \begin{bmatrix} 2\\-4\\-4 \end{bmatrix}
|
||||
$$\\
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Plot 2d plandes in a 3d space using e.g. matplotlib
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\end{document}
|
||||
Binary file not shown.
|
|
@ -0,0 +1,2 @@
|
|||
set table "INF1004_W9_Tutorial_2200624_Solution.pgf-plot.table"; set format "%.5f"
|
||||
set format "%.7e";; set samples 100; set dummy x; plot [x=-5:5] 2x+1.5;
|
||||
|
|
@ -0,0 +1,630 @@
|
|||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
\usepackage{framed}
|
||||
|
||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
|
||||
%
|
||||
|
||||
\topmargin=-0.45in
|
||||
\evensidemargin=0in
|
||||
\oddsidemargin=0in
|
||||
\textwidth=6.5in
|
||||
\textheight=9.0in
|
||||
\headsep=0.25in
|
||||
|
||||
\linespread{1.1}
|
||||
|
||||
\pagestyle{fancy}
|
||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
|
||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
|
||||
\cfoot{\thepage}
|
||||
|
||||
\renewcommand\headrulewidth{0.4pt}
|
||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#9\ Solutions}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Draw these affine spaces
|
||||
$$
|
||||
\begin{align*}
|
||||
\begin{bmatrix} 2\\-1 \end{bmatrix} \cdot x + 1.5 &= 0\\
|
||||
\begin{bmatrix} -5\\-1 \end{bmatrix} \cdot x + 6 &= 0\\
|
||||
\begin{bmatrix} -5\\-1 \end{bmatrix} \cdot x - 6 &= 0\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
Draw $x$ where $w\cdot x = 0$\\
|
||||
$\begin{pmatrix}2&-1\end{pmatrix} \cdot\begin{pmatrix}x_0&x_1\end{pmatrix} = 0\quad \text{thus } x_1=2x_0$
|
||||
|
||||
Find $\frac{b}{w\cdot w}w = \frac{-1.5}{2^2+(-1)^2}\begin{pmatrix}2&1\end{pmatrix} = -0.3\begin{pmatrix}2&-1\end{pmatrix} = \begin{pmatrix}-0.6&0.3\end{pmatrix}$\\
|
||||
Shift the original line by $\frac{b}{w\cdot w}w$\\
|
||||
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[xshift=9cm,
|
||||
xmin=-3,xmax=3,
|
||||
ymin=-3,ymax=3,
|
||||
grid=both,
|
||||
axis lines=middle,
|
||||
minor tick num=5,
|
||||
enlargelimits={abs=0.5},
|
||||
axis line style={latex-latex},
|
||||
ticklabel style={font=\tiny,fill=white},
|
||||
xlabel style={at={(ticklabel* cs:1)},anchor=north west},
|
||||
ylabel style={at={(ticklabel* cs:1)},anchor=south west}
|
||||
]
|
||||
% Draw line y = 2x + 1.5 with function
|
||||
\addplot[mark=none,thick,green] {2*x+1.5};
|
||||
\addplot[mark=none,thick,blue] {2*x};
|
||||
\addplot[mark=none,black] {-0.5 * x};
|
||||
% add an arrow at from 0,0, to (2,-1)
|
||||
\node (source) at (axis cs:0,0) {};
|
||||
\node (target) at (axis cs:2,-1) {};
|
||||
\draw[->,thick,red] (source) -- (target);
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
Verify $\begin{pmatrix}2,-1\end{pmatrix} \cdot \begin{pmatrix}x_0,x_1\end{pmatrix} + 1.5$ is drawn\\
|
||||
$2 x_0-x_1 + 1.5 = 0$ or $x_1 = 2x_0 + 1.5$\\
|
||||
Looks like $y = 2x + 1.5$\\
|
||||
|
||||
\pagebreak
|
||||
\part
|
||||
|
||||
Draw $x$ where $w\cdot x = 0$\\
|
||||
$\begin{pmatrix}-5&-1\end{pmatrix} \cdot\begin{pmatrix}x_0&x_1\end{pmatrix} = 0\quad \text{thus } x_1=-5x_0$\\
|
||||
|
||||
Find $\frac{b}{w\cdot w}w = \frac{-6}{(-5)^2+(-1)^2}\begin{pmatrix}-5&-1\end{pmatrix} = -\frac{6}{26}\begin{pmatrix}-5&-1\end{pmatrix} = \begin{pmatrix}\frac{15}{13} & \frac{3}{13}\end{pmatrix}$\\
|
||||
|
||||
Shift the original line by $\frac{b}{w\cdot w}w$\\
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[xshift=9cm,
|
||||
xmin=-6,xmax=4,
|
||||
ymin=-2,ymax=6,
|
||||
grid=both,
|
||||
axis lines=middle,
|
||||
minor tick num=5,
|
||||
enlargelimits={abs=0.5},
|
||||
axis line style={latex-latex},
|
||||
ticklabel style={font=\tiny,fill=white},
|
||||
xlabel style={at={(ticklabel* cs:1)},anchor=north west},
|
||||
ylabel style={at={(ticklabel* cs:1)},anchor=south west}
|
||||
]
|
||||
|
||||
\addplot[mark=none,thick,red] {-5 * x + 6};
|
||||
\addplot[mark=none,thick,blue] {-5 * x};
|
||||
\addplot[mark=none,black] {0.2 * x};
|
||||
% add an arrow at from 0,0, to (-5,-1)
|
||||
\node (source) at (axis cs:0,0) {};
|
||||
\node (target) at (axis cs:-5,-1) {};
|
||||
\draw[->,thick,red] (source) -- (target);
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
Verify $\begin{pmatrix}-5,-1\end{pmatrix} \cdot \begin{pmatrix}x_0,x_1\end{pmatrix} + 6$ is drawn\\
|
||||
$-5 x_0-x_1 + 6 = 0$ or $x_1 = -5x_0 + 6$\\
|
||||
Looks like $y = -5x + 6$\\
|
||||
|
||||
|
||||
\part
|
||||
|
||||
Draw $x$ where $w\cdot x = 0$\\
|
||||
$\begin{pmatrix}-5&-1\end{pmatrix} \cdot\begin{pmatrix}x_0&x_1\end{pmatrix} = 0\quad \text{thus } x_1=-5x_0$\\
|
||||
|
||||
Find $\frac{b}{w\cdot w}w = \frac{6}{(-5)^2+(-1)^2}\begin{pmatrix}-5&-1\end{pmatrix} = \frac{6}{26}\begin{pmatrix}-5&-1\end{pmatrix} = \begin{pmatrix}-\frac{15}{13} & -\frac{3}{13}\end{pmatrix}$\\
|
||||
|
||||
Shift the original line by $\frac{b}{w\cdot w}w$\\
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[xshift=9cm,
|
||||
xmin=-6,xmax=2,
|
||||
ymin=-6,ymax=2,
|
||||
grid=both,
|
||||
axis lines=middle,
|
||||
minor tick num=5,
|
||||
enlargelimits={abs=0.5},
|
||||
axis line style={latex-latex},
|
||||
ticklabel style={font=\tiny,fill=white},
|
||||
xlabel style={at={(ticklabel* cs:1)},anchor=north west},
|
||||
ylabel style={at={(ticklabel* cs:1)},anchor=south west}
|
||||
]
|
||||
|
||||
\addplot[mark=none,thick,red] {-5 * x - 6};
|
||||
\addplot[mark=none,thick,blue] {-5 * x};
|
||||
\addplot[mark=none,black] {0.2 * x};
|
||||
% add an arrow at from 0,0, to (-5,-1)
|
||||
\node (source) at (axis cs:0,0) {};
|
||||
\node (target) at (axis cs:-5,-1) {};
|
||||
\draw[->,thick,red] (source) -- (target);
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
Verify $\begin{pmatrix}-5,-1\end{pmatrix} \cdot \begin{pmatrix}x_0,x_1\end{pmatrix} - 6$ is drawn\\
|
||||
$-5 x_0-x_1 - 6 = 0$ or $x_1 = -5x_0 - 6$\\
|
||||
Looks like $y = -5x - 6$\\
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
Find two non-parallel vectors $x$ solving
|
||||
$$
|
||||
\begin{align*}
|
||||
&w \cdot x= 3\\
|
||||
w &= \begin{bmatrix} 1\\-2\\4 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
\solution
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
w\cdot x = 3 &\implies \begin{bmatrix} 1\\-2\\4 \end{bmatrix} \cdot \begin{bmatrix}x_0\\x_1\\x_2\end{bmatrix} = 3\\
|
||||
&\implies x_0 + 2x_1 + 4x_2 = 3\\
|
||||
&\implies x_0 =3+2x_1+4x_2 (=3+2s-4t)\\
|
||||
&\implies x_1 = x_1 (=s)\\
|
||||
&\implies x_2 = x_2 (=t)\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$\begin{bmatrix}x_0\\x_1\\x_2\end{bmatrix} = \begin{bmatrix}3\\0\\0\end{bmatrix} + s\begin{bmatrix}2\\1\\0\end{bmatrix} - t\begin{bmatrix}-4\\0\\1\end{bmatrix}$$
|
||||
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
s = 1, t = 0 &: \begin{bmatrix}x_0\\x_1\\x_2\end{bmatrix} = \begin{bmatrix}3\\0\\0\end{bmatrix} + \begin{bmatrix}2\\1\\0\end{bmatrix} = \begin{bmatrix}5\\1\\0\end{bmatrix}\\
|
||||
s = 0, t = 1 &: \begin{bmatrix}x_0\\x_1\\x_2\end{bmatrix} = \begin{bmatrix}3\\0\\0\end{bmatrix} - \begin{bmatrix}4\\0\\1\end{bmatrix} = \begin{bmatrix}-1\\0\\-1\end{bmatrix}\\
|
||||
\end{align*}
|
||||
$$
|
||||
They are clearly linear independent. There is no $c$ such that
|
||||
$$
|
||||
\begin{bmatrix}5\\1\\0\end{bmatrix} = c\begin{bmatrix}-1\\0\\-1\end{bmatrix}
|
||||
$$
|
||||
|
||||
$$
|
||||
\boxed{
|
||||
\text{Answer: vectors are: } \begin{bmatrix}5\\1\\0\end{bmatrix} \text{ and } \begin{bmatrix}-1\\0\\-1\end{bmatrix}
|
||||
}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
Show that the line given by
|
||||
$$
|
||||
\begin{align*}
|
||||
f(t) &= \begin{bmatrix} 2\\3\\1 \end{bmatrix} + t \begin{bmatrix} -1\\4\\2 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
does not intersect the plane given by
|
||||
$$
|
||||
2x+z=9
|
||||
$$
|
||||
Note:
|
||||
$$
|
||||
f(t) = \begin{bmatrix} x(t)\\y(t)\\z(t) \end{bmatrix}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
\begin{center}
|
||||
\fbox{\begin{minipage}{40em}
|
||||
Steps:
|
||||
\begin{itemize}
|
||||
\item Plug in $f(t)$ into your equation (system).
|
||||
\item Write the resulting equations as an affine system in $t: at = b$.
|
||||
\item Solve it for $t$.
|
||||
\end{itemize}
|
||||
\end{minipage}}
|
||||
\end{center}
|
||||
|
||||
$2x+z=9$ take the $x$ and $z$ componenets
|
||||
$$
|
||||
\begin{align*}
|
||||
f(t) &= \begin{bmatrix} 2 + t(-1)\\3 + t(4)\\1 + t(2) \end{bmatrix}\\
|
||||
x(t) &=2+t(-1)\\
|
||||
z(t) &=1+2t\\
|
||||
\implies 2x+z &= 2(2-t)+(1+2t) = 4-2t+1+2t = 5\\
|
||||
\end{align*}
|
||||
$$
|
||||
Cannot satisfy $5=9$ for any $t$. It is an affine equation.
|
||||
|
||||
Note: for the equation $2x+z = 5$ any $t$ solves.\\
|
||||
Therefore: The line lies inside the plane given by $2x+z=5$.
|
||||
\begin{center}
|
||||
\fbox{\begin{minipage}{25em}
|
||||
Answer: Showed that the line does not intersect plane.
|
||||
\end{minipage}}
|
||||
\end{center}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
|
||||
Show that the line given by
|
||||
$$
|
||||
\begin{align*}
|
||||
f(t) &= \begin{bmatrix} 1\\-3\\2 \end{bmatrix} + t \begin{bmatrix} 2\\3\\-5 \end{bmatrix}
|
||||
\end{align*}
|
||||
$$
|
||||
does not intersect the plane given by
|
||||
$$
|
||||
3x-2y+2z = 18
|
||||
$$
|
||||
Note:
|
||||
$$
|
||||
f(t) = \begin{bmatrix} x(t)\\y(t)\\z(t) \end{bmatrix}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
3x-2y+2z &= 18 \text{ take the x, y and z components}\\
|
||||
f(t) &= \begin{bmatrix} x(t)\\y(t)\\z(t) \end{bmatrix}\\
|
||||
x(t) &= 1+2t\\
|
||||
y(t) &= -3+3t\\
|
||||
z(t) &= 2-5t\\
|
||||
\implies 3x-2y+2z &= 3(1+2t)-2(-3+3t)+2(2-5t) = 3+6t+6-6t+4-10t = 13-10t =18\\
|
||||
\end{align*}
|
||||
$$
|
||||
Solve $13-10t = 18$ for $t$. It is an affine equation.\\
|
||||
$-10t=5$\\
|
||||
$t = -0.5$
|
||||
|
||||
$$
|
||||
\implies t = -0.5 \implies \text{ plug it in }f(-0.5) = \begin{bmatrix} 1\\-3\\2\end{bmatrix} - 0.5 \begin{bmatrix} 2\\3\\-5\end{bmatrix} = \begin{bmatrix} 0\\-4.5\\4.5\end{bmatrix}
|
||||
$$
|
||||
|
||||
Verify: $3x-2y+2z = 3\cdot 0-2\cdot(-4.5)+2\cdot4.5 = 18$
|
||||
|
||||
\begin{center}
|
||||
\fbox{\begin{minipage}{25em}
|
||||
Answer: Showed that the line intersects plane at $t=-0.5$.
|
||||
\end{minipage}}
|
||||
\end{center}
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Check whether the plane given by
|
||||
$$
|
||||
f(s,t) = \begin{bmatrix} 0\\1\\-3 \end{bmatrix} + t \begin{bmatrix} -2\\0\\1 \end{bmatrix} + s \begin{bmatrix} 1\\2\\-1 \end{bmatrix}
|
||||
$$
|
||||
has an intersection with the line given by
|
||||
$$
|
||||
\begin{align*}
|
||||
x+2y-z &= 3\\
|
||||
2x-y+z &= 6
|
||||
\end{align*}
|
||||
$$
|
||||
Note:
|
||||
$$
|
||||
f(s,t) = \begin{bmatrix} x(s,t)\\y(s,t)\\z(s,t) \end{bmatrix}
|
||||
$$
|
||||
|
||||
\answer
|
||||
|
||||
\begin{center}
|
||||
\fbox{\begin{minipage}{40em}
|
||||
Steps:
|
||||
\begin{itemize}
|
||||
\item Plug in $f(s,t)$ into your equation (system).
|
||||
\item Write the resulting equations as an affine system in vector $\begin{bmatrix} s\\t \end{bmatrix}: A\begin{bmatrix} s\\t \end{bmatrix} = b$.
|
||||
\item Solve it for $t$.
|
||||
\end{itemize}
|
||||
\end{minipage}}
|
||||
\end{center}
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
f(s,t) &= \begin{bmatrix} x(t)\\y(t)\\z(t) \end{bmatrix}\\
|
||||
x(s,t) &= 0+t(-2)+s(1)\\
|
||||
y(s,t) &= 1+t(0)+s(2)\\
|
||||
z(s,t) &= -3+t(1)+s(-1)\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{align*}
|
||||
x+2y-z &= -2t+s+2(1+2s)-(-3+t-s)=5-2t-t+s+4s+s=5+6s-3t\\
|
||||
2x-y+z &= 2(-2t+s)-(1+2s)+-3+t-s = -4-4t+t+2s-2s-s=-4-s-3t\\
|
||||
5+6s - 3t &=3\\
|
||||
-4-s-3t &=6\\
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\left [
|
||||
\begin{array}{rr|r}
|
||||
6&-3&-2\\
|
||||
-1&-3&10
|
||||
\end{array}
|
||||
\right ] \rightarrow
|
||||
\left [
|
||||
\begin{array}{rr|r}
|
||||
1&3&-10\\
|
||||
6&-3&-2
|
||||
\end{array}
|
||||
\right ] \rightarrow
|
||||
\left [
|
||||
\begin{array}{rr|r}
|
||||
1&3&-10\\
|
||||
0 & -21 & 58
|
||||
\end{array}
|
||||
\right ] \rightarrow
|
||||
\left [
|
||||
\begin{array}{rr|r}
|
||||
1&3&-10\\
|
||||
0 & 1 & -58/21
|
||||
\end{array}
|
||||
\right ] \rightarrow
|
||||
\left [
|
||||
\begin{array}{rr|r}
|
||||
1&0&-\frac{12}{7}\\
|
||||
0 & 1 & -\frac{58}{21}
|
||||
\end{array}
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\right ]
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|
||||
$$
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|
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\begin{center}
|
||||
\fbox{\begin{minipage}{25em}
|
||||
{\large Answer: Intersects where $s = -\frac{12}{7}$ and $t = -\frac{58}{21}$.}
|
||||
\end{minipage}}
|
||||
\end{center}
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Convert the plane equation into the form $Ax=b$ for
|
||||
$$
|
||||
\begin{bmatrix}x_0\\y_0\\z_0 \end{bmatrix} = \begin{bmatrix} 5\\-2\\-3 \end{bmatrix}
|
||||
+ s \begin{bmatrix} 1\\2\\3 \end{bmatrix} + t \begin{bmatrix} 2\\-1\\2 \end{bmatrix}
|
||||
$$\\
|
||||
Steps:
|
||||
\begin{itemize}
|
||||
\item What is the dimensionality of the whole vector space in which these equations are defined?
|
||||
\item What is the dimensionality of the affine space spanned by the plane equation?
|
||||
\item How does the matrix B look like for which we seek solutions $x$ such that $Bx = 0$ ?
|
||||
\item Conclude based on the dimensionality of the whole vector space and the dimensionality of the plane, what is the dimensionality of solutions $x$ which we are searching for ?
|
||||
\item Find a basis for these solutions. Turn it into a matrix $A$
|
||||
\item Get the correct bias vector $b$ based the $A$ which you found
|
||||
\end{itemize}
|
||||
|
||||
\part
|
||||
|
||||
The whole space is 3 dim
|
||||
|
||||
\part
|
||||
|
||||
$v_0 = \begin{bmatrix}1\\2\\3\end{bmatrix}$ and $v_1 = \begin{bmatrix}2\\-1\\2\end{bmatrix}$ \\
|
||||
are linearly independent, therefore they span a 2-dim space (k=2)
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
Need $v_0 \cdot w = 0, v_1 \cdot w = 0$ \\
|
||||
$\begin{pmatrix}1&2&3\end{pmatrix} \cdot w = 0$ \\
|
||||
$\begin{pmatrix}2&-1&2\end{pmatrix} \cdot w = 0$ \\
|
||||
$$
|
||||
\begin{align*}
|
||||
x+2y+3z = 0\\
|
||||
2x+-y+2z = 0
|
||||
\end{align*}
|
||||
$$
|
||||
$$
|
||||
\begin{bmatrix}1&2&3\\2&-1&2\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}\\
|
||||
$$
|
||||
$$
|
||||
\left[
|
||||
\begin{array}{rrr|r}
|
||||
1&2&3&0\\
|
||||
2&-1&2&0
|
||||
\end{array}
|
||||
\right] \rightarrow
|
||||
\left[
|
||||
\begin{array}{rrr|r}
|
||||
1&0&\frac{7}{5}&0\\
|
||||
0&1&\frac{4}{5}&0
|
||||
\end{array}
|
||||
\right]
|
||||
|
||||
$$
|
||||
$$
|
||||
\begin{align*}
|
||||
x + \frac{7}{5}z &= 0\\
|
||||
y + \frac{4}{5}z &= 0
|
||||
\end{align*}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{bmatrix}x\\y\\z\end{bmatrix} = z\begin{bmatrix}-\frac{7}{5}\\-\frac{4}{5}\\1\end{bmatrix}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
We need to find one orthogonal vector $w = \begin{bmatrix}x\\y\\z\end{bmatrix}$ which is orthorgonal because then we have 3 vectors. 3 vectors span a 3-dim space. ($e-k=1$)
|
||||
|
||||
\part
|
||||
|
||||
Let $z=1$:
|
||||
|
||||
$$
|
||||
w = \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-\frac{7}{5}\\-\frac{4}{5}\\1\end{bmatrix}
|
||||
$$
|
||||
$Ax =b$ \\
|
||||
To find A and B
|
||||
|
||||
$$
|
||||
A = \begin{bmatrix}-\frac{7}{5}&-\frac{4}{5}&1\end{bmatrix}
|
||||
$$
|
||||
\part
|
||||
|
||||
$$
|
||||
b = A\begin{bmatrix}5\\-2\\-3\end{bmatrix} = \begin{bmatrix}-\frac{7}{5} & -\frac{4}{5} & 1\end{bmatrix} \begin{bmatrix}5\\-2\\-3\end{bmatrix} = -8.4
|
||||
$$ \part
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||||
|
||||
$$
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||||
b = A\begin{bmatrix}5\\-2\\-3\end{bmatrix} = \begin{bmatrix}-\frac{7}{5} & -\frac{4}{5} & 1\end{bmatrix} \begin{bmatrix}5\\-2\\-3\end{bmatrix} = -8.4
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
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||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Convert the plane equation into the form $Ax=b$ for
|
||||
$$
|
||||
\begin{bmatrix}x_0\\y_0\\z_0 \end{bmatrix} = \begin{bmatrix} -1\\0\\1 \end{bmatrix}
|
||||
+ s \begin{bmatrix} -3\\1\\6 \end{bmatrix} + t \begin{bmatrix} 2\\-4\\-4 \end{bmatrix}
|
||||
$$\\
|
||||
|
||||
\answer
|
||||
|
||||
We have a 2 dimensinal affine space, spanned by:
|
||||
$$
|
||||
\begin{bmatrix}-3\\1\\6\end{bmatrix} \quad \begin{bmatrix}2\\-4\\-4\end{bmatrix}
|
||||
$$
|
||||
|
||||
There can be in 3 dimensionsonly one orthogonal vector $w$ which is orthogonal to both these vectors
|
||||
|
||||
$
|
||||
\begin{bmatrix}-3&1&6\\2&-4&-4\end{bmatrix} \cdot w = 0
|
||||
$\\
|
||||
$
|
||||
\begin{bmatrix}-3&1&6\\2&-4&-4\end{bmatrix} \rightarrow
|
||||
\begin{bmatrix}1&0&-2\\0&1&0\end{bmatrix}
|
||||
$\\
|
||||
|
||||
$ y = 0$\\
|
||||
$x-2z = 0$
|
||||
|
||||
$$w = \begin{bmatrix}x\\y\\z\end{bmatrix} = z\begin{bmatrix}2\\0\\-1\end{bmatrix}$$
|
||||
|
||||
$w = \begin{bmatrix}x\\y\\z\end{bmatrix} \text{ and thus } A = \begin{bmatrix}2&0&-1\end{bmatrix}$\\
|
||||
$ b= \begin{bmatrix}2&0&-1\end{bmatrix} \begin{bmatrix}-1\\0\\1\end{bmatrix} = -1$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Plot 2d plandes in a 3d space using e.g. matplotlib
|
||||
|
||||
\solution
|
||||
|
||||
\end{homeworkProblem}
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||||
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||||
\pagebreak
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\end{document}
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[]
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||||
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[6 <./images//T1Q3.png>]
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\OML/cmm/m/it/10 P\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 B\OMS/cmsy/m/n/10 j[]\OT1/cmr/m/n/10 ) = [] = [] = 0\OML/cmm/m/it/10 :\OT1/cmr/m/n/10 0204168439 \OMS/cmsy/m/n/10 ^^Y \OT1/cmr/m/n/10 0\OML/cmm/m/it/10 :\OT1/cmr/m/n/10 0204
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[12] (./INF1004_W1_W2_Tutorial_2200624.aux) )
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\documentclass[a4paper]{article}
|
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|
||||
\usepackage{fancyhdr}
|
||||
\usepackage{extramarks}
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||||
\usepackage{amsmath}
|
||||
\usepackage{amsthm}
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||||
\usepackage{amsfonts}
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||||
\usepackage{tikz}
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||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
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||||
\graphicspath{ {./images/} }
|
||||
|
||||
\usetikzlibrary{automata,positioning}
|
||||
|
||||
%
|
||||
% Basic Document Settings
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||||
%
|
||||
|
||||
\topmargin=-0.45in
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||||
\evensidemargin=0in
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||||
\oddsidemargin=0in
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||||
\textwidth=6.5in
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||||
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||||
\linespread{1.1}
|
||||
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||||
\pagestyle{fancy}
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||||
\lhead{\hmwkAuthorName}
|
||||
\chead{\hmwkClass\ : \hmwkTitle}
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||||
\rhead{\firstxmark}
|
||||
\lfoot{\lastxmark}
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||||
\cfoot{\thepage}
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||||
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||||
\renewcommand\headrulewidth{0.4pt}
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||||
\renewcommand\footrulewidth{0.4pt}
|
||||
|
||||
\setlength\parindent{0pt}
|
||||
|
||||
%
|
||||
% Create Problem Sections
|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
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||||
\stepcounter{#1}
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\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
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||||
}
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||||
\setcounter{secnumdepth}{0}
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||||
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|
||||
\newcounter{homeworkProblemCounter}
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||||
\setcounter{homeworkProblemCounter}{1}
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||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
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||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#1 \&\ \#2}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
In a sample of 20 people the times taken, in seconds, to solve a simple numerical puzzle were as follows:
|
||||
\\
|
||||
\\
|
||||
\(\begin{matrix}
|
||||
17 & 19 & 22 & 26 & 28 & 31 & 34 & 36 & 38 & 39 \\
|
||||
41 & 42 & 43 & 47 & 50 & 51 & 53 & 55 & 57 & 58 \\
|
||||
\end{matrix}\)
|
||||
\\
|
||||
|
||||
\begin{enumerate}
|
||||
\item Calculate the mean and standard deviation of these times.
|
||||
\item In fact, 23 people attempted to solve this puzzle. However, 3 of them failed to solve it within the allotted time of 60 seconds, taking 62, 65 and 97 seconds respectively. Calculate the interquartile range (IQR) of the times taken by all 23 people.
|
||||
\item Are there any outliers in the dataset?
|
||||
\end{enumerate}
|
||||
|
||||
|
||||
\solution
|
||||
|
||||
\part{1}
|
||||
\\
|
||||
\textbf{Mean}
|
||||
\[
|
||||
\begin{split}
|
||||
\bar{x} &= \frac{1}{n} \sum_{i=1}^{n} x_i \\
|
||||
&= \frac{\left( 17 + 19 + 22 + 26 + 28 + 31 + 34 + 36 + 38 + 39 + 41 + 42 + 43 + 47 + 50 + 51 + 53 + 55 + 57 + 58 \right)}{20} \\
|
||||
&= \frac{787}{20}\\
|
||||
&= 39.35
|
||||
\end{split}
|
||||
\]
|
||||
\\
|
||||
\textbf{Standard Deviation}
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
\sigma &= \sqrt{\frac{1}{N} \sum_{i=1}^{n} (x_i - \bar{x})^2} \\
|
||||
&= \sqrt{\frac{1}{20} \sum_{i=1}^{20} (x_i - 39.35)^2} \\
|
||||
&= \sqrt{\frac{1}{20} \left( (17 - 39.35)^2 + (19 - 39.35)^2 + (22 - 39.35)^2 + (26 - 39.35)^2 + (28 - 39.35)^2 + (31 - 39.35)^2 + (34 - 39.35)^2 + (36 - 39.35)^2 + (38 - 39.35)^2 + (39 - 39.35)^2 + (41 - 39.35)^2 + (42 - 39.35)^2 + (43 - 39.35)^2 + (47 - 39.35)^2 + (50 - 39.35)^2 + (51 - 39.35)^2 + (53 - 39.35)^2 + (55 - 39.35)^2 + (57 - 39.35)^2 + (58 - 39.35)^2 \right)} \\
|
||||
&= 12.35829681 \approx 12.4
|
||||
\end{split}
|
||||
\]
|
||||
The mean is \framebox[1.5\width]{39.35} and the standard deviation is \framebox[1.5\width]{12.4}.
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part{2}
|
||||
\textbf{Interquartile Range}
|
||||
|
||||
Rearranging the data in ascending order, we get:\\\\
|
||||
17, 19, 22, 26, 28, 31, 34, 36, 38, 39, 41, 42, 43, 47, 50, 51, 53, 55, 57, 58, 62, 65, 97.\\
|
||||
|
||||
Hence, Q1 = 31 and Q3 = 55.\\
|
||||
\[
|
||||
\begin{split}
|
||||
\text{IQR} &= \text{Q3} - \text{Q1} \\
|
||||
&= 55 - 31 \\
|
||||
&= 24
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
\part{3}
|
||||
\textbf{Outliers}
|
||||
|
||||
The interquartile range is 24.\\
|
||||
$1.5 \times \text{IQR} = 36$.\\
|
||||
$Q3+1.5 \times \text{IQR} = 55 + 36 = 91$.\\
|
||||
$Q1-1.5 \times \text{IQR} = 31 - 36 = -5$.\\
|
||||
Hence, the outlier is 97, since it is greater than $1.5 \times \text{IQR}$ \\
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Consider the below tree diagram.
|
||||
\\
|
||||
\includegraphics{T1Q2.png}
|
||||
|
||||
\begin{enumerate}[label=\roman*.]
|
||||
\item The probability $x$ represents (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $P(A_1)$
|
||||
\item $P(A_1|B_2)$
|
||||
\item $P(B_2|A_1)$
|
||||
\item $P(C_1|B_2 \cap A_1)$
|
||||
\end{enumerate}
|
||||
\item The probability $y$ represents (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $P(B_2)$
|
||||
\item $P(A_1|B_2)$
|
||||
\item $P(B_2|A_1)$
|
||||
\item $P(C_1|B_2 \cap A_1)$
|
||||
\end{enumerate}
|
||||
\item The probability $z$ represents (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $P(C_1)$
|
||||
\item $P(B_2|C_1)$
|
||||
\item $P(C_1|B_2)$
|
||||
\item $P(C_1|B_2 \cap A_1)$
|
||||
\end{enumerate}
|
||||
\item The circled node represents the event (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $C_1$
|
||||
\item $B_2\cap C_1$
|
||||
\item $A_1 \cap B_2 \cap C_1$
|
||||
\item $C_1|B_2\cap A_1$
|
||||
\end{enumerate}
|
||||
\text{[v. and vi. do not refer to the tree diagram above]}
|
||||
\item Let A and B be two events. Suppose that the probability that neither event occurs is 3/8. What is the probability that at least one of the events occurs?
|
||||
\item Let C and D be two events. Suppose P(C) = 0.5, $P(C \cap D) = 0.2$ and $P((C \cup D)\ complement) = 0.4$. What is P(D)?
|
||||
\end{enumerate}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\solution
|
||||
|
||||
\part{1}
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item $P(A_1) = \frac{1}{2}$
|
||||
\item $P(A_1|B_2) = \frac{1}{2}$
|
||||
\item $P(B_2|A_1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
|
||||
\item $P(C_1|B_2 \cap A_1) = (\frac{1}{2})^3 = \frac{1}{8}$
|
||||
\end{enumerate}
|
||||
|
||||
\part{2}
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item $P(B_2) = 2(\frac{1}{2} \times \frac{1}{2}) = \frac{1}{2}$
|
||||
\item $P(A_1|B_2) = \frac{1}{4}$
|
||||
\item $P(B_2|A_1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
|
||||
\item $P(C_1|B_2 \cap A_1) = (\frac{1}{2} +\frac{1}{2}) \times \frac{1}{2} = \frac{1}{2}$
|
||||
\end{enumerate}
|
||||
|
||||
\part{3}
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item $P(C_1) = 4((\frac{1}{2})^3) = \frac{4}{8} = \frac{1}{2}$
|
||||
\item $P(B_2|C_1) = \frac{1}{2} \times \frac{1}{2}$
|
||||
\item $P(C_1|B_2) = 2(\frac{1}{2})^3 = \frac{1}{4}$
|
||||
\item $P(C_1|B_2 \cap A_1) = (\frac{1}{2})^3 = \frac{1}{8}$
|
||||
\end{enumerate}
|
||||
|
||||
\part{4}
|
||||
|
||||
The circled node represents the event (MCQ)?
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item $C_1 = \frac{1}{8}$
|
||||
\item $B_2\cap C_1 = \frac{}{}$
|
||||
\item $A_1 \cap B_2 \cap C_1$
|
||||
\item $C_1|B_2\cap A_1$
|
||||
\end{enumerate}
|
||||
|
||||
\part{5}
|
||||
|
||||
Since $P(A^c \cap B^c) = \frac{3}{8}$ and $P(A) +P(B) + P(A \cap B) = 1$,
|
||||
$P(\text{At least 1 event occurs}) = 1 - \frac{3}{8}$
|
||||
\\\\
|
||||
\part{6}
|
||||
|
||||
Given:\\
|
||||
$P(C) = 0.5$ \\
|
||||
$P(C \cap D) = 0.2$ \\
|
||||
$P((C \cup D)^c) = 0.4$\\
|
||||
\\
|
||||
$P(C \cup D) = 1-0.4 = 0.6$
|
||||
$P(D) = P(C \cup D) - P(C) = 0.6 - 0.5 = 0.1$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
In a survey of 1929 students, the following data were obtained on “students’ first reason for application to the university in which they matriculated.”
|
||||
\\
|
||||
\includegraphics{T1Q3.png}
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that university quality is the first reason for a student to choose a university?
|
||||
\item For a full-time student, what is the probability that university quality is the first reason for choosing a university?
|
||||
\item Let A denote the event that a student is full-time, and let B denote the event that the student lists university quality as the first reason for applying. Are events A and B independent?
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
\part{a}
|
||||
|
||||
$P(\text{quality}) = \frac{821}{1929}$
|
||||
\\\\
|
||||
\part{b}
|
||||
|
||||
$P(\text{Full-Time} \cap \text{quality}) = \frac{421}{890}$
|
||||
\\\\
|
||||
\part{c}
|
||||
|
||||
A: A student is full-time\\
|
||||
B: the student lists university quality as the first reason for applying.\\
|
||||
Are events A and B independent?
|
||||
\\
|
||||
No
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A rare blood disease is found in 2\% of a certain population. A new blood test can correctly
|
||||
identify 96\% of the people with the disease and 94\% of the people without the disease.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that a person who is tested positive by the blood test actually has the disease?
|
||||
\item What is the probability that a person who is tested negative by the blood test actually does not have the disease?
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
|
||||
$P(B) = 0.02$\\
|
||||
$P(B^c) = 0.98$\\
|
||||
$P(\text{Correct}|B) = 0.96$\\
|
||||
$P(\text{Correct}|B^c) = 0.94$\\
|
||||
|
||||
\part{a}
|
||||
|
||||
$$P(B|\text{Correct}) = \frac{P(\text{Correct}|B)P(B)}{P(\text{Correct}|B)P(B) + P(\text{Correct}|B^c)P(B^c)} = \frac{0.96 \times 0.02}{0.96 \times 0.02 + 0.94 \times 0.98} = 0.0204168439 \approx 0.0204$$
|
||||
|
||||
|
||||
\part{b}
|
||||
$$P(B^c|\text{Correct}) = \frac{P(\text{Correct}|B^c)P(B^c)}{P(\text{Correct}|B)P(B) + P(\text{Correct}|B^c)P(B^c)} = \frac{0.94 \times 0.98}{0.96 \times 0.02 + 0.94 \times 0.98} = 0.979583156 \approx 0.9796$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A woman is pregnant with male twins. Twins may be either identical or fraternal (non-
|
||||
identical). In general, $\frac{1}{3}$ twins born are identical. Obviously, identical twins must be of the
|
||||
same sex; fraternal twins may or may not be. Assume that identical twins are equally likely
|
||||
to be both boys or both girls, while for fraternal twins all possibilities are equally likely:
|
||||
boy-girl, girl-boy, boy-boy, girl-girl. Given the above information, what is the probability
|
||||
that the woman’s male twins are identical?
|
||||
\\
|
||||
\\
|
||||
\solution\\
|
||||
|
||||
$$P(\text{identical}|BB) = \frac{P(BB|\text{identical})P(\text{identical})}{P(BB)} =
|
||||
\frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}\cdot \frac{1}{3} + \frac{1}{4}\cdot\frac{2}{3}} = \frac{1}{2}$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
If a family had three kids, named Alice, Bob, and Carl. Assume that each is equally likely
|
||||
to be born; I.E., $\frac{1}{3}$ chance for each of them to be born first etc.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that Alice is older than Bob, given that Alice is older than Carl.
|
||||
\item Is event “Alice is older than Bob” independent from event “Alice is older than C”?
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
|
||||
A: Alice older than Bob\\
|
||||
B: Alice older than Carl
|
||||
|
||||
$$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}[7]
|
||||
A boy receives a school report card each week. He is given a special treat whenever his
|
||||
report indicates “Good behavior” \textbf{AND} “Excellent Homework”. His behavior is good with
|
||||
probability 0.6. When his behavior is good, he has a probability of 0.8 of doing excellent
|
||||
homework. When his behavior is not good, his probability of doing excellent homework
|
||||
is only 0.5.
|
||||
|
||||
\begin{enumerate}[label=\alph*)]
|
||||
\item For a random week, calculate the probability that he will be given a special treat
|
||||
\item For a random Week, calculate the probability that his homework will be excellent but he will not be given a special treat
|
||||
\item For a random Week, calculate the probability that his homework will be excellent
|
||||
\item Given that his homework is excellent, calculate the conditional probability that he is NOT given a special treat.
|
||||
\end{enumerate}
|
||||
|
||||
|
||||
\solution
|
||||
\\
|
||||
$P(\text{Good Behaviour}) = 0.6$\\
|
||||
$P(\text{Bad Behaviour}) = 0.4$\\
|
||||
$P(\text{Excellent Homework}|\text{Good Behaviour}) = 0.8$\\
|
||||
$P(\text{Excellent Homework}|\text{Bad Behaviour}) = 0.5$\\
|
||||
|
||||
\part{1}
|
||||
\[
|
||||
\begin{split}
|
||||
P(\text{Special Treat}) &= P(\text{Good Behavior} \cap \text{Excellent Homework}) \\
|
||||
&= P(\text{Good Behavior}) \times P(\text{Excellent Homework} | \text{Good Behavior}) \\
|
||||
&= 0.6 \times 0.8 \\
|
||||
&= 0.48
|
||||
\end{split}
|
||||
\]
|
||||
\\
|
||||
\\
|
||||
\part{2}
|
||||
\[
|
||||
\begin{split}
|
||||
P(\text{Excellent Homework} \cap \text{No Special Treat}) &= P(\text{Excellent Homework}) - P(\text{Excellent Homework} \cap \text{Special Treat}) \\
|
||||
&= 0.8 - 0.48 \\
|
||||
&= 0.32
|
||||
\end{split}
|
||||
\]
|
||||
\\
|
||||
\\
|
||||
\part{3}
|
||||
\[
|
||||
\begin{split}
|
||||
P(\text{Excellent Homework}) &= P(\text{Excellent Homework} \cap \text{Good Behavior}) + P(\text{Excellent Homework} \cap \text{Good Behavior}^c) \\
|
||||
&= 0.48 + 0.32 \\
|
||||
&= 0.8
|
||||
\end{split}
|
||||
\]
|
||||
\\
|
||||
\\
|
||||
\part{4}
|
||||
\[
|
||||
\begin{split}
|
||||
P(\text{Good Behavior}^c | \text{Excellent Homework}) &= \frac{P(\text{Excellent Homework} \cap \text{Good Behavior}^c)}{P(\text{Excellent Homework})} \\
|
||||
&= \frac{0.32}{0.8} \\
|
||||
&= 0.4
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}[8]
|
||||
In an attempt to find the mean number of hours his tutorial classmates spent per day preparing for tutorials,
|
||||
John collected data from 10 of his friends in the tutorial group and found that the sample mean is 2.4 hours with a sample standard deviation of 0.8 hours.
|
||||
However, a day later he felt that the sample size is too small.
|
||||
So he collected data from another 5 of his friends and found that the sample mean is 2.0 hours with a sample standard deviation of 1.2 hours.
|
||||
\\
|
||||
Find the sample mean and sample standard deviation \underline{when these 2 sets of data are combined}.
|
||||
\\
|
||||
\\
|
||||
\solution
|
||||
\\
|
||||
\part*{Sample Mean}
|
||||
\[
|
||||
\begin{split}
|
||||
\bar{x} &= \frac{2.4 \times 10 + 2.0 \times 5}{15} \\
|
||||
&= \frac{24 + 10}{15} \\
|
||||
&= 2.27
|
||||
\end{split}
|
||||
\]
|
||||
\\
|
||||
\\
|
||||
\part*{Sample Standard Deviation}
|
||||
\[
|
||||
\begin{split}
|
||||
s &= \frac{n_1 \times (s_1 + \bar{x}|1^2) + n2 \times (s_2 + \bar{x}|2^2)}{n_1 + n_2} - \bar{x}^2 \\
|
||||
&= \frac{10\times(0.8 + 2.4^2) + 5\times(1.2 + 2.0^2)}{10 + 5} - 2.27^2 \\
|
||||
&= \frac{65.5+26}{15} - 5.14\\
|
||||
&= 0.97
|
||||
\end{split}
|
||||
\]
|
||||
%\[
|
||||
% \begin{split}
|
||||
% s &= \sqrt{\frac{0.8^2 \times 10 + 1.2^2 \times 5}{15}} \\
|
||||
% &= \sqrt{\frac{0.64 \times 10 + 1.44 \times 5}{15}} \\
|
||||
% &= \sqrt{\frac{6.4 + 7.2}{15}} \\
|
||||
% &= \sqrt{\frac{13.6}{15}} \\
|
||||
% &= \sqrt{0.908} \\
|
||||
% &= 0.952
|
||||
% \end{split}
|
||||
%\]
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}[9]
|
||||
A certain family has 6 children, consisting of 3 boys and 3 girls.
|
||||
Assuming that all birth orders are equally likely,
|
||||
what is the probability that the 3 eldest children are the 3 girls?
|
||||
\\
|
||||
\\
|
||||
\solution
|
||||
\\
|
||||
\\
|
||||
Probability that the 3 eldest are the 3 girls = $\frac{1}{\binom{3}{6}} = \frac{1}{20} = 0.05$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
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\end{document}
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\documentclass[a4paper]{article}
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}{
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\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
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\newcommand{\hmwkTitle}{Tutorial\ \#1 \&\ \#2\ Solutions}
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\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
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\newcommand{\hmwkStudentID}{\textbf{2200624}}
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\title{
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\vspace{2in}
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\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
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\vspace{3in}
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}
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\author{\hmwkAuthorName \\ \hmwkStudentID}
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\date{\today}
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\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
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%
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% Probability commands: Expectation, Variance, Covariance, Bias
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\newcommand{\E}{\mathrm{E}}
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\newcommand{\Var}{\mathrm{Var}}
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\begin{document}
|
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\maketitle
|
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\pagebreak
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||||
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||||
\begin{homeworkProblem}
|
||||
In a sample of 20 people the times taken, in seconds, to solve a simple numerical puzzle were as follows:
|
||||
\\
|
||||
\\
|
||||
\(\begin{matrix}
|
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17 & 19 & 22 & 26 & 28 & 31 & 34 & 36 & 38 & 39 \\
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41 & 42 & 43 & 47 & 50 & 51 & 53 & 55 & 57 & 58 \\
|
||||
\end{matrix}\)
|
||||
\\
|
||||
|
||||
\begin{enumerate}
|
||||
\item Calculate the mean and standard deviation of these times.
|
||||
\item In fact, 23 people attempted to solve this puzzle. However, 3 of them failed to solve it within the allotted time of 60 seconds, taking 62, 65 and 97 seconds respectively. Calculate the interquartile range (IQR) of the times taken by all 23 people.
|
||||
\item Are there any outliers in the dataset?
|
||||
\end{enumerate}
|
||||
|
||||
|
||||
\solution
|
||||
|
||||
\part{1}
|
||||
\\
|
||||
\textbf{Mean}
|
||||
\[
|
||||
\begin{split}
|
||||
\bar{x} &= \frac{1}{n} \sum_{i=1}^{n} x_i \\
|
||||
&= \frac{\left( 17 + 19 + 22 + 26 + 28 + 31 + 34 + 36 + 38 + 39 + 41 + 42 + 43 + 47 + 50 + 51 + 53 + 55 + 57 + 58 \right)}{20} \\
|
||||
&= \frac{787}{20}\\
|
||||
&= 39.35
|
||||
\end{split}
|
||||
\]
|
||||
\\
|
||||
\textbf{Standard Deviation}
|
||||
|
||||
\[
|
||||
\begin{split}
|
||||
s^2 &= \frac{\sum (x_i - \bar{x})^2}{n-1} =\frac{1}{n-1} \Bigr[ \sum x_i^2 - \frac{(\sum x_i)^2}{n} \Bigr]\\
|
||||
&= \frac{1}{19} \Bigr[ \sum x_i^2 - \frac{(\sum x_i)^2}{20} \Bigr]\\
|
||||
&= \frac{1}{19} \Bigr[ \sum x_i^2 - \frac{787^2}{20} \Bigr]\\
|
||||
&= \frac{1}{19} \Bigr[ (17^2 + 19^2 + \ldots + 58^2 ) - \frac{787^2}{20} \Bigr]\\
|
||||
&= 12.86
|
||||
\end{split}
|
||||
\]
|
||||
The mean is \framebox[1.5\width]{39.35} and the standard deviation is \framebox[1.5\width]{12.68}.
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part{2}
|
||||
\textbf{Interquartile Range}
|
||||
|
||||
Rearranging the data in ascending order, we get:\\\\
|
||||
17, 19, 22, 26, 28, 31, 34, 36, 38, 39, 41, 42, 43, 47, 50, 51, 53, 55, 57, 58, 62, 65, 97.\\
|
||||
|
||||
Hence, Q1 = 31 and Q3 = 55.\\
|
||||
\[
|
||||
\begin{split}
|
||||
\text{IQR} &= \text{Q3} - \text{Q1} \\
|
||||
&= 55 - 31 \\
|
||||
&= 24
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
\part{3}
|
||||
\textbf{Outliers}
|
||||
|
||||
The interquartile range is 24.\\
|
||||
$1.5 \times \text{IQR} = 36$.\\
|
||||
$Q3+1.5 \times \text{IQR} = 55 + 36 = 91$.\\
|
||||
$Q1-1.5 \times \text{IQR} = 31 - 36 = -5$.\\
|
||||
Hence, the outlier is 97, since it is greater than $1.5 \times \text{IQR}$ \\
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Consider the below tree diagram.
|
||||
\\
|
||||
\includegraphics{T1Q2.png}
|
||||
|
||||
\begin{enumerate}[label=\roman*.]
|
||||
\item The probability $x$ represents (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $P(A_1)$
|
||||
\item $P(A_1|B_2)$
|
||||
\item $P(B_2|A_1)$
|
||||
\item $P(C_1|B_2 \cap A_1)$
|
||||
\end{enumerate}
|
||||
\item The probability $y$ represents (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $P(B_2)$
|
||||
\item $P(A_1|B_2)$
|
||||
\item $P(B_2|A_1)$
|
||||
\item $P(C_1|B_2 \cap A_1)$
|
||||
\end{enumerate}
|
||||
\item The probability $z$ represents (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $P(C_1)$
|
||||
\item $P(B_2|C_1)$
|
||||
\item $P(C_1|B_2)$
|
||||
\item $P(C_1|B_2 \cap A_1)$
|
||||
\end{enumerate}
|
||||
\item The circled node represents the event (MCQ)?
|
||||
\begin{enumerate}
|
||||
\item $C_1$
|
||||
\item $B_2\cap C_1$
|
||||
\item $A_1 \cap B_2 \cap C_1$
|
||||
\item $C_1|B_2\cap A_1$
|
||||
\end{enumerate}
|
||||
\text{[v. and vi. do not refer to the tree diagram above]}
|
||||
\item Let A and B be two events. Suppose that the probability that neither event occurs is 3/8. What is the probability that at least one of the events occurs?
|
||||
\item Let C and D be two events. Suppose P(C) = 0.5, $P(C \cap D) = 0.2$ and $P((C \cup D)\ complement) = 0.4$. What is P(D)?
|
||||
\end{enumerate}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\solution
|
||||
|
||||
\begin{enumerate}[label=\roman*.]
|
||||
\item The probability $x$ represents (MCQ)? $x = P(A_1)$
|
||||
\item The probability $y$ represents (MCQ)? $y = P(B_2|A_1)$
|
||||
\item The probability $z$ represents (MCQ)? $z =P(C_1|B_2 \cap A_1)$
|
||||
\item The circled node represents the event (MCQ)? $A_1 \cap B_2 \cap C_1$
|
||||
\item Since $P(A^c \cap B^c) = \frac{3}{8}$ and $P(A) +P(B) + P(A \cap B) = 1$,
|
||||
$P(\text{At least 1 event occurs}) = 1 - \frac{3}{8} = \frac{5}{8}$
|
||||
\end{enumerate}
|
||||
|
||||
Given:\\
|
||||
$P(C) = 0.5$ \\
|
||||
$P(C \cap D) = 0.2$ \\
|
||||
$P((C \cup D)^c) = 0.4$\\
|
||||
$P(C \cup D) = 1-0.4 = 0.6$\\\\
|
||||
|
||||
$P(C \cup D) = P(C)+P(D)-P(C\cap D)$\\
|
||||
$\rightarrow 0.6 = 0.5 + P(D) - 0.2 $\\
|
||||
$\rightarrow P(D) = 0.3$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
In a survey of 1929 students, the following data were obtained on “students’ first reason for application to the university in which they matriculated.”
|
||||
\\
|
||||
\includegraphics{T1Q3.png}
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that university quality is the first reason for a student to choose a university?
|
||||
\item For a full-time student, what is the probability that university quality is the first reason for choosing a university?
|
||||
\item Let A denote the event that a student is full-time, and let B denote the event that the student lists university quality as the first reason for applying. Are events A and B independent?
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
\part{a}
|
||||
|
||||
$P(\text{quality}) = \frac{821}{1929} \approx 0.426$
|
||||
\\\\
|
||||
\part{b}
|
||||
|
||||
$P(\text{quality}|\text{full-time}) = \frac{P(\text{quality} \cap \text{full-time})}{P(\text{full-time})} = \frac{\frac{421}{1929}}{\frac{890}{1929}} \approx 0.473$
|
||||
\\\\
|
||||
\part{c}
|
||||
|
||||
A: A student is full-time\\
|
||||
B: the student lists university quality as the first reason for applying.\\
|
||||
Are events A and B independent?
|
||||
\\
|
||||
First, check if $P(B) = P(B|A)$\\
|
||||
$$
|
||||
\begin{align}
|
||||
P(B) &= P(\text{University Quality}) = \frac{821}{1929} = 0.426 \\
|
||||
P(B|A) &= P(\text{University Quality}|\text{full time student}) = \frac{421}{890} = 0.473\\
|
||||
\end{align}
|
||||
$$
|
||||
\\
|
||||
Since $P(B|!A) \neq P(B)$, the events are not independent.
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A rare blood disease is found in 2\% of a certain population. A new blood test can correctly
|
||||
identify 96\% of the people with the disease and 94\% of the people without the disease.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item What is the probability that a person who is tested positive by the blood test actually has the disease?
|
||||
\item What is the probability that a person who is tested negative by the blood test actually does not have the disease?
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
Let D = the person has the disease\\
|
||||
Let T = the person tests positive\\
|
||||
|
||||
$P(D) = 0.02$\\
|
||||
$P(T|D) = 0.96$\\
|
||||
$P(T^c|D^c) = 0.94$\\
|
||||
|
||||
\part{a}
|
||||
$$
|
||||
\begin{align}
|
||||
P(D|T) &= \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D^c)P(D^c)}\\
|
||||
&=\frac{0.96 \times 0.02}{0.96 \times 0.02 + 0.06 \times 0.98}\\
|
||||
&= 0.246
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part{b}
|
||||
$$
|
||||
\begin{align}
|
||||
P(D^c|T^c) &= \frac{P(T^c|D^c)P(D^c)}{P(T^c|D^c)P(D^c) + P(T^c|D)P(D)}\\
|
||||
&=\frac{0.94 \times 0.98}{0.94 \times 0.98 + 0.04 \times 0.02}\\
|
||||
&= 0.999
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A woman is pregnant with male twins. Twins may be either identical or fraternal (non-
|
||||
identical). In general, $\frac{1}{3}$ twins born are identical. Obviously, identical twins must be of the
|
||||
same sex; fraternal twins may or may not be. Assume that identical twins are equally likely
|
||||
to be both boys or both girls, while for fraternal twins all possibilities are equally likely:
|
||||
boy-girl, girl-boy, boy-boy, girl-girl. Given the above information, what is the probability
|
||||
that the woman’s male twins are identical?
|
||||
\\
|
||||
\\
|
||||
\solution\\
|
||||
|
||||
$$P(\text{identical}|BB) = \frac{P(BB|\text{identical})P(\text{identical})}{P(BB)} =
|
||||
\frac{\frac{1}{2}\cdot \frac{1}{3}}{\frac{1}{2}\cdot \frac{1}{3} + \frac{1}{4}\cdot\frac{2}{3}} = \frac{1}{2}$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
If a family had three kids, named Alice, Bob, and Carl. Assume that each is equally likely
|
||||
to be born; I.E., $\frac{1}{3}$ chance for each of them to be born first etc.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that Alice is older than Bob, given that Alice is older than Carl.
|
||||
\item Is event “Alice is older than Bob” independent from event “Alice is older than C”?
|
||||
\end{enumerate}
|
||||
|
||||
\solution\\
|
||||
A)\\
|
||||
Let A represent the event of Alice born\\
|
||||
Let B represent the event of Bob born\\
|
||||
Let C represent the event of Carl born
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(A>B|A>C) &= \frac{P(A>B \cap A>C)}{P(A>C)}\\
|
||||
&= \frac{\frac{1}{3}}{\frac{1}{2}}\\
|
||||
&= \frac{2}{3}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
B)\\
|
||||
They are not independent because $P(A>B|A>C) \neq P(A>B)$\\
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}[7]
|
||||
A boy receives a school report card each week. He is given a special treat whenever his
|
||||
report indicates “Good behavior” \textbf{AND} “Excellent Homework”. His behavior is good with
|
||||
probability 0.6. When his behavior is good, he has a probability of 0.8 of doing excellent
|
||||
homework. When his behavior is not good, his probability of doing excellent homework
|
||||
is only 0.5.
|
||||
|
||||
\begin{enumerate}[label=\alph*)]
|
||||
\item For a random week, calculate the probability that he will be given a special treat
|
||||
\item For a random Week, calculate the probability that his homework will be excellent but he will not be given a special treat
|
||||
\item For a random Week, calculate the probability that his homework will be excellent
|
||||
\item Given that his homework is excellent, calculate the conditional probability that he is NOT given a special treat.
|
||||
\end{enumerate}
|
||||
|
||||
|
||||
\solution
|
||||
\\
|
||||
Let G represent the event of good behavior\\
|
||||
Let E represent the event of excellent homework\\
|
||||
$E^c$ is the complement of E and $G^c$ is the complement of G\\
|
||||
|
||||
A)\\
|
||||
"Good Behavior" and "Excellent Homework"\\
|
||||
$$P(G\cap E) = P(G)P(E|G) = 0.6 \times 0.8 = 0.48$$
|
||||
|
||||
B)\\
|
||||
"Not Good behavior" AND "Excellent Homework"\\
|
||||
$$P(G^c \cap E) = P(G^c)P(E|G^c) = 0.4 \times 0.5 = 0.2$$
|
||||
|
||||
C)\\
|
||||
"Excellent Homework"\\
|
||||
$$P(E) = P(G)P(E|G) + P(G^c)P(E|G^c) = 0.6 \times 0.8 + 0.4 \times 0.5 = 0.68$$
|
||||
|
||||
D)\\
|
||||
"Not Good Behavior" given "Excellent Homework"\\
|
||||
$$P(G^c|E) = \frac{P(G^c \cap E)}{P(E)} = \frac{0.2}{0.68} = 0.294$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}[8]
|
||||
In an attempt to find the mean number of hours his tutorial classmates spent per day preparing for tutorials,
|
||||
John collected data from 10 of his friends in the tutorial group and found that the sample mean is 2.4 hours with a sample standard deviation of 0.8 hours.
|
||||
However, a day later he felt that the sample size is too small.
|
||||
So he collected data from another 5 of his friends and found that the sample mean is 2.0 hours with a sample standard deviation of 1.2 hours.
|
||||
\\
|
||||
Find the sample mean and sample standard deviation \underline{when these 2 sets of data are combined}.
|
||||
\\
|
||||
\\
|
||||
\solution\\
|
||||
Given:
|
||||
\[
|
||||
\begin{split}
|
||||
X &= \{x_1, x_2, \dots, x_{10}\} \\
|
||||
Y &= \{y_1, y_2, \dots, y_{5}\}\\\\
|
||||
\bar{x} &= \frac{\sum x_i}{10} = 2.4\\
|
||||
\sum x_i &= 10 \times 2.4 = 24 \\
|
||||
\bar{y} &= \frac{\sum y_i}{5} = 2.0 \\
|
||||
\sum y_i &= 5 \times 2.0 = 10 \\
|
||||
\text{new mean}\ \bar{z} &= \frac{\sum x_i + \sum y_i}{n_x + n_y} = \frac{24 + 10}{10+5} = 2.267\\ \\
|
||||
\end{split}
|
||||
\]
|
||||
\\
|
||||
Sample Standard Deviation can be used to figure out the missing values\\
|
||||
\[
|
||||
\begin{split}
|
||||
S_x^2 &= \frac{1}{10-1} \Bigr[\sum x_i^2 - \frac{(\sum x_i)^2}{10}\Bigl] = 0.8^2\text{, so } \sum x_i^2 = 9 \times 0.8^2 + \frac{24^2}{10} = 63.36\\
|
||||
S_y^2 &= \frac{1}{5-1} \Bigr[\sum y_i^2 - \frac{(\sum y_i)^2}{5}\Bigl] = 1.2^2\text{, so } \sum y_i^2 = 4 \times 1.2^2 + \frac{10^2}{5} = 25.76\\
|
||||
\text{new variance } s_z^2 &= \frac{1}{10+5-1} \Bigr[ \sum x_i^2 + \sum y_i^2 - \frac{(\sum x_i + \sum y_i)^2}{10+5} \Bigl] \\
|
||||
&= \frac{1}{14} \Bigr[63.36 + 25.76 - \frac{(24+10)^2}{15} \Bigl] = 0.86\\
|
||||
\text{new Standard Deviation } s_z &= \sqrt{s_z^2} = \sqrt{0.86} = 0.927
|
||||
\end{split}
|
||||
\]
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}[9]
|
||||
A certain family has 6 children, consisting of 3 boys and 3 girls.
|
||||
Assuming that all birth orders are equally likely,
|
||||
what is the probability that the 3 eldest children are the 3 girls?
|
||||
\\
|
||||
\\
|
||||
\solution
|
||||
\\
|
||||
$$
|
||||
\begin{align}
|
||||
\text{Probability that the 3 eldest are the 3 girls} &= \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4}\\
|
||||
&= \frac{6}{120}\\
|
||||
&= 0.05
|
||||
\end{align}
|
||||
$$
|
||||
Note: Only one possible way the 3 girls can be the oldest
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
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\end{document}
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$
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$
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[9] (./INF1004_W3_Tutorial_2200624.aux) )
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\documentclass[a4paper]{article}
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\title{
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}
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\author{\hmwkAuthorName \\ \hmwkStudentID}
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\begin{document}
|
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\maketitle
|
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|
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\pagebreak
|
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|
||||
\begin{homeworkProblem}
|
||||
A spam filter is designed by looking at commonly occurring phrases in spam. Suppose
|
||||
that 80\% of email is spam. In 10\% of the spam emails, the phrase “free money” is used.
|
||||
Whereas the phrase “free money” is used in only 1\% of non-spam emails. A new email
|
||||
has just arrived, which does mention “free money”. What is the probability that it is
|
||||
spam?
|
||||
\\\\
|
||||
\solution
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(\text{spam} \mid \text{free money}) &= \frac{P(\text{free money} \mid \text{spam}) \cdot P(\text{spam})}{P(\text{free money})} \\
|
||||
&= \frac{0.1 \cdot 0.8}{0.8 \cdot 0.1 + 0.2 \cdot 0.01} \\
|
||||
&= \frac{40}{41}\\
|
||||
&\approx 0.9756097561
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A sample of people, who commute regularly from a town in
|
||||
surrey into London, were asked for an estimate of the time
|
||||
taken on their most recent journey. The replies are
|
||||
summarized in the given table.
|
||||
\\\\
|
||||
\begin{tabular}{|c|c|}
|
||||
\hline
|
||||
Time in minutes & Frequency \\
|
||||
\hline
|
||||
35 to 45 & 12\\
|
||||
45 to 55 & 54\\
|
||||
55 to 65 & 68\\
|
||||
65 to 85 & 41\\
|
||||
85 to 105 & 23\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
|
||||
\begin{enumerate}[label=\alph*)]
|
||||
\item Calculate the estimate of the mean and the standard deviation of these times.
|
||||
\item What is the median?
|
||||
\item What is the mode?
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
\\
|
||||
\part \\
|
||||
|
||||
Using the midpoints of the intervals, we can calculate the mean and standard deviation as follows:\\\\
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
Time in minutes & Frequency & $x_if_i$ & $(x_if_i)^2$\\
|
||||
\hline
|
||||
40& 12 & 480 & 230400\\
|
||||
50& 54 & 2700 & 729000\\
|
||||
60& 68 & 4080 & 16646400\\
|
||||
70& 41 & 2870 & 8236900\\
|
||||
95& 23 & 2185 & 4774225\\
|
||||
\hline
|
||||
total & 198 & 12315 & 30616925\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\\\
|
||||
$$
|
||||
\begin{align}
|
||||
\bar{x} &= \frac{40 \cdot 12 + 50 \cdot 54 + 60 \cdot 68 + 70 \cdot 41 + 95 \cdot 23}{12 + 54 + 68 + 41 + 23} \\
|
||||
&= \frac{12735}{198} \\
|
||||
&= 62.19697 \text{ minutes}
|
||||
\\\\
|
||||
s &= \sqrt{\frac{1}{198 - 1} \sum x_i^2f_i-\frac{(\sum x_if_i)^2 }{n}}\\
|
||||
&= 14.517505\\
|
||||
&\approx 14.52
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\pagebreak
|
||||
|
||||
\part
|
||||
|
||||
The median is the middle value of the data set. Since there are 198 data points, the median is the average of the 99th and 100th data points.\\
|
||||
|
||||
Since the 99th and 100th data points lie in the 55 to 65 interval,
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
Q &=L_m +\Bigr( \frac{\frac{n}{2} -cf_{m-1}}{f_m}\Bigr)w_m\\
|
||||
&= 55 + \frac{\frac{198}{2} - 66}{68} \cdot 10\\
|
||||
&= 59.85294\\
|
||||
&\approx 59.85 \text{ minutes}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
The modal class is the class "55 to 65" with a frequency of 68.
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Maurice works at home. At 2 pm he decides to take a break to buy a copy of the chronicle
|
||||
newspaper. There are three nearby newsagents: Arif, Bob and Carol. However, by 2 pm
|
||||
they may have sold all their chronicles and so have none available. The independent
|
||||
probabilities that they have a chronicle available at 2 pm are: Arif 0.4, Bob 0.7, Carol 0.25
|
||||
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that none of the three newsagents have the chronicles at 2 pm.?
|
||||
\item Maurice decides to visit the newsagent in turn until he obtains a chronicle or until he has visited all three. He tosses a coin. If it lands heads he will visit the three newsagents in the order Bob, Arif, Carol. If it lands tail, he will visit them in order Carol, Arif, Bob. Find the probability that he will obtain a chronicle from Arif.
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{none available}) = 0.6 \cdot 0.3 \cdot 0.75 = 0.045$$
|
||||
|
||||
\part
|
||||
|
||||
$$P(\text{Arif}) = 0.5\cdot 0.3 \cdot 0.4 \cdot 0.75 + 0.5 \cdot 0.75 \cdot 0.4 \cdot 0.0.3 = 0.09$$
|
||||
\\\\
|
||||
First statement: "Head", "bob", "arif", "carol"\\
|
||||
Second statement: "Tail", "carol", "arif", "bob"\\
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
The following table summarizes the number of late arrivals of all students who attend a
|
||||
certain school on 5 mornings of particular week. (Show your steps for all of the following.)
|
||||
\\
|
||||
\begin{tabular}{|c|c|}
|
||||
\hline
|
||||
Number of late arrivals & Number of Pupils\\
|
||||
\hline
|
||||
0 & 275\\
|
||||
1 & 111\\
|
||||
2 & 33\\
|
||||
3 & 12\\
|
||||
4 & 13\\
|
||||
5 & 16\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Let random variable X=number of late arrivals. Is the random variable discrete or continuous?
|
||||
\item Draw the probability distribution for X. Calculate the mean and the standard deviation of the data in the table.
|
||||
\item Calculate the expected value.
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
The random variable is \boxed{discrete}.
|
||||
\\\\
|
||||
\part
|
||||
|
||||
\\
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
Number of late arrivals & Number of Pupils &$P(X)$ & $P(X) \cdot x$\\
|
||||
\hline
|
||||
0 & 275 & $\frac{265}{460} \approx 0.598$ & 0\\
|
||||
1 & 111 & $\frac{111}{460} \approx 0.241$ & 0.241\\
|
||||
2 & 33 & $\frac{33}{460} \approx 0.072$ & 0.143\\
|
||||
3 & 12 & $\frac{12}{460} \approx 0.026$ & 0.5217\\
|
||||
4 & 13 & $\frac{13}{460} \approx 0.028$ & 0.0565\\
|
||||
5 & 16 & $\frac{16}{460} \approx 0.035$ & 0.0695\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\\\
|
||||
$$\text{mean} = \frac{275 \cdot 0 + 111 \cdot 1 + 33 \cdot 2 + 12 \cdot 3 + 13 \cdot 4 + 16 \cdot 5}{460} \approx 0.75$$
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
s &= \sqrt{\frac{1}{460 - 1} \sum x_i^2p_i-\frac{(\sum x_ip_i)^2 }{n}}\\
|
||||
&= 1.23515
|
||||
\end{align}
|
||||
$$
|
||||
\part
|
||||
|
||||
$$E(x) = 0.75$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A Terrence walks to school every morning. The probability that he arrives late is 0.15 and
|
||||
independent of whether he arrives late on any other morning. For a week, in which he
|
||||
decides to walk to school on five mornings, find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Probability he arrives late on two or fewer mornings.
|
||||
\item Probability he arrives late on three or more mornings.
|
||||
\item Find the mean and standard deviation of the number of mornings on which he arrives late.?
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
Let X be the number of days he arrives late.\\
|
||||
n = 5\\
|
||||
p(X) = 0.15\\
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(X \leq 2) &= P(X = 0) + P(X = 1) + P(X = 2)\\
|
||||
&= \binom{5}{2}(o.15)^2(1-0.15)^{5-2} + \binom{5}{1}(0.15)(1-0.15)^{5-1} + \binom{5}{0}(1-0.15)^5\\
|
||||
&= 0.973388125\\
|
||||
&\approx 0.9734
|
||||
\end{align}
|
||||
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(X \geq 3) &= P(X = 3) + P(X = 4) + P(X = 5)\\
|
||||
&= \binom{5}{3}(0.15)^3(1-0.15)^{5-3} + \binom{5}{4}(0.15)^4(1-0.15)^{5-4} + \binom{5}{5}(0.15)^5\\
|
||||
&= 0.026611875\\
|
||||
&\approx 0.0266
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
E(x) &= 5 \cdot 0.15 = 0.75\\
|
||||
Var(X) &= 0.75(1-0.15) = 0.6375\\
|
||||
s &= \sqrt{0.6375} \approx 0.799
|
||||
\end{align}
|
||||
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Bob is a student who travels to and from University by bus. He believes that the
|
||||
probability of having to wait more than six minutes to catch a bus is 0.4 and is
|
||||
independent of the time of day and direction of travel.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Assuming Bob beliefs are correct. Calculate values for the mean and standard deviation of the number of times he has to wait more than six minutes to catch a bus in a week when he catches 10 buses.
|
||||
\item During a thirteen-week period, the number of times (out of 10) he had to wait more than six minutes to catch a bus were as follows:
|
||||
$$4, 8, 8, 9, 3, 2, 2, 7, 0, 1, 5, 2, 0$$
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item Calculate the mean and standard deviation of this data.
|
||||
\item Does the answer in (b)i support Bob’s beliefs that the probability of having to wait more than six minutes to catch a bus is constant as 0.4 regardless of any factors?
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
\\
|
||||
\textbf{Given}:\\
|
||||
n = 10\\
|
||||
p(X) = 0.4\\
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
E(X) &= 10 \cdot 0.4 = 4\\
|
||||
Var(X) &= 10 \cdot 0.4 \cdot 0.6 = 2.4\\
|
||||
s &= \sqrt{2.4} \approx 1.549
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
E(X) &= \frac{4+8+8+9+3+2+2+7+0+1+5+2+0}{10} \approx 3.923076923\\
|
||||
s &= \sqrt{\frac{1}{10-1} \sum (x_i - 3.6923)^2} \approx 3.17441
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
The Mean of the data is 3.6923 which is not equal to 0.4. Therefore, Bob's belief is incorrect.
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A shop sells two types of bath cubes: relaxing and invigorating. The owner of a hotel
|
||||
buys 50 cubes from this shop. For each cube, there is independently a probability of 0.4
|
||||
that it will be relaxing.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that more than 90\% of the 50 cubes will be relaxing cubes.
|
||||
\item The 50 cubes included 22 relaxing cubes. Give a reason, whether or not a
|
||||
binomial distribution will provide an appropriate model for the random variable,
|
||||
R, in each of the following cases:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item A guest at the hotel randomly selects one of the 50 cubes. If it is not a
|
||||
relaxing cube, he replaces it and again selects a cube at random. He
|
||||
continues this procedure until he obtains a relaxing cube. The random
|
||||
variable R denotes the number of cubes he selects until he obtains a
|
||||
relaxing cube.
|
||||
\item The owner randomly selects 20 of the 50 cubes and places them in a
|
||||
bowl. The random variable R denotes the number of relaxing cubes
|
||||
placed in the bowl; each being evaluated as one trial.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
|
||||
\part
|
||||
|
||||
Let X= the number of relaxing cubes.\\
|
||||
\textbf{Given}:\\
|
||||
p(X) = 0.4\\
|
||||
|
||||
90\% of 50 = 45\\
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
p(X > 45) &= p(X = 46) + p(X = 47) + p(X = 48) + p(X = 49) + p(X = 50)\\
|
||||
&= \binom{50}{46}(0.4)^{46}(0.6)^{50-46} + \binom{50}{47}(0.4)^{47}(0.6)^{50-47}\\
|
||||
&+ \binom{50}{48}(0.4)^{48}(0.6)^{50-48} + \binom{50}{49}(0.4)^{49}(0.6)^{50-49}\\
|
||||
&+ \binom{50}{50}(0.4)^{50}(0.6)^{50-50}\\
|
||||
&=
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
As p(R) is not an independent event, the binomial distribution will not be an appropriate model for the random variable R.
|
||||
The variable n is also not fixed, therefore, the binomial distribution will not be an appropriate model for the random variable R.
|
||||
|
||||
\part
|
||||
|
||||
As the probability is not constant, the binomial distribution will not be an appropriate model for the random variable R.
|
||||
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\end{document}
|
||||
|
|
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|
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|
||||
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|
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|
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|
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./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Extra }, or forgotten $.
|
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<template> \unskip \hfil }
|
||||
\hskip \tabcolsep \hskip -.5\arrayrulewidth \vrule...
|
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l.281 x & f & $P(x) = \frac{f}{460} &
|
||||
$P(x)\cdot x$ & (x-\mu)^2 & (x-\mu)^2 ...
|
||||
I've deleted a group-closing symbol because it seems to be
|
||||
spurious, as in `$x}$'. But perhaps the } is legitimate and
|
||||
you forgot something else, as in `\hbox{$x}'. In such cases
|
||||
the way to recover is to insert both the forgotten and the
|
||||
deleted material, e.g., by typing `I$}'.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.281 x & f & $P(x) = \frac{f}{460} &
|
||||
$P(x)\cdot x$ & (x-\mu)^2 & (x-\mu)^2 ...
|
||||
I've inserted a begin-math/end-math symbol since I think
|
||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.281 x & f & $P(x) = \frac{f}{460} &
|
||||
$P(x)\cdot x$ & (x-\mu)^2 & (x-\mu)^2 ...
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.281 ... = \frac{f}{460} & $P(x)\cdot x$ & (x-\mu
|
||||
)^2 & (x-\mu)^2 \cdot P(x)\\
|
||||
I've inserted a begin-math/end-math symbol since I think
|
||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Extra }, or forgotten $.
|
||||
<template> \unskip \hfil }
|
||||
\hskip \tabcolsep \hskip -.5\arrayrulewidth \vrule...
|
||||
l.281 ...rac{f}{460} & $P(x)\cdot x$ & (x-\mu)^2 &
|
||||
(x-\mu)^2 \cdot P(x)\\
|
||||
I've deleted a group-closing symbol because it seems to be
|
||||
spurious, as in `$x}$'. But perhaps the } is legitimate and
|
||||
you forgot something else, as in `\hbox{$x}'. In such cases
|
||||
the way to recover is to insert both the forgotten and the
|
||||
deleted material, e.g., by typing `I$}'.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.281 ...rac{f}{460} & $P(x)\cdot x$ & (x-\mu)^2 &
|
||||
(x-\mu)^2 \cdot P(x)\\
|
||||
I've inserted a begin-math/end-math symbol since I think
|
||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.281 ...rac{f}{460} & $P(x)\cdot x$ & (x-\mu)^2 &
|
||||
(x-\mu)^2 \cdot P(x)\\
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:281: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.281 ...460} & $P(x)\cdot x$ & (x-\mu)^2 & (x-\mu
|
||||
)^2 \cdot P(x)\\
|
||||
I've inserted a begin-math/end-math symbol since I think
|
||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:282: Extra }, or forgotten $.
|
||||
<template> \unskip \hfil }
|
||||
\hskip \tabcolsep \hskip -.5\arrayrulewidth \vrule...
|
||||
l.282 \hline
|
||||
|
||||
I've deleted a group-closing symbol because it seems to be
|
||||
spurious, as in `$x}$'. But perhaps the } is legitimate and
|
||||
you forgot something else, as in `\hbox{$x}'. In such cases
|
||||
the way to recover is to insert both the forgotten and the
|
||||
deleted material, e.g., by typing `I$}'.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:282: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.282 \hline
|
||||
|
||||
I've inserted a begin-math/end-math symbol since I think
|
||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:282: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.282 \hline
|
||||
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: LaTeX Error: \begin{tabular} on input line 279 ended by \end{tabluar}.
|
||||
|
||||
See the LaTeX manual or LaTeX Companion for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.292 \end{tabluar}
|
||||
|
||||
Your command was ignored.
|
||||
Type I <command> <return> to replace it with another command,
|
||||
or <return> to continue without it.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.292 \end{tabluar}
|
||||
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.292 \end{tabluar}
|
||||
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: Missing \cr inserted.
|
||||
<inserted text>
|
||||
\cr
|
||||
l.292 \end{tabluar}
|
||||
|
||||
I'm guessing that you meant to end an alignment here.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: Missing { inserted.
|
||||
<inserted text>
|
||||
{
|
||||
l.292 \end{tabluar}
|
||||
|
||||
I've put in what seems to be necessary to fix
|
||||
the current column of the current alignment.
|
||||
Try to go on, since this might almost work.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.292 \end{tabluar}
|
||||
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.292 \end{tabluar}
|
||||
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:292: Missing } inserted.
|
||||
<inserted text>
|
||||
}
|
||||
l.292 \end{tabluar}
|
||||
|
||||
I've inserted something that you may have forgotten.
|
||||
(See the <inserted text> above.)
|
||||
With luck, this will get me unwedged. But if you
|
||||
really didn't forget anything, try typing `2' now; then
|
||||
my insertion and my current dilemma will both disappear.
|
||||
|
||||
|
||||
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||||
|
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|
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|
||||
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./INF1004_W3_Tutorial_2200624_Solutions.tex:335: Package amsmath Error: Erroneous nesting of equation structures;
|
||||
(amsmath) trying to recover with `aligned'.
|
||||
|
||||
See the amsmath package documentation for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.335 \end{align}
|
||||
|
||||
Try typing <return> to proceed.
|
||||
If that doesn't work, type X <return> to quit.
|
||||
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||||
|
||||
Underfull \hbox (badness 10000) in paragraph at lines 339--340
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[]
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|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:350: Package amsmath Error: Erroneous nesting of equation structures;
|
||||
(amsmath) trying to recover with `aligned'.
|
||||
|
||||
See the amsmath package documentation for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.350 \end{align}
|
||||
|
||||
Try typing <return> to proceed.
|
||||
If that doesn't work, type X <return> to quit.
|
||||
|
||||
|
||||
Underfull \hbox (badness 10000) in paragraph at lines 354--355
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||||
|
||||
[]
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||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:360: Package amsmath Error: Erroneous nesting of equation structures;
|
||||
(amsmath) trying to recover with `aligned'.
|
||||
|
||||
See the amsmath package documentation for explanation.
|
||||
Type H <return> for immediate help.
|
||||
...
|
||||
|
||||
l.360 \end{align}
|
||||
|
||||
Try typing <return> to proceed.
|
||||
If that doesn't work, type X <return> to quit.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:361: Missing $ inserted.
|
||||
<inserted text>
|
||||
$
|
||||
l.361
|
||||
|
||||
I've inserted a begin-math/end-math symbol since I think
|
||||
you left one out. Proceed, with fingers crossed.
|
||||
|
||||
./INF1004_W3_Tutorial_2200624_Solutions.tex:361: Display math should end with $$.
|
||||
<to be read again>
|
||||
\tex_par:D
|
||||
l.361
|
||||
|
||||
The `$' that I just saw supposedly matches a previous `$$'.
|
||||
So I shall assume that you typed `$$' both times.
|
||||
|
||||
[7]
|
||||
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[8]
|
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|
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|
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[9] (./INF1004_W3_Tutorial_2200624_Solutions.aux) )
|
||||
Here is how much of TeX's memory you used:
|
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</usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmbx10.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmbx12.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmex10.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmmi10.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmmi7.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmr10.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmr12.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmr7.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmsy10.pfb></usr/share/texmf-dist/fonts/type1/public/amsfonts/cm/cmsy7.pfb>
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Output written on INF1004_W3_Tutorial_2200624_Solutions.pdf (9 pages, 249340 bytes).
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|
||||
|
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|
||||
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|
||||
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|
||||
\usepackage{amsthm}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{tikz}
|
||||
\usepackage[plain]{algorithm}
|
||||
\usepackage{algpseudocode}
|
||||
\usepackage{enumitem}
|
||||
|
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\graphicspath{ {./images/} }
|
||||
|
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\usetikzlibrary{automata,positioning}
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||||
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|
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|
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|
||||
\linespread{1.1}
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|
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\pagestyle{fancy}
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\lhead{\hmwkAuthorName}
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\chead{\hmwkClass\ : \hmwkTitle}
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\rhead{\firstxmark}
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|
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|
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\setlength\parindent{0pt}
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%
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|
||||
%
|
||||
|
||||
\newcommand{\enterProblemHeader}[1]{
|
||||
\nobreak\extramarks{}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
}
|
||||
|
||||
\newcommand{\exitProblemHeader}[1]{
|
||||
\nobreak\extramarks{Question \arabic{#1} (continued)}{Question \arabic{#1} continued on next page\ldots}\nobreak{}
|
||||
\stepcounter{#1}
|
||||
\nobreak\extramarks{Question \arabic{#1}}{}\nobreak{}
|
||||
}
|
||||
|
||||
\setcounter{secnumdepth}{0}
|
||||
\newcounter{partCounter}
|
||||
\newcounter{homeworkProblemCounter}
|
||||
\setcounter{homeworkProblemCounter}{1}
|
||||
\nobreak\extramarks{Question \arabic{homeworkProblemCounter}}{}\nobreak{}
|
||||
|
||||
\newenvironment{homeworkProblem}[1][-1]{
|
||||
\ifnum#1>0
|
||||
\setcounter{homeworkProblemCounter}{#1}
|
||||
\fi
|
||||
\section{Question \arabic{homeworkProblemCounter}}
|
||||
\setcounter{partCounter}{1}
|
||||
\enterProblemHeader{homeworkProblemCounter}
|
||||
}{
|
||||
\exitProblemHeader{homeworkProblemCounter}
|
||||
}
|
||||
|
||||
\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
|
||||
\newcommand{\hmwkTitle}{Tutorial\ \#3\ Solutions}
|
||||
\newcommand{\hmwkAuthorName}{\textbf{Woon Jun Wei}}
|
||||
\newcommand{\hmwkStudentID}{\textbf{2200624}}
|
||||
|
||||
\title{
|
||||
\vspace{2in}
|
||||
\textmd{\textbf{\hmwkClass \\ \hmwkTitle}}\\
|
||||
\vspace{3in}
|
||||
}
|
||||
|
||||
\author{\hmwkAuthorName \\ \hmwkStudentID}
|
||||
\date{\today}
|
||||
|
||||
\renewcommand{\part}[1]{\textbf{\large Part \arabic{partCounter}}\stepcounter{partCounter}\\}
|
||||
|
||||
%
|
||||
% Various Helper Commands
|
||||
%
|
||||
|
||||
% Useful for algorithms
|
||||
\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
|
||||
|
||||
% For derivatives
|
||||
\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
|
||||
|
||||
% For partial derivatives
|
||||
\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
|
||||
|
||||
% Integral dx
|
||||
\newcommand{\dx}{\mathrm{d}x}
|
||||
|
||||
% Alias for the Solution section header
|
||||
\newcommand{\solution}{\textbf{\large My Solution\\}}
|
||||
\newcommand{\answer}{\textbf{\large Sample Solutions\\}}
|
||||
|
||||
% Probability commands: Expectation, Variance, Covariance, Bias
|
||||
\newcommand{\E}{\mathrm{E}}
|
||||
\newcommand{\Var}{\mathrm{Var}}
|
||||
\newcommand{\Cov}{\mathrm{Cov}}
|
||||
\newcommand{\Bias}{\mathrm{Bias}}
|
||||
|
||||
\begin{document}
|
||||
|
||||
\maketitle
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A spam filter is designed by looking at commonly occurring phrases in spam. Suppose
|
||||
that 80\% of email is spam. In 10\% of the spam emails, the phrase “free money” is used.
|
||||
Whereas the phrase “free money” is used in only 1\% of non-spam emails. A new email
|
||||
has just arrived, which does mention “free money”. What is the probability that it is
|
||||
spam?
|
||||
\\\\
|
||||
\answer
|
||||
|
||||
Let S be the event that an email is spam\\
|
||||
Let F be the event that an email has the "free money" phrase\\
|
||||
$P(S) = 0.8$\\
|
||||
$P(S^c) = 0.2$\\
|
||||
$P(F \mid S) = 0.1$ (Free money given spam)\\
|
||||
$P(F \mid S^c) = 0.01$ (Free money given non-spam)\\
|
||||
$P(S \mid F)$ Probability ofspam given free money phrase?\\
|
||||
|
||||
Using bayes rule
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(S \mid F) &= \frac{P(F \mid S) \cdot P(S)}{P(F)} \\
|
||||
&= \frac{0.1 \cdot 0.8}{0.8 \cdot 0.1 + 0.2 \cdot 0.01} \\
|
||||
&= \frac{40}{41}\\
|
||||
&\approx 0.9756097561
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A sample of people, who commute regularly from a town in
|
||||
surrey into London, were asked for an estimate of the time
|
||||
taken on their most recent journey. The replies are
|
||||
summarized in the given table.
|
||||
\\\\
|
||||
\begin{tabular}{|c|c|}
|
||||
\hline
|
||||
Time in minutes & Frequency \\
|
||||
\hline
|
||||
35 to 45 & 12\\
|
||||
45 to 55 & 54\\
|
||||
55 to 65 & 68\\
|
||||
65 to 85 & 41\\
|
||||
85 to 105 & 23\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
|
||||
\begin{enumerate}[label=\alph*)]
|
||||
\item Calculate the estimate of the mean and the standard deviation of these times.
|
||||
\item What is the median?
|
||||
\item What is the mode?
|
||||
\end{enumerate}
|
||||
|
||||
\answer
|
||||
|
||||
\begin{tabular}{|c|c|c|c|c|c|}
|
||||
\hline
|
||||
Time in minutes & $f$ & Midpoint = x & $x_if_i$ & $x_i^2f_i$ & cf\\
|
||||
\hline
|
||||
35 to 45 & 12 & 40 & 480 & 19200 & 12\\
|
||||
45 to 55 & 54 & 50 & 2700 & 135000 & 66\\
|
||||
55 to 65 & 68 & 60 & 4080 & 244800 & 134\\
|
||||
65 to 85 & 41 & 75 & 3075 & 230625 & 175\\
|
||||
85 to 105 & 23 & 95 & 2185 & 207575 & 198\\
|
||||
\hline
|
||||
Sum: & 198 & & 12520 & 837200 & \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\\\
|
||||
$$
|
||||
\begin{aligned}
|
||||
\bar{x} &= \frac{\sum x_if_i}{n} = \frac{12520}{198} = \boxed{63.23}\\\\
|
||||
s^2 &= \frac{1}{n-1} \Bigr [\sum x_i^2f_i - \frac{(\sum x_if_i)^2}{n}\Bigr]\\
|
||||
&= \frac{1}{198-1} \Bigr [837200 - \frac{12520^2}{198}\Bigr] = 231.1234\\\\
|
||||
s &= \sqrt{231.1234} = \boxed{15.202}\\\\
|
||||
Q &=L_m +\Bigr( \frac{\frac{n}{2} -cf_{m-1}}{f_m}\Bigr)w_m\\
|
||||
&= 55 + \frac{\frac{198}{2} - 66}{68} \cdot (65-55)\\
|
||||
&= 59.85294\\
|
||||
&\approx \boxed{59.85 \text{ minutes}}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
b)\\
|
||||
$\frac{198}{2} = 99$\\
|
||||
Median class: $\boxed{55-65}$\\
|
||||
c)\\
|
||||
Mode class: $\boxed{55-65}$\\
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Maurice works at home. At 2 pm he decides to take a break to buy a copy of the chronicle
|
||||
newspaper. There are three nearby newsagents: Arif, Bob and Carol. However, by 2 pm
|
||||
they may have sold all their chronicles and so have none available. The independent
|
||||
probabilities that they have a chronicle available at 2 pm are: Arif 0.4, Bob 0.7, Carol 0.25
|
||||
|
||||
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that none of the three newsagents have the chronicles at 2 pm.?
|
||||
\item Maurice decides to visit the newsagent in turn until he obtains a chronicle or until he has visited all three. He tosses a coin. If it lands heads he will visit the three newsagents in the order Bob, Arif, Carol. If it lands tail, he will visit them in order Carol, Arif, Bob. Find the probability that he will obtain a chronicle from Arif.
|
||||
\end{enumerate}
|
||||
|
||||
\answer
|
||||
|
||||
Let the event of Ariv having the chronicles at 2pm be A\\
|
||||
Let the event of Bob having the chronicles at 2pm be B\\
|
||||
Let the event of Carol having the chronicles at 2pm be C\\
|
||||
$P(A) = 0.4, P(B) = 0.7, P(C) = 0.25$\\
|
||||
$P(A^c) = 0.6, P(B^c) = 0.3, P(C^c) = 0.75$\\
|
||||
|
||||
a)\\
|
||||
None of the three news agents have the chronicles at 2pm\\
|
||||
$$P(A^c \cap B^c \cap C^c) = P(A^c) \cdot P(B^c) \cdot P(C^c) = 0.6 \cdot 0.3 \cdot 0.75 = \frac{27}{200} = \boxed{0.135}$$
|
||||
|
||||
b)\\
|
||||
Find the probability that he will obtain a chronicle from Arif\\
|
||||
$$0.5 \cdot 0.3 \cdot 0.4 + 0.5 \cdot 0.75 \cdot 0.4 = \boxed{0.21}$$
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
The following table summarizes the number of late arrivals of all students who attend a
|
||||
certain school on 5 mornings of particular week. (Show your steps for all of the following.)
|
||||
\\
|
||||
\begin{tabular}{|c|c|}
|
||||
\hline
|
||||
Number of late arrivals & Number of Pupils\\
|
||||
\hline
|
||||
0 & 275\\
|
||||
1 & 111\\
|
||||
2 & 33\\
|
||||
3 & 12\\
|
||||
4 & 13\\
|
||||
5 & 16\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\\
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Let random variable X=number of late arrivals. Is the random variable discrete or continuous?
|
||||
\item Draw the probability distribution for X. Calculate the mean and the standard deviation of the data in the table.
|
||||
\item Calculate the expected value.
|
||||
\end{enumerate}
|
||||
|
||||
\answer
|
||||
|
||||
\part
|
||||
|
||||
a)\\
|
||||
The random variable is \boxed{discrete}.\\
|
||||
|
||||
b)\\
|
||||
\includegraphics{Q4.png}
|
||||
\\\\
|
||||
\begin{tabular}{|c|c|c|c|c|c|}
|
||||
\hline
|
||||
x & f & $P(x) = \frac{f}{460} & $P(x)\cdot x$ & (x-\mu)^2 & (x-\mu)^2 \cdot P(x)\\
|
||||
\hline
|
||||
0 & 275 & 0.598 & 0 & 0.5625 & 0.336\\
|
||||
1 & 111 & 0.241 & 0.241 & 0.0625 & 0.01508\\
|
||||
2 & 33 & 0.072 & 0.143 & 1.5625 & 0.11209\\
|
||||
3 & 12 & 0.026 & 0.0782 & 5.0625 & 0.132065\\
|
||||
4 & 13 & 0.028 & 0.1130 & 10.5625 & 0.29850\\
|
||||
5 & 16 & 0.035 & 0.173913 & 18.0625 & 0.62826\\
|
||||
\hline
|
||||
sum & 460 & 1 & 0.75 & & 1.5222\\
|
||||
\hline
|
||||
\end{tabluar}
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
E(X) &= \mu = \sum x P(X=x) = \boxed{0.75}\\\\
|
||||
Var(X) &= \sigma^2 = \sum (x-\mu)^2 P(X=x) = \boxed{1.52228}\\
|
||||
\sigma &= \sqrt{1.52228} = \boxed{1.23}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
c)\\
|
||||
$$E(X) = \boxed{0.75}$$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A Terrence walks to school every morning. The probability that he arrives late is 0.15 and
|
||||
independent of whether he arrives late on any other morning. For a week, in which he
|
||||
decides to walk to school on five mornings, find:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Probability he arrives late on two or fewer mornings.
|
||||
\item Probability he arrives late on three or more mornings.
|
||||
\item Find the mean and standard deviation of the number of mornings on which he arrives late.?
|
||||
\end{enumerate}
|
||||
|
||||
\answer
|
||||
|
||||
Let X be the number of days he arrives late.\\
|
||||
n = 5\\
|
||||
p(X) = 0.15\\
|
||||
|
||||
$$X \sim B(5, 0.15)$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(X \leq 2) &= P(X = 0) + P(X = 1) + P(X = 2)\\
|
||||
&= \binom{5}{2}(o.15)^2(1-0.15)^{5-2} + \binom{5}{1}(0.15)(1-0.15)^{5-1} + \binom{5}{0}(1-0.15)^5\\
|
||||
&= 0.973388125\\
|
||||
&\approx \boxed{0.9734}\\
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
P(X \geq 3) &= P(X = 3) + P(X = 4) + P(X = 5)\\
|
||||
&= \binom{5}{3}(0.15)^3(1-0.15)^{5-3} + \binom{5}{4}(0.15)^4(1-0.15)^{5-4} + \binom{5}{5}(0.15)^5\\
|
||||
&= 0.026611875\\
|
||||
&\approx \boxed{0.0266}\\\\
|
||||
P(X \geq 3) &= 1-P(X \leq 2)\\
|
||||
&= 1-0.9734\\
|
||||
&= 0.026611875\\
|
||||
&\approx \boxed{0.0266}
|
||||
\end{align}
|
||||
$$
|
||||
|
||||
\part
|
||||
|
||||
$$
|
||||
\begin{align}
|
||||
E(x) &= 5 \cdot 0.15 = \boxed{0.75}\\
|
||||
Var(X) &= 0.75(1-0.15) = 0.6375\\
|
||||
s &= \sqrt{0.6375} \approx \boxed{0.799}\ (\text{answer given was 0.7984})\\
|
||||
\end{align}
|
||||
|
||||
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
Bob is a student who travels to and from University by bus. He believes that the
|
||||
probability of having to wait more than six minutes to catch a bus is 0.4 and is
|
||||
independent of the time of day and direction of travel.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Assuming Bob beliefs are correct. Calculate values for the mean and standard deviation of the number of times he has to wait more than six minutes to catch a bus in a week when he catches 10 buses.
|
||||
\item During a thirteen-week period, the number of times (out of 10) he had to wait more than six minutes to catch a bus were as follows:
|
||||
$$4, 8, 8, 9, 3, 2, 2, 7, 0, 1, 5, 2, 0$$
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item Calculate the mean and standard deviation of this data.
|
||||
\item Does the answer in (b)i support Bob’s beliefs that the probability of having to wait more than six minutes to catch a bus is constant as 0.4 regardless of any factors?
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
|
||||
\answer
|
||||
|
||||
a)\\
|
||||
Binomial n = 10, p = 0.4\\
|
||||
Mean is $np = 10 \cdot 0.4 = \boxed{4}$\\
|
||||
Variance is $np(1-p) = 10 \cdot 0.4 \cdot 0.6 = \boxed{2.4}$\\
|
||||
Standard deviation is $s = \sqrt{2.4} \approx \boxed{1.549}$\\
|
||||
|
||||
b)\\
|
||||
$$4,8,8,9,3,2,2,7,0,1,5,2,0$$
|
||||
\\
|
||||
\begin{tabular}{|c|c|c|c|}
|
||||
\hline
|
||||
Number waits per week (X) & Frequency & $x_if_i$ & $x_i^2f_i$\\
|
||||
\hline
|
||||
0 & 2 & 0 & 0\\
|
||||
1 & 1 & 1 & 1\\
|
||||
2 & 3 & 6 & 12\\
|
||||
3 & 1 & 3 & 9\\
|
||||
4 & 1 & 4 & 16\\
|
||||
5 & 1 & 5 & 25\\
|
||||
6 & 0 & 0 & 0\\
|
||||
7 & 1 & 7 & 49\\
|
||||
8 & 2 & 16 & 128\\
|
||||
9 & 1 & 9 & 81\\
|
||||
\hline
|
||||
total & 13 & 51 & 341\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
|
||||
|
||||
i)\\
|
||||
|
||||
Mean = $\frac{51}{13} = \boxed{3.923076923}$\\
|
||||
Variance = $\frac{1}{13-1} \sum (x_i - 3.6923)^2 = \boxed{10.07692308}$\\
|
||||
Standard deviation = $\sqrt{10.07692308} \approx \boxed{3.17441}$\\
|
||||
ii)\\
|
||||
The mean is similar to binomial's mean, but the standard deviation is significantly higher which
|
||||
both indicate that 'p' isn't constant. This doesn't support Bob's belief.
|
||||
\end{homeworkProblem}
|
||||
|
||||
\pagebreak
|
||||
|
||||
\begin{homeworkProblem}
|
||||
A shop sells two types of bath cubes: relaxing and invigorating. The owner of a hotel
|
||||
buys 50 cubes from this shop. For each cube, there is independently a probability of 0.4
|
||||
that it will be relaxing.
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item Find the probability that more than 90\% of the 50 cubes will be relaxing cubes.
|
||||
\item The 50 cubes included 22 relaxing cubes. Give a reason, whether or not a
|
||||
binomial distribution will provide an appropriate model for the random variable,
|
||||
R, in each of the following cases:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item A guest at the hotel randomly selects one of the 50 cubes. If it is not a
|
||||
relaxing cube, he replaces it and again selects a cube at random. He
|
||||
continues this procedure until he obtains a relaxing cube. The random
|
||||
variable R denotes the number of cubes he selects until he obtains a
|
||||
relaxing cube.
|
||||
\item The owner randomly selects 20 of the 50 cubes and places them in a
|
||||
bowl. The random variable R denotes the number of relaxing cubes
|
||||
placed in the bowl; each being evaluated as one trial.
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
|
||||
\solution
|
||||
$$X \sim B(50, 0.4)$$
|
||||
n = 50, p = 0.4. Let X be the number of cubes of the 50 which are relaxing cubes
|
||||
|
||||
a)\\
|
||||
90\% of 50 = 45\\
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(x>45) &= P(x=46) + P(x=47) + P(x=48) + P(x=49) + P(x=50)\\
|
||||
&= \binom{50}{46}(0.4)^{46}(0.6)^{50-46} + \binom{50}{47}(0.4)^{47}(0.6)^{50-47}\\
|
||||
&+ \binom{50}{48}(0.4)^{48}(0.6)^{50-48} + \binom{50}{49}(0.4)^{49}(0.6)^{50-49}\\
|
||||
&+ \binom{50}{50}(0.4)^{50}(0.6)^{50-50}\\
|
||||
&= \boxed{1.65639 \cdot 10^{-14}}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
b)\\
|
||||
i)\\
|
||||
R = number of cubes selected until a relaxing cube.
|
||||
$\boxed{\text{Not binomial as n is not fixed}}$
|
||||
|
||||
ii)\\
|
||||
R = number of relaxing cubes placed in a bowl.\\
|
||||
$\text{Perhaps not enough information; }\boxed{\text{but not binomial as p is not constant}}$
|
||||
|
||||
\end{homeworkProblem}
|
||||
|
||||
|
||||
\end{document}
|
||||
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Reference in New Issue