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\newcommand{\hmwkClass}{INF 1004 Mathematics 2}
\newcommand{\hmwkTitle}{Tutorial\ \#8}
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\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

$$
\begin{aligned}
    3x+4y&=1\\
    2x+3y&=12
\end{aligned}
$$
\\
$$
\begin{aligned}
    3x-2y&=4\\
    -6x+4y&=7
\end{aligned}
$$
\\
$$
\begin{aligned}
    2x+y+z-6&=0\\
    4y+z+x&=5\\
    2x+z+3y&=7
\end{aligned}
$$

\pagebreak
\solution

\begin{multicols}{2}

\textbf{Part 1}\\

$$
\begin{aligned}
    x+2y-z&=3\\
    2x-3y+2z&=5\\
    -3x+y+5z&=13\\
\end{aligned}
$$
Transformed into augmented matrix form:
$$
\left[\begin{array}{rrr|r}
1 & 2 & -1 & 3\\
2 & -3 & 2 & 5\\
-3 & 1 & 5 & 13
\end{array}\right]
$$
$3\rho_1+\rho_3$:
$
\left[\begin{array}{rrr|r}
1 & 2 & -1 & 3\\
2 & -3 & 2 & 5\\
0 & 7 & 2 & 22
\end{array}\right]
$\\
$2\rho_1 - \rho_2$:
$
\left[\begin{array}{rrr|r}
1 & 2 & -1 & 3\\
0&7&-4&1\\
0 & 7 & 2 & 22
\end{array}\right]
$\\
$\rho_2 - \rho_3$:
$
\left[\begin{array}{rrr|r}
1 & 2 & -1 & 3\\
0&7&-4&1\\
0 & 0 & -6 & -21
\end{array}\right]
$\\
$\frac{1}{6}\rho_3$:
$
\left[\begin{array}{rrr|r}
1 & 2 & -1 & 3\\
0&7&-4&1\\
0 & 0 & -1 & \frac{21}{6}
\end{array}\right]
$\\
$\frac{1}{7}(4\rho_3 + \rho_2)$:
$
\left[\begin{array}{rrr|r}
1 & 2 & -1 & 3\\
0&1&0&\frac{1}{7}(1+4\cdot\frac{21}{6}) = \frac{15}{7}\\
0 & 0 & 1 & \frac{21}{6}
\end{array}\right]
$\\
$\rho_1+\rho_3-2\rho_2$:
$
\left[\begin{array}{rrr|r}
        1&0&0&3+\frac{21}{6}-2\cdot\frac{15}{7} = \frac{31}{14}\\
        0&1&0&\frac{15}{7}\\
        0&0&1&\frac{21}{6}
\end{array}\right]
$\\
\textbf{Answer:}\\
$$
\left[\begin{array}{rrr|r}
        1&0&0&\frac{31}{14}\\
        0&1&0&\frac{15}{7}\\
        0&0&1&\frac{21}{6}
\end{array}\right]
$$

\columnbreak

\textbf{Part 2}\\

$$
\begin{aligned}
    3x+4y&=1\\
    2x+3y&=12
\end{aligned}
$$
Transformed into augmented matrix form:
$$
\left[\begin{array}{rr|r}
3 & 4 & 1\\
2 & 3 & 12
\end{array}\right]
$$
$\rho_1-\rho_2$:
$
\left[\begin{array}{rr|r}
3 & 4 & 1\\
1&1&-11
\end{array}\right]
$\\
$4\rho_2-\rho_1$:
$
\left[\begin{array}{rr|r}
        1&0&-45\\
        1&1&-11
\end{array}\right]
$\\
$-\rho_1+\rho_2$:
$
\left[\begin{array}{rr|r}
        1&0&45\\
        0&1&34
\end{array}\right]
$\\

\textbf{Answer:}\\
$$
\left[\begin{array}{rr|r}
        1&0&45\\
        0&1&34
\end{array}\right]
$$

\textbf{Part 3}\\

$$
\begin{aligned}
    3x-2y&=4\\
    -6x+4y&=7
\end{aligned}
$$
Transformed into augmented matrix form:
$$
\left[\begin{array}{rr|r}
3 & -2 & 4\\
-6 & 4 & 7
\end{array}\right]
$$
$2\rho_1+\rho_2$:
$
\left[\begin{array}{rr|r}
3 & -2 & 4\\
0&0&15
\end{array}\right]
$\\
This equation does not have a solution.
\end{multicols}
\pagebreak
\textbf{Part 4}
$$
\begin{aligned}
    2x+y+z&=6\\
    x+4y+z&=5\\
    3x+2y+5z&=7
\end{aligned}
$$
Transformed into augmented matrix form:
$$
\left[\begin{array}{rrr|r}
2 & 1 & 1 & 6\\
1 & 4 & 1 & 5\\
3 & 2 & 5 & 7
\end{array}\right]
$$
$\rho_1 - \rho_3$:
$
\left[\begin{array}{rrr|r}
        2&1&1&6\\
        1&4&1&5\\
        0&-2&0&-1
\end{array}\right]
$\\
Swap $\rho_3$ and $\rho_2$, $-\frac{1}{2}\rho_2$:
$
\left[\begin{array}{rrr|r}
        2&1&1&6\\
        0&1&0&\frac{1}{2}\\
        1&4&1&5
\end{array}\right]
$\\
$-4\rho_2+\rho_3$:
$
\left[\begin{array}{rrr|r}
        2&1&1&6\\
        0&1&0&\frac{1}{2}\\
        0&0&1&3
\end{array}\right]
$\\
$\rho_3-\rho_1$:
$
\left[\begin{array}{rrr|r}
        -1&-1&0&-3\\
        0&1&0&\frac{1}{2}\\
        1&0&1&3
\end{array}\right]
$\\
$-(\rho_2+\rho_1)$:
$
\left[\begin{array}{rrr|r}
        1&0&0&2.5\\
        0&1&0&0.5\\
        1&0&1&3
\end{array}\right]
$\\
$\rho_1-\rho_3$:
$
\left[\begin{array}{rrr|r}
        1&0&0&2.5\\
        0&1&0&0.5\\
        0&0&1&0.5 
\end{array}\right]
$\\
\textbf{Answer:}\\
$$
\left[\begin{array}{rrr|r}
        1&0&0&2.5\\
        0&1&0&0.5\\
        0&0&1&0.5 
\end{array}\right]
$$

\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

$$
\begin{align}
    x+y&=2\cdot 45091.50\\
    x+y&=90183\\
    x-y&=5231\\
\end{align}
$$

In augmented matrix form:
$$
\left[\begin{array}{rr|r}
        1&1&90183\\
        1&-1&5231
\end{array}\right]
$$
$\rho_2 + \rho_1$:
$
\left[\begin{array}{rr|r}
        2&0&95414\\
        1&-1&90183
\end{array}\right]
$\\
$\rho_1 - \rho_2$:
$
\left[\begin{array}{rr|r}
        1&0&47707\\
        1&-1&5231
\end{array}\right]
$\\
$\rho_1 - \rho_2$:
$
\left[\begin{array}{rr|r}
        1&0&47707\\
        0&1&42476
\end{array}\right]
$\\
\textbf{Answer:}\\
$$
\left[\begin{array}{rr|r}
        1&0&47707\\
        0&1&42476
\end{array}\right]
$$

\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

Let
$\begin{cases}
    x = \text{volume of 2\% solution in $cm^3$}\\
    y = \text{volume of 7\% solution in $cm^3$}\\
\end{cases}
$

$$
\begin{align*}
    x+y&=40\\
    \frac{2x+7y}{40}&=3.2\\
    2x+7y&=128\\
\end{align*}
$$

In augmented matrix form:
$$
\left[\begin{array}{rr|r}
        1&1&40\\
        2&7&128
\end{array}\right]
$$
$\rho_2 - 2\rho_1$:
$
\left[\begin{array}{rr|r}
        0&5&48\\
        2&7&128
\end{array}\right]
$\\
$\rho_2 - \rho_1$:
$
\left[\begin{array}{rr|r}
        0&5&48\\
        2&2&80
\end{array}\right]
$\\
$\frac{1}{2}\rho_2, \frac{1}{5}\rho_1$:
$
\left[\begin{array}{rr|r}
        0&1&\frac{48}{5}\\
        1&1&40
\end{array}\right]
$\\
$\rho_2 - \rho_1$:
$
\left[\begin{array}{rr|r}
        0&1&\frac{48}{5}\\
        1&0&30.4
\end{array}\right]
$\\
\textbf{Answer:}\\
$$
\left[\begin{array}{rr|r}
        0&1&\frac{48}{5}\\
        1&0&30.4
\end{array}\right]
$$

\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

\begin{multicols}{2}
\begin{align*}
    -20x+50y+10z&=320\\
    40x+10y+10z&=280\\
    x+y+z&=16
\end{align*}
\columnbreak

In augmented matrix form:
$$
\left[\begin{array}{rrr|r}
        -20&50&10&320\\
        40&10&10&280\\
        1&1&1&16
\end{array}\right]
$$
\end{multicols}

\begin{multicols}{2}
$\frac{1}{10}\rho_1, \frac{1}{10}\rho_2$:
$
\left[\begin{array}{rrr|r}
        -2&5&1&32\\
        4&1&1&28\\
        1&1&1&16
\end{array}\right]
$\\
$-4\rho_3+\rho_2$:
$
\left[\begin{array}{rrr|r}
        -2&5&1&32\\
        0&-3&-3&-36\\
        1&1&1&16
\end{array}\right]
$\\
$\frac{1}{3}\rho_2$:
$
\left[\begin{array}{rrr|r}
        -2&5&1&32\\
        0&-1&-1&-12\\
        1&1&1&16
\end{array}\right]
$\\
$\rho_3 + \rho_2$:
$
\left[\begin{array}{rrr|r}
        -2&5&1&32\\
        0&-1&-1&-12\\
        1&0&0&4
\end{array}\right]
$
\columnbreak

$\rho_2+\rho_1$:
$
\left[\begin{array}{rrr|r}
        -2&4&0&20\\
        0&-1&-1&-12\\
        1&0&0&4
\end{array}\right]
$\\
$2\rho_3 + \rho_1$:
$
\left[\begin{array}{rrr|r}
        0&4&0&28\\
        0&-1&-1&-12\\
        1&0&0&4
\end{array}\right]
$\\
$\frac{1}{4}\rho_1, -\rho_2$:
$
\left[\begin{array}{rrr|r}
        0&1&0&7\\
        0&1&1&12\\
        1&0&0&4
\end{array}\right]
$\\
$\rh0_2 - \rho_1$:
$
\left[\begin{array}{rrr|r}
        0&1&0&7\\
        0&0&1&5\\
        1&0&0&4
\end{array}\right]
$
\end{multicols}

\textbf{Answer:}\\
$$
\left[\begin{array}{rrr|r}
        1&0&0&4\\
        0&1&0&7\\
        0&0&1&5
\end{array}\right]
$$

\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

\begin{multicols}{2}
    \part

$$
\begin{align*}
    x +y+z&=1\\
    2x-y+z&=-1\\
    x+3y-z&=7
\end{align*}
$$

In augmented matrix form:
$$
\left[\begin{array}{rrr|r}
        1&1&1&1\\
        2&-1&1&-1\\
        1&3&-1&7
\end{array}\right]
$$

$\rho_1-\rho_3$:
$
\left[\begin{array}{rrr|r}
        1&1&1&1\\
        2&-1&1&-1\\
        0&-2&2&-6
\end{array}\right]
$\\
$\frac{1}{2}\rho_3$:
$
\left[\begin{array}{rrr|r}
        1&1&1&1\\
        2&-1&1&-1\\
        0&-1&1&-3
\end{array}\right]
$\\
$2\rho_1-\rho_2$:
$
\left[\begin{array}{rrr|r}
        1&1&1&1\\
        0&3&1&3\\
        0&-1&1&-3
\end{array}\right]
$\\
$\rho_2+3\rho_3$:
$
\left[\begin{array}{rrr|r}
        1&1&1&1\\
        0&3&1&3\\
        0&0&4&-6
\end{array}\right]
$\\
$\frac{1}{4}\rho_3$:
$
\left[\begin{array}{rrr|r}
        1&1&1&1\\
        0&3&1&3\\
        0&0&1&-\frac{3}{2}
\end{array}\right]
$\\
$\frac{1}{3}(\rho_2-\rho_3)$:
$
\left[\begin{array}{rrr|r}
        1&1&1&1\\
        0&1&0&\frac{1}{3}(3-\frac{3}{2})\\
        0&0&1&-\frac{3}{2}
\end{array}\right]
$\\
$\rho_1-\rho_2-\rho_3$:
$
\left[\begin{array}{rrr|r}
        1&0&0&1-\frac{3}{2}-(-\frac{3}{2}) = 1\\
        0&1&0&\frac{3}{2}\\
        0&0&1&-\frac{3}{2}
\end{array}\right]
$\\

\textbf{Answer:}
$$
\left[\begin{array}{rrr|r}
        1&0&0&1\\
        0&1&0&\frac{3}{2}\\
        0&0&1&-\frac{3}{2}
\end{array}\right]
$$

\columnbreak

\part
$$
\begin{align*}
    3x-4y &=8\\
    x+y+z&=2\\
    2x-5y-z &=6
\end{align*}
$$

In augmented matrix form:
$$
\left[\begin{array}{rrr|r}
        3&-4&0&8\\
        1&1&1&2\\
        2&-5&-1&6
\end{array}\right]
$$

$2\rho_2-\rho_3$:
$
\left[\begin{array}{rrr|r}
        3&-4&0&8\\
        1&1&1&2\\
        0&7&3&-2
\end{array}\right]
$\\
Swap $\rho_2$ and $\rho_1$:
$
\left[\begin{array}{rrr|r}
        1&1&1&2\\
        3&-4&0&8\\
        0&7&3&-2
\end{array}\right]
$\\
$3\rho_1-\rho_2$:
$
\left[\begin{array}{rrr|r}
        1&1&1&2\\
        0&7&3&2\\
        0&7&3&-2
\end{array}\right]
$\\
$\rho_2-\rho_3$:
$
\left[\begin{array}{rrr|r}
        1&1&1&2\\
        0&7&3&2\\
        0&0&0&4
\end{array}\right]
$\\
There are no solutions to these equations

\end{multicols}

\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\\

    $$
    \begin{align*}
        2x_0 - 9x_1 -6x_2 + 2x_3 &=5\\
        -2x_0 + 3x_1 + 4x_2 - 2x_3 &=-2\\\\
        2x_0-6x_1-6x_2+3x_3&=5\\
        -x_0-2x_1+3x_2-2x_3&=-2\\
        2x_0+4x_1-6x_2+4x_3&=7
    \end{align*}
    $$

    \solution

    \begin{multicols}{2}

    \part

    $$
    \begin{align*}
        2x_0 - 9x_1 -6x_2 + 2x_3 &=5\\
        -2x_0 + 3x_1 + 4x_2 - 2x_3 &=-2
    \end{align*}
    $$
    In augmented matrix form:
    $$
    \left[\begin{array}{rrrr|r}
        2&-9&-6&2&5\\
        -2&3&4&-2&-2
    \end{array}\right]
    $$
    $\rho_1+\rho_2$:
    $
    \left[\begin{array}{rrrr|r}
        2&-9&-6&2&5\\
        0&-6&-2&0&3
    \end{array}\right]
    $\\
    $3\rho_2+\rho_1$:
    $
    \left[\begin{array}{rrrr|r}
        2&9&0&2&-4\\
        0&-6&-2&0&3
    \end{array}\right]
    $\\
    There will be more than one solution, because the last row is not zero.\\

    \columnbreak

    \part

    $$
    \begin{align*}
        2x_0-6x_1-6x_2+3x_3&=5\\
        -x_0-2x_1+3x_2-2x_3&=-2\\
        2x_0+4x_1-6x_2+4x_3&=7
    \end{align*}
    $$
    In augmented matrix form:
    $$
    \left[\begin{array}{rrrr|r}
        2&-6&-6&3&5\\
        -1&-2&3&-2&-2\\
        2&4&-6&4&7
    \end{array}\right]
    $$
    $2\rho_2+\rho_3$:
    $
    \left[\begin{array}{rrrr|r}
        2&-6&-6&3&5\\
        -1&-2&3&-2&-2\\
        0&0&0&0&3
    \end{array}\right]
    $\\
    This equation has no solution, because the last row is zero.
    \end{multicols}
\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}
