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\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}
