\begin{enumerate}
    \item
    \text{Let} {A} = \left(
    \begin{array}{cccc}
    a_{11} & a_{12} & ... & a_{1m} \\
    a_{21} & a_{22} & ... & a_{2m} \\
    a_{31}& a_{32} & ... & a_{3m} \\
    \vdots & \vdots & \ddots & \vdots \\
    a_{n1} & a_{n2} & \cdots & a_{nm}\\
\end{array}
\right)
\\
\\
\\
    \text{Let} {B} = \left(
    \begin{array}{cccc}
    b_{11} & b_{12} & ... & b_{1m} \\
    b_{21} & b_{22} & ... & b_{2m} \\
    b_{31}& b_{32} & ... & b_{3m} \\
    \vdots & \vdots & \ddots & \vdots \\
    b_{n1} & b_{n2} & \cdots & b_{nm}\\
\end{array}
\right).
\\
\\
\text{Then}\\
A^T + B^T = \left(
    \begin{array}{cccc}
    a_{11} + b_{11} & a_{21} + b_{21} & ... & a_{n1} + b_{n1} \\
    a_{12} + b_{12} & a_{22} + b_{22} & ... & a_{n2} + b_{n2} \\
    a_{31} + b_{13}& a_{23} + b_{23} & ... & a_{n3} + b_{n3} \\
    \vdots & \vdots & \ddots & \vdots \\
    a_{1m} + b_{1m} & au_{n2} & \cdots & a_{nm} + b_{nm}\\
\end{array}
\right)
\\
\\
 A + B = \left(\begin{array}{cccc}
    a_{11} + b_{11} & a_{21} + b_{21} & ... & a_{1m} + b_{1m} \\
    a_{21} + b_{21} & a_{22} + b_22 & ... & a_{2m} + b_{2m} \\
    a_{31} + b_{31}& a_{23} + b_{23} & ... & a_{3m} + b_{3m} \\
    \vdots & \vdots & \ddots & \vdots \\
    a_{1m} + b_{1m} & a_{n2} & \cdots & a_{nm} + b_{nm}\\
\end{array}
\right)

\text{The Transpose if this sum is clearly the same as}  A^T + B^T

\item
Let \textit{A} be an arbitrary matrix that is symmetric and invertible.\\
A property of the identity matrix is\\
    I = I^T \\
    \textit{  Since any matrix times its inverse is the identity:}\\
    AA^{-1} = (AA^{-1})^T\\
    A^{-1}A = (A^{-1})^TA^T\\
    \textit{Since A is symmetric:} \\
    A^{-1}A = (A^{-1})^TA\\
    A^{-1}AA^{-1} = (A^{-1})^TAA^{-1}\\
     A^{-1}I = (A^{-1})^TI\\
     Therefore A^{-1} = (A^{-1})^T \textit{when A is symmetric}\\

\item
We know that (AB)^{-1}AB = I\\
B^{-1}A^{-1}AB = B^{-1}IB = B^{-1}B=I\\
\text{Since} I = (AB)^{-1}AB = B^{-1}A^{-1}AB \\
(AB)^{-1}AB = B^{-1}A^{-1}AB \textit{if we divide by AB}
\end{enumerate}
\item
\begin{enumerate}
\item
\begin{enumerate}
    \item
    $f(x_1,x_2,x_3)=1$ when $x_1=1, x_2=0, x_3=1$ \\
    This means that $\theta\cdot x_1 + \theta \cdot x_3 > 0$ and $\theta_1 + \theta_3 > 0$.\\
    Also, $f(x_1,x_2,x_3)=0$ when $x_1=1, x_2=0, x_3=0$ and $x_1=0, x_2=0, x_3=1$
    This suggests $\theta_1 < 0$ and $\theta_3 < 0$ respectively.\\
    This contradicts our original inequality since the sum of two negatives cannot be positive. Therefore it is not possible to learn such a $\bar{\theta}$\\
    \item
    We can simply add $b$ to each of the inequalities derived from part (i).\\
    (1) $\theta_1 + \theta_3 + b > 0$\\
    (2) $\theta_1 + b < 0$\\
    (3) $\theta_3 + b < 0$\\
    If we combine inequalities (2) and (3), we get $\theta_1 +\theta_3 < -2b$ and rearranging (1) gives us $\theta_1 + \theta_3 > -b$. Now we know that there is no solution since $\theta_1 + \theta_3$ cannot be less than $-2b$, but also greater than $-b$
\end{enumerate}
\item
\begin{enumerate}
    \item Yes. If $\theta = [2,1]^T$ it correctly classifies the points.
    \item Yes. If $\theta = [2,1]^T$ and $b=-1$, the points are correctly classified.
    \item No. If the square is formed by $s$ just greater than $1$, it is the minimum size to include the point $[0,1]$, but it doesn't correctly classify the data, since $[2,-2]$ is not included. $s$ must be greater than $2$ for $[2,-2]$ to be correctly classified, but $[-2,-1]$ will be classified as positive, and thus misclassified.
    \item Yes. Centered at $[2,0]$ with $s=5$
\end{enumerate}
\end{enumerate}
\end{enumerate}