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\begin{coverpages}
\begin{center}\huge MATH 3362 - EXAM 1 FALL 2017\\
SOLUTION 
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday 15 September 2017 \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}


\end{coverpages}

\begin{questions}

\question[8] Circle $\textbf{L}$ for linear and $\textbf{N}$ for nonlinear.
\begin{parts}
\part[2] \circled{\textbf{L}}\hspace{5pt} \textbf{N} \hspace{5pt}  $0.1x_1-0.5x_2=0$
\part[2] \textbf{L}\hspace{5pt} \circled{\textbf{N}} \hspace{5pt}  $5\cos(x)+y+3z=5$
\part[2] \circled{\textbf{L}}\hspace{5pt} \textbf{N} \hspace{5pt}  $2x+3y+z=0$
\part[2] \circled{\textbf{L}}\hspace{5pt} \textbf{N} \hspace{5pt}  $\cos(5)x+y+3z=5$
\end{parts}

\question[8] Compute the linear combination.
\begin{parts}
\part[4] (Field: $\mathbb{R}$) $2 \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} + 3 \begin{bmatrix} \pi \\ -5 \\ \sqrt{2} \end{bmatrix}$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=2*%7B%7B1%7D,%7B0%7D,%7B2%7D%7D%2B3%7B%7Bpi%7D,%7B-5%7D,%7Bsqrt%7B2%7D%7D%7D}{Compute}
$$\begin{array}{ll}
2\begin{bmatrix}
1 \\ 
0 \\
2 \end{bmatrix} + 3 \begin{bmatrix} \pi \\ -5 \\ \sqrt{2} \end{bmatrix} &= \begin{bmatrix} 2 \\ 0 \\ 4 \end{bmatrix} + \begin{bmatrix} 3\pi \\ -15 \\ 3\sqrt{2} \end{bmatrix} \\
&= \begin{bmatrix} 2+3\pi \\ -15 \\ 4+3\sqrt{2} \end{bmatrix}.
\end{array}$$

%\part[1] (Field: $\mathbb{Z}_5$) $4\begin{bmatrix} 1 \\ 3 \end{bmatrix} + 3 \begin{bmatrix} 2 \\ 4 \end{bmatrix}$
\part[4] (Field: $\mathbb{C}$) $i \begin{bmatrix} 2i & 3 \\ 8 & -i \end{bmatrix} + 3 \begin{bmatrix} 0 & 2 \\ 4 & 0 \end{bmatrix}$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=i*%7B%7B2*i,3%7D,%7B8,-i%7D%7D%2B3*%7B%7B0,2%7D,%7B4,0%7D%7D}{Compute}
$$\begin{array}{ll}
i \begin{bmatrix} 2i & 3 \\ 8 & -i \end{bmatrix} + 3 \begin{bmatrix} 0 & 2 \\ 4 & 0 \end{bmatrix} &= \begin{bmatrix} -2 & 3i \\ 8i & 1 \end{bmatrix} + \begin{bmatrix} 0 & 6 \\ 12 & 0 \end{bmatrix} \\
&= \begin{bmatrix} -2 & 6+3i \\ 12+8i & 1 \end{bmatrix}.
\end{array}$$
\end{parts}


\question[21] Put the following matrices into reduced echelon form.
\begin{parts}
\part[7] (Field: $\mathbb{R}$) $\begin{bmatrix} 1 & 0 & 2 \\ 0&10 &1  \\ 1 & 5 & -1 \end{bmatrix}$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B1,0,2%7D,%7B0,10,1%7D,%7B1,5,-1%7D%7D}{Compute} 
$$\begin{array}{ll}
\begin{bmatrix} 1 & 0 & 2 \\ 0&10 &1  \\ 1 & 5 & -1 \end{bmatrix} &\stackrel{r_3^*=r_3-r_1}{\sim} \begin{bmatrix} 1 & 0 & 2 \\ 0&10&1 \\ 0&5&-3  \end{bmatrix} \\
&\stackrel{r_3^*=r_3-\frac{1}{2}r_2}{\sim} \begin{bmatrix} 1&0&2 \\ 0&10&1 \\ 0&0&-\frac{7}{2} \end{bmatrix} \\
&\stackrel{r_2^*=\frac{1}{10}r_2, r_3^*=-\frac{2}{7}r_3}{\sim} \begin{bmatrix} 1&0&2 \\ 0&1&\frac{1}{10} \\ 0&0&1\end{bmatrix} \\
&\stackrel{r_2^*=r_2-\frac{1}{10}r_3,r_1^*=r_1-2r_3}{\sim} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}.
\end{array}$$
\part[7] (Field: $\mathbb{Z}_3$) $\begin{bmatrix} 0&2&1 \\ 1&1&2 \\ 0&1&2 \end{bmatrix}$ \\
\textit{Solution:} Compute
$$\begin{array}{ll}
\begin{bmatrix} 0&2&1 \\ 1&1&2 \\ 0&1&2 \end{bmatrix} &\stackrel{r_1 \leftrightarrow r_2}{\sim} \begin{bmatrix} 1&1&2 \\ 0&2&1 \\ 0&1&2 \end{bmatrix} \\
&\stackrel{r_3^*=r_3-2r_2}{\sim} \begin{bmatrix} 1&1&2 \\ 0&2&1 \\ 0&0&0  \end{bmatrix}    \\
&\stackrel{r_1=r_1+r_2}{\sim} \begin{bmatrix} 1&0&0 \\ 0&2&1 \\ 0&0&0 \end{bmatrix} \\
&\stackrel{r_2^*=2r_2}{\sim} \begin{bmatrix} 1&0&0 \\ 0&1&2 \\ 0&0&0 \end{bmatrix}
\end{array}$$
\part[7] (Field: $\mathbb{C}$) $\begin{bmatrix} i & 1+i & 0 \\ 1&2i & 1 \end{bmatrix}$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=rref+%7B%7Bi,1%2Bi,0%7D,%7B1,2*i,1%7D%7D}{Compute}
$$\begin{array}{ll}
\begin{bmatrix} i & 1+i & 0 \\ 1 & 2i & 1 \end{bmatrix} &\stackrel{r_2^*=r_2+ir_1}{\sim} \begin{bmatrix} i & 1+i & 0 \\ 0 & 3i-1 & 1 \end{bmatrix} \\
&\stackrel{r_2^*=\frac{1}{3i-1}r_2}{\sim} \begin{bmatrix} i & 1+i & 0 \\ 0 & 1 & \frac{1}{3i-1} \end{bmatrix} \\
&\stackrel{r_1^*=r_1-(1+i)r_2}{\sim} \begin{bmatrix} i & 0 & -\frac{1+i}{3i-1} \\ 0&1& \frac{1}{3i-1} \end{bmatrix} \\
&\stackrel{r_1^*=\frac{1}{i}r_1}{\sim} \begin{bmatrix} 1 & 0 & -\frac{1}{i} \left( \frac{1+i}{3i-1} \right) \\
0&1&\frac{1}{3i-1} \end{bmatrix} \\
&\stackrel{r_1^*=\frac{1}{i}r_1}{\sim} \begin{bmatrix} 1 & 0 & \frac{1+i}{i+3}  \\
0&1&\frac{1}{3i-1} \end{bmatrix} \\
\end{array}$$
Note: the above is an acceptable answer. However, the complex numbers could also be put into ``standard form $a+bi$" by computing
$$\dfrac{1+i}{3+i} = \dfrac{1+i}{3+i} \left(\dfrac{3-i}{3-i}\right) = \dfrac{3-i+3i-i^2}{9+1} = \dfrac{4+2i}{10} = \dfrac{2}{5} + \dfrac{1}{5}i$$
and
$$\dfrac{1}{-1+3i} = \dfrac{1}{-1+3i} \left( \dfrac{-1-3i}{-1-3i} \right) = \dfrac{-1-3i}{1+9} = \dfrac{-1-3i}{10} = -\dfrac{1}{10} - \dfrac{3}{10}i,$$
yielding
$$\begin{bmatrix} i & 1+i & 0 \\ 1 & 2i & 1 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 &  \frac{2}{5} + \frac{1}{5}i \\ 0&1& -\frac{1}{10} - \frac{3}{10}i \end{bmatrix}.$$
\end{parts}

\question[21] Answer the following questions. \\
\begin{parts}
\part[7] (Field: $\mathbb{R}$) Write the vector $\vec{b} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ as a linear combination of the vectors $\vec{v_1} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$ and $\vec{v_2} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}$ \textbf{or} explain why that is not possible. \\
\textit{Solution:} We can determine the answer by solving the vector equation
$$c_1 \vec{v}_1 + c_2 \vec{v}_2 = \vec{b},$$
i.e.
$$c_1 \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.$$
Write the augmented matrix and row reduce: \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B-1,2,1%7D,%7B0,1,0%7D,%7B1,0,1%7D%7D}{compute}
$$\begin{array}{ll}
\begin{bmatrix} -1&2&1 \\ 0&1&0 \\ 1&0&1 \end{bmatrix} &\stackrel{r_3^*=r_3+r_1}{\sim} \begin{bmatrix} -1&2&1 \\ 0&1&0 \\ 0&2&2 \end{bmatrix} \\
&\stackrel{r_3^*=r_3-2r_2}{\sim} \begin{bmatrix} -1&2&1 \\ 0&1&0 \\ 0&0&2 \end{bmatrix} \\
&\stackrel{r_1^*=-r_1, r_3^*=\frac{1}{2}r_2}{\sim} \begin{bmatrix} 1&-2&-1 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \\
&\stackrel{r_1^*=r_1+2r_2}{\sim} \begin{bmatrix} 1 & 0 & -1 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} \\
&\stackrel{r_1^*=r_1+r_3}{\sim} \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}.
\end{array}$$
From this we conclude that there is no solution because it encodes the following system of equations:
$$\left\{\begin{array}{lll}
c_1 & & = 0 \\
& c_2 & = 0 \\
& 0 &= 1.
\end{array} \right.$$
The third equation is always false, and hence no solution exists.
\part[7] (Field: $\mathbb{R}$) Is the vector $\vec{b} = \begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}$ in the span of the columns of the matrix $A = \begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix}$? \\
\textit{Solution:} Recall that the span of a set of vectors is the set of all linear combinations of those vectors. Therefore, this question is asking whether or not the following vector equation has a solution: 
$$c_1 \begin{bmatrix} 1\\4\\7 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix} + c_3 \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}.$$
To answer the question, set up the augmented matrix and \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B1,2,3,1%7D,%7B4,5,6,1%7D,%7B7,8,9,1%7D%7D}{compute}
$$\begin{array}{ll}
\begin{bmatrix} 1&2&3&1 \\ 4&5&6&1 \\ 7&8&9&1 \end{bmatrix} & \stackrel{r_2^*=r_2-4r_1,r_3^*=r_3-7r_1}{\sim} \begin{bmatrix} 1&2&3&1 \\ 0&-3&-6&-3 \\ 0 & -6 & -12 & -6\end{bmatrix} \\
&\stackrel{r_2^*-\frac{1}{3}r_2, r_3^*=-\frac{1}{6}r_3}{\sim} \begin{bmatrix} 1&2&3&1 \\ 0&1&2&1 \\ 0&1&2&1 \end{bmatrix} \\
&\stackrel{r_3^*=r_3-r_2}{\sim} \begin{bmatrix} 1&2&3&1 \\ 0&1&2&1 \\ 0&0&0&0 \end{bmatrix} \\
&\stackrel{r_1^*=r_1-2r_2}{\sim} \begin{bmatrix} 1&0&-1&-1 \\ 0&1&2&1 \\ 0&0&0&0 \end{bmatrix}.
\end{array}$$

\part[7] (Field: $\mathbb{R}$) Prove that $\mathbb{R}^{2 \times 1} = \mathrm{span} \left\{ \begin{bmatrix} 1 \\ 0\end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix}\right\}$. \\
\textit{Solution:} Recall that a set $X$ equals a set $Y$ if and only if $X \subseteq Y$ and $Y \subseteq X$. ``Clearly",
$$\mathrm{span} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1\\1 \end{bmatrix} \right\} \subseteq \mathbb{R}^{2\times 1},$$
because any linear combination of vectors in $\mathbb{R}^{2 \times 1}$ (with weights in the field $\mathbb{R}$) remains in $\mathbb{R}^{2 \times 1}$. It remains to show that
$$\mathbb{R}^{2 \times 1} \subseteq \mathrm{span} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1\\1 \end{bmatrix} \right\}.$$
Let $\begin{bmatrix} a \\ b \end{bmatrix} \in \mathbb{R}^{2 \times 1}$ be an arbitrary vector. We must show that there is a linear combination lying in the span that equals $\begin{bmatrix} a \\ b \end{bmatrix}$. In other words, we must solve the vector equation
$$c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix}.$$
Set up the augmented matrix and \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B1,1,a%7D,%7B0,1,b%7D%7D}{compute}
$$\begin{array}{ll}
\begin{bmatrix} 1&1&a \\ 0&1&b \end{bmatrix} &\stackrel{r_1^*=r_1-r_2}{\sim} \begin{bmatrix} 1&0&a-b \\ 0&1&b \end{bmatrix}.
\end{array}$$
This shows that
$$(a-b) \begin{bmatrix} 1 \\ 0 \end{bmatrix} + b \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix},$$
i.e. $\begin{bmatrix} a\\b \end{bmatrix}$ may in fact be written as a linear combination of $\begin{bmatrix} 1&0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$, proving that $\mathbb{R}^{2 \times 1} \subseteq \mathrm{span} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1\\1 \end{bmatrix} \right\},$ completing the proof. $\blacksquare$
\end{parts}

\question[14] Are the following sets independent or dependent?
\begin{parts}
\part[7] (Field: $\mathbb{R}$)  $\left\{ \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix}, \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \right\}$ \\
\textit{Solution:} We must seek a solution to the matrix equation
$$c_1 \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + c_2\begin{bmatrix} 1 & 0 \\ 2 & 0 \end{bmatrix} + c_3\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0&0 \end{bmatrix}.$$
Using matrix algebra, we see that the left-hand side becomes
$$\begin{bmatrix} c_1+c_2+2c_3 & 0 \\ c_1+2c_2+c_3 & c_1+c_3 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}.$$
Hence we are being asked to solve the following system of linear equations:
$$\left\{ \begin{array}{llll}
c_1 &+ c_2 &+ 2c_3 &= 0 \\
0&&&= 0 \\
c_1 &+ 2c_2 &+ c_3 &= 0 \\
c_1 & &+c_3 &= 0.
\end{array} \right.$$
Set up an augmented matrix and \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B1,1,2,0%7D,%7B0,0,0,0%7D,%7B1,2,1,0%7D,%7B1,0,1,0%7D%7D}{compute}
$$\begin{array}{ll}
\begin{bmatrix} 1&1&2&0 \\ 0&0&0&0 \\ 1&2&1&0 \\ 1&0&1&0 \end{bmatrix} &\stackrel{r_2 \leftrightarrow r_4}{\sim} \begin{bmatrix} 1&1&2&0 \\ 1&0&1&0 \\ 1&2&1&0 \\ 0&0&0&0 \end{bmatrix} \\
&\stackrel{r_2 \leftrightarrow r_3}{\sim} \begin{bmatrix} 1&1&2&0 \\ 1&2&1&0 \\ 1&0&1&0 \\ 0&0&0&0 \end{bmatrix} \\
&\stackrel{r_2^*=r_2-r_1,r_3^*=r_3-r_1}{\sim} \begin{bmatrix} 1&1&2&0 \\ 0&1&-1&0 \\ 0&-1&-1&0 \\ 0&0&0&0 \end{bmatrix} \\
&\stackrel{r_3^*=r_3+r_2}{\sim} \begin{bmatrix} 1&1&2&0 \\ 0&1&-1&0 \\ 0&0&-2&0 \\ 0&0&0&0 \end{bmatrix} \\
&\stackrel{r_3^*=-\frac{1}{2}r_3}{\sim} \begin{bmatrix} 1&1&2&0 \\ 0&1&-1&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{bmatrix} \\
&\stackrel{r_2^*=r_2+r_3, r_1^*=r_1-2r_3}{\sim} \begin{bmatrix} 1&1&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{bmatrix} \\
&\stackrel{r_1^*=r_1-r_2}{\sim} \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{bmatrix}.
\end{array}$$
This shows that the only solution to the matrix equation is $\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix},$
and therefore we see that the given set of matrices is \textbf{independent}.

\part[7] (Field: $\mathbb{R}$) $\left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1  \end{bmatrix}, \begin{bmatrix}2 \\ 0  \end{bmatrix}, \begin{bmatrix} 5 \\ 10 \end{bmatrix}, \begin{bmatrix} \pi \\ \sqrt{2} \end{bmatrix}\right\}$ \\
\textit{Solution:} Dependent. There are more vectors in the set than the length of each vector, so we may say it is dependent by a theorem.
\end{parts}

\question[28] Perform the computation or state that it is not well-defined. 
\begin{parts}
\part[7] (Field: $\mathbb{R}$) $\begin{bmatrix} 1&2&3 \\ 4&5&6 \end{bmatrix} - 2\begin{bmatrix} 0 & 1 & 4 \\ 3&0&1 \\ 3 & 1 & 0 \end{bmatrix}$ \\
\textit{Solution:} Not well-defined. We can only add matrices that are of the same size. 
\part[7] (Field: $\mathbb{R}$) $\begin{bmatrix} 1&2&3 \\ 4&5&6 \end{bmatrix} + 3 \begin{bmatrix} 1&2 \\ 3&0 \\ 5&0 \end{bmatrix}^T$  \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=%7B%7B1,2,3%7D,%7B4,5,6%7D%7D+%2B+3*+transpose(%7B%7B1,2%7D,%7B3,0%7D,%7B5,0%7D%7D)}{Compute}
$$\begin{array}{ll}
\begin{bmatrix} 1&2&3 \\ 4&5&6 \end{bmatrix} + 3 \begin{bmatrix} 1&2 \\ 3&0 \\ 5&0 \end{bmatrix}^T &= \begin{bmatrix} 1&2&3 \\ 4&5&6 \end{bmatrix} + 3 \begin{bmatrix} 1&3&5 \\ 2&0&0 \end{bmatrix} \\
&= \begin{bmatrix} 1&2&3 \\ 4&5&6 \end{bmatrix} + \begin{bmatrix} 3&9&15 \\ 6&0&0 \end{bmatrix} \\
&= \begin{bmatrix} 4 & 11 & 18 \\ 10 & 5 & 6 \end{bmatrix}.
\end{array}$$
\part[7] (Field: $\mathbb{R}$) $\begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix} \begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}$ \\
\textit{Solution:} \href{http://www.wolframalpha.com/input/?i=%7B%7B1,2%7D,%7B3,4%7D%7D+*+%7B%7B2,1,1%7D,%7B1,1,0%7D%7D}{Compute}
$$\begin{bmatrix} 1&2 \\ 3&4 \end{bmatrix} \begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 2+2 & 1+2 & 1+0 \\ 6+4 & 3+4 & 3+0 \end{bmatrix} = \begin{bmatrix} 4 & 3 & 1 \\ 10 & 7 & 3 \end{bmatrix}.$$
\part[7] (Field: $\mathbb{R}$) $\begin{bmatrix} 1 & 1 & 0 \\ 2 & 6 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 0 & 2 & 5 \end{bmatrix}$\\
\textit{Solution:} Not well-defined. We may not multiply a matrix of size $2 \times 3$ on the left of a matrix of size $2 \times 3$. 
\end{parts}

\hrule
\noindent Recall that the trace of a square matrix $A \in \mathbb{F}^{n \times n}$ with representation $A=\{a_{ij}\}$ is the number 
$$\mathrm{tr}(A)=a_{11} + a_{22} + \ldots + a_{nn}.$$
For instance, in $\mathbb{F}=\mathbb{R}$, $\mathrm{tr} \left( \begin{bmatrix} 1 & 2 & 3  \\ 4 & 5 &6 \\ 7 & 8 & 9 \end{bmatrix} \right) = 1+5+9 = 15$.
\question[4] (Bonus) Prove that $\mathrm{tr}(A)=\mathrm{tr}(A^T)$. \\
\textit{Solution:} Since the transpose makes the columns of $A$ into rows and vice versa, if $A = \{a_{ij}\}$, then $A^T = \{a_{ji}\}$. In particular, observe that the diagonal is not affected by a transpose (of an $n \times n$ matrix). Therefore
$$\mathrm{tr}(A)=a_{11}+\ldots+a_{nn} = \mathrm{tr}(A^T).$$
\end{questions}

\end{document}
