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\begin{document}

\begin{center}
\textbf{Math 3013.62667}\\[0pt]
SOLUTIONS TO SECOND EXAM \\[0pt]
October 22, 2021
\end{center}

1. Complete the following mathematical definitions

(a) (5 pts) A \textbf{subspace} of a vector space $V$ is $\ldots $

\begin{itemize}
\item a subset $W$ of $V$ such that

\begin{itemize}
\item[(i)] $\lambda \in \mathbb{R}\ ,\ \mathbf{w}\in W\quad \Rightarrow
\quad \left( \lambda \mathbf{w}\right) \in W$

\item[(ii)] $\mathbf{w}_{1},\mathbf{w}_{2}\in W\quad \Rightarrow \quad
\left( \mathbf{w}_{1}+\mathbf{w}_{2}\right) \in W$
\end{itemize}
\end{itemize}

(b) (5 pts) A \textbf{basis} for a subspace $W$ is $\ldots $

\begin{itemize}
\item a set of vectors $B=\left\{ \mathbf{b}_{1},\ldots ,\mathbf{b}%
_{k}\right\} $ such that every vector $\mathbf{w}\in W$ can be expressed as%
\begin{equation*}
\mathbf{w}=c_{1}\mathbf{b}_{1}+c_{2}\mathbf{b}_{2}+\cdots +c_{k}\mathbf{b}%
_{k}
\end{equation*}%
for exactly one choice of coefficients $c_{1},c_{2},\ldots ,c_{k}$.
\end{itemize}

(c) (5 pts) A set of vectors $\left\{ \mathbf{v}_{1},\ldots ,\mathbf{v}%
_{k}\right\} $ is \textbf{linearly independent} if $\ldots $

\begin{itemize}
\item the only solution of%
\begin{equation*}
x_{1}\mathbf{v}_{1}+x_{2}\mathbf{v}_{2}+\cdots +x_{k}\mathbf{v}_{k}=\mathbf{0%
}
\end{equation*}%
is%
\begin{equation*}
x_{1}=0\ ,\ x_{2}=0\ ,\ \ldots \ ,\ x_{k}=0
\end{equation*}
\end{itemize}

(d) (5 pts) A function $T:\mathbb{R}^{m}\rightarrow \mathbb{R}^{n}$ is a 
\textbf{linear transformation} if $\ldots $

\begin{itemize}
\item 
\begin{itemize}
\item[(i)] $T\left( \lambda \mathbf{x}\right) =\lambda T\left( \mathbf{x}%
\right) $ for all $\lambda \in \mathbb{R}$ and all $\mathbf{x\in }\mathbb{R}%
^{m}$

\item[(ii)] $T\left( \mathbf{x}_{1}+\mathbf{x}_{2}\right) =T\left( \mathbf{x}%
_{1}\right) +T\left( \mathbf{x}_{2}\right) $ for all $\mathbf{x}_{1},\mathbf{%
x}_{2}\in \mathbb{R}^{m}$
\end{itemize}
\end{itemize}

\newpage

2. Consider the vectors $\left\{[0,1,-1,2],[1,1,1,1],[2,1,3,0] \right\} \in 
\mathbb{R}^{4}$

(a) (5 pts) Determine if these vectors are linearly independent.

\begin{itemize}
\item We'll form a matrix using the given vectors as rows, and then row
reduce that matrix to Row Echelon Form%
\begin{eqnarray*}
&&\mathbf{A} =\left[ 
\begin{array}{cccc}
0 & 1 & -1 & 2 \\ 
1 & 1 & 1 & 1 \\ 
2 & 1 & 3 & 0%
\end{array}%
\right] \ \underrightarrow{\ R_{1}\longleftrightarrow R_{2}\ }\ \left[ 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
0 & 1 & -1 & 2 \\ 
2 & 1 & 3 & 0%
\end{array}%
\right] \ \  \\
&&\underrightarrow{\ R_{3}\rightarrow R_{3}-2R_{1}\ }\ \left[ 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
0 & 1 & -1 & 2 \\ 
0 & -1 & 1 & -2%
\end{array}%
\right] \ \underrightarrow{\ R_{3}\rightarrow R_{3}+R_{2}\ }\ \left[ 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
0 & 1 & -1 & 2 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right]
\end{eqnarray*}%
Since we ended up with a zero row, the original set of vectors \textbf{are
not linearly independent}.
\end{itemize}

(b) (5 pts) What is the dimension of the subspace generated by these vectors?

\begin{itemize}
\item The row space of the matrix $\mathbf{A}$ constructed in part (a) has
exactly two basis vectors $\left( \left[ 1,1,1,1\right] \text{ and }\left[
0,1,-1,2\right] \right) $ and so%
\begin{equation*}
W=span\left( [0,1,-1,2],[1,1,1,1],[2,1,3,0]\right) =RowSp\left( \mathbf{A}%
\right)
\end{equation*}
is $2$-dimensional.
\end{itemize}

3. Given that the following matrix: \quad $\displaystyle\mathbf{A}=\left[ 
\begin{array}{cccc}
1 & 0 & 1 & 3 \\ 
2 & 0 & 2 & 5 \\ 
1 & 0 & 1 & 3%
\end{array}%
\right] $ row reduces to $\left[ 
\begin{array}{cccc}
1 & 0 & 1 & 0 \\ 
0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right] $ \quad

(a) (5 pts) Find a basis for the row space of $\mathbf{A}$.

\begin{itemize}
\item A basis for $RowSp\left( \mathbf{A}\right) $ is given by the non-zero
rows of $R.E.F.\left( \mathbf{A}\right) $. Hence,%
\begin{equation*}
\text{basis for }RowSp\left( \mathbf{A}\right) =\left\{ \left[ 1,0,1,0\right]
,\left[ 0,0,0,1\right] \right\}
\end{equation*}
\end{itemize}

(b) (5 pts) Find a basis for the column space of $\mathbf{A}$.

\begin{itemize}
\item A basis for $ColSp\left( \mathbf{A}\right) $ is given by the columns
of $\mathbf{A}$ that correspond to the columns of $R.E.F.\left( \mathbf{A}%
\right) $ that contain pivots.

Since the pivots of $R.E.F.\left( \mathbf{A}\right) $ occur in the $1^{st}$
and $4^{th}$ columns,%
\begin{equation*}
\text{basis for }ColSp\left( \mathbf{A}\right) =\left\{ \left[ 
\begin{array}{c}
1 \\ 
2 \\ 
1%
\end{array}%
\right] ,\left[ 
\begin{array}{c}
3 \\ 
5 \\ 
3%
\end{array}%
\right] \right\}
\end{equation*}
\end{itemize}

(c) (5 pts) Find a basis for the null space of $\mathbf{A}$.

\begin{itemize}
\item We need to solve $\mathbf{Ax}=\mathbf{0}$. \ From the row reduction of 
$\mathbf{A}$ we can infer%
\begin{equation*}
\left[ \mathbf{A}|\mathbf{0}\right] \quad \underrightarrow{\text{ row
reduction }}\ \left[ 
\begin{array}{cccc|c}
1 & 0 & 1 & 0 & 0 \\ 
0 & 0 & 0 & 1 & 0 \\ 
0 & 0 & 0 & 0 & 0%
\end{array}%
\right] \quad \Rightarrow \quad \left\{ 
\begin{array}{r}
x_{1}+x_{3}=0 \\ 
x_{4}=0 \\ 
0=0%
\end{array}%
\right.
\end{equation*}%
Since columns $2$ and $3$ of the R.R.E.F. of $\left[ \mathbf{A}|\mathbf{0}%
\right] $ don't have pivots, $x_{2}$ and $x_{3}$ will be the free variable
of the solution:%
\begin{equation*}
\mathbf{x}=\left[ 
\begin{array}{c}
-x_{3} \\ 
x_{2} \\ 
x_{3} \\ 
0%
\end{array}%
\right] =x_{2}\left[ 
\begin{array}{c}
0 \\ 
1 \\ 
0 \\ 
0%
\end{array}%
\right] +x_{3}\left[ 
\begin{array}{c}
-1 \\ 
0 \\ 
1 \\ 
0%
\end{array}%
\right] \quad \Rightarrow \quad \text{basis for }NullSp\left( \mathbf{A}%
\right) =\left\{ \left[ 
\begin{array}{c}
0 \\ 
1 \\ 
0 \\ 
0%
\end{array}%
\right] \ ,\ \left[ 
\begin{array}{c}
-1 \\ 
0 \\ 
1 \\ 
0%
\end{array}%
\right] \right\}
\end{equation*}
\end{itemize}

(d) (5 pts)\ What is the rank of $\mathbf{A}$?

\begin{itemize}
\item Since the row space (and column space) of $\mathbf{A}$ has $2$ basis
vectors%
\begin{equation*}
Rank\left( \mathbf{A}\right) =2
\end{equation*}
\end{itemize}

\newpage

4. Consider the following linear transformation: \newline
$T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}:$ $T\left( \left[
x_{1},x_{2},x_{3}\right] \right) =\left[
x_{1}+x_{2},x_{1}-x_{3},x_{1}+2x_{2}+x_{3}\right] $.

(a) (5 pts) Find the matrix $\mathbf{A}_{T}$ such that $\mathbf{A}_{T}%
\mathbf{x}=T\left( \mathbf{x}\right) $.

\begin{itemize}
\item 
\begin{equation*}
\mathbf{A}_{T}=\left[ 
\begin{array}{ccc}
\uparrow & \uparrow & \uparrow \\ 
T\left( \left[ 1,0,0\right] \right) & T\left( \left[ 0,1,0\right] \right) & 
T\left( \left[ 0,0,1\right] \right) \\ 
\downarrow & \downarrow & \downarrow%
\end{array}%
\right] =\left[ 
\begin{array}{ccc}
1 & 1 & 0 \\ 
1 & 0 & -1 \\ 
1 & 2 & 1%
\end{array}%
\right]
\end{equation*}
\end{itemize}

(b) (5 pts) Find a basis for the range of $T$.

\begin{itemize}
\item Let's first row reduce $\mathbf{A}_{T}$ to its R.R.E.F.%
\begin{eqnarray*}
&&\mathbf{A}_{T} =\left[ 
\begin{array}{ccc}
1 & 1 & 0 \\ 
1 & 0 & -1 \\ 
1 & 2 & 1%
\end{array}%
\right] \ \underrightarrow{%
\begin{array}{c}
\ R_{2}\rightarrow R_{2}-R_{1} \\ 
R_{3}\rightarrow R_{3}-R_{1}%
\end{array}%
}\ \left[ 
\begin{array}{ccc}
1 & 1 & 0 \\ 
0 & -1 & -1 \\ 
0 & 1 & 1%
\end{array}%
\right] \  \\
&& \underrightarrow{\ R_{3}\rightarrow R_{3}-R_{2}}\ \left[ 
\begin{array}{ccc}
1 & 1 & 0 \\ 
0 & -1 & -1 \\ 
0 & 0 & 0%
\end{array}%
\right] \underrightarrow{%
\begin{array}{c}
\ R_{1}\rightarrow R_{1}-R_{2} \\ 
R_{2}\rightarrow -R_{2}%
\end{array}%
}\ \left[ 
\begin{array}{ccc}
1 & 0 & -1 \\ 
0 & 1 & 1 \\ 
0 & 0 & 0%
\end{array}%
\right] =R.R.E.F.\left( \mathbf{A}_{T}\right)
\end{eqnarray*}%
Since $Range\left( T\right) =ColSp\left( \mathbf{A}_{T}\right) $ and a basis
for $ColSp\left( \mathbf{A}_{T}\right) $ is given by the first two columns
of $\mathbf{A}_{T}$ (as these are the columns of $\mathbf{A}_{T}$ that
correspond to the columns of $R.R.E.F.\left( \mathbf{A}_{T}\right) $ that
have pivots). 
\begin{equation*}
\text{basis for }Range\left( T\right) =\left\{ \left[ 
\begin{array}{c}
1 \\ 
1 \\ 
1%
\end{array}%
\right] \ ,\ \left[ 
\begin{array}{c}
1 \\ 
0 \\ 
2%
\end{array}%
\right] \right\}
\end{equation*}
\end{itemize}

(c) (5 pts) Find a basis for the kernel of $T$.

\begin{itemize}
\item We have $\ker \left( T\right) =NullSp\left( \mathbf{A}_{T}\right) =$
solution set of $\mathbf{A}_{T}\mathbf{x}=\mathbf{0}$. \ From the R.R.E.F.
of $\mathbf{A}_{T}$, we see that solutions of $\mathbf{A}_{T}\mathbf{x}=0$
must satisfy%
\begin{equation*}
\left. 
\begin{array}{r}
x_{1}-x_{3}=0 \\ 
x_{2}+x_{3}=0 \\ 
0=0%
\end{array}%
\right\} \quad \Rightarrow \quad \left\{ 
\begin{array}{c}
x_{1}=x_{3} \\ 
x_{2}=-x_{3}%
\end{array}%
\right. \quad \Rightarrow \quad \mathbf{x}=\left[ 
\begin{array}{c}
x_{3} \\ 
-x_{3} \\ 
x_{3}%
\end{array}%
\right] =x_{3}\left[ 
\begin{array}{c}
1 \\ 
-1 \\ 
1%
\end{array}%
\right]
\end{equation*}%
and so%
\begin{equation*}
\text{basis for }\ker \left( T\right) =\left\{ \left[ 1,-1,1\right] \right\}
\end{equation*}
\end{itemize}

5. Consider the matrix $\mathbf{A}=\left[ 
\begin{array}{cccc}
0 & 0 & 0 & 3 \\ 
0 & 1 & -3 & 0 \\ 
1 & 0 & 0 & 0 \\ 
0 & 0 & -2 & 0%
\end{array}%
\right] $

(a) 7 pts) Compute $\det \left( \mathbf{A}\right) $ via a cofactor expansion
along the third row.

\begin{itemize}
\item 
\begin{eqnarray*}
\det \left( \mathbf{A}\right) &=&\left( 1\right) \left( -1\right) ^{3+1}\det
\left( \mathbf{M}_{13}\right) +0+0+0 \\
&=&\left( 1\right) \left( -1\right) ^{4}\det \left( \left[ 
\begin{array}{ccc}
0 & 0 & 3 \\ 
1 & -3 & 0 \\ 
0 & -2 & 0%
\end{array}%
\right] \right) \\
&=&\left( 0+0+\left( 3\right) \left( -1\right) ^{1+3}\det \left( \left[ 
\begin{array}{cc}
1 & -3 \\ 
0 & -2%
\end{array}%
\right] \right) \right) \\
&=&\left( 3\right) \left( -1\right) ^{4}\left( -2+0\right) \\
&=&-6
\end{eqnarray*}
\end{itemize}

(b) (8 pts) Compute $\det \left( \mathbf{A}\right) $ by row reducing $%
\mathbf{A}$ to an upper triangular matrix.

\begin{itemize}
\item 
\begin{equation*}
\left[ 
\begin{array}{cccc}
0 & 0 & 0 & 3 \\ 
0 & 1 & -3 & 0 \\ 
1 & 0 & 0 & 0 \\ 
0 & 0 & -2 & 0%
\end{array}%
\right] \ \underrightarrow{\ R_{1}\longleftrightarrow R_{3}\ }\ \left[ 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & 1 & -3 & 0 \\ 
0 & 0 & 0 & 3 \\ 
0 & 0 & -2 & 0%
\end{array}%
\right] \ \underrightarrow{\ R_{3}\longleftrightarrow R_{4}\ }\ \left[ 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & 1 & -3 & 0 \\ 
0 & 0 & 3 & 0 \\ 
0 & 0 & 0 & -2%
\end{array}%
\right] =R.E.F.\left( \mathbf{A}\right)
\end{equation*}%
Since we performed 2 row interchanges, and $0$ row rescalings, in our row
reduction to R.E.F.,%
\begin{equation*}
\det \left( \mathbf{A}\right) =\left( -1\right) ^{2}\det \left( R.E.F.\left( 
\mathbf{A}\right) \right) =\det \left( \left[ 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & 1 & -3 & 0 \\ 
0 & 0 & 3 & 0 \\ 
0 & 0 & 0 & -2%
\end{array}%
\right] \right) =\left( 1\right) \left( 1\right) \left( 3\right) \left(
-2\right) =-6
\end{equation*}

\newpage
\end{itemize}

6. (10 pts) Use Cramer's Rule to solve 
\begin{eqnarray*}
x_{1}+2x_{2} &=&3 \\
-x_{1}+x_{2} &=&3
\end{eqnarray*}

\begin{itemize}
\item We have%
\begin{eqnarray*}
\mathbf{A} &=&\left[ 
\begin{array}{cc}
1 & 2 \\ 
-1 & 1%
\end{array}%
\right] \quad \Rightarrow \quad \det \left( \mathbf{A}\right) =\left(
1\right) \left( 1\right) -\left( 2\right) \left( -1\right) =3 \\
\mathbf{B}_{1} &=&\left[ 
\begin{array}{cc}
3 & 2 \\ 
3 & 1%
\end{array}%
\right] \quad \Rightarrow \quad \det \left( \mathbf{B}_{1}\right) =\left(
3\right) \left( 1\right) -\left( 2\right) \left( 3\right) =-3 \\
\mathbf{B}_{2} &=&\left[ 
\begin{array}{cc}
1 & 3 \\ 
-1 & 3%
\end{array}%
\right] \quad \Rightarrow \quad \det \left( \mathbf{B}_{2}\right) =\left(
1\right) \left( 3\right) -\left( 3\right) \left( -1\right) =6
\end{eqnarray*}%
Hence, Cramer's Rule tells us that the components of a solution vector are%
\begin{eqnarray*}
x_{1} &=&\frac{\det \left( \mathbf{B}_{1}\right) }{\det \left( \mathbf{A}%
\right) }=\frac{-3}{3}=-1 \\
x_{2} &=&\frac{\det \left( \mathbf{B}_{2}\right) }{\det \left( \mathbf{A}%
\right) }=\frac{6}{3}=2
\end{eqnarray*}
\end{itemize}

7. (10 pts) Find all the cofactors of $\mathbf{A}=\left[ 
\begin{array}{cc}
2 & 2 \\ 
3 & 4%
\end{array}%
\right] $ and then use these cofactors to compute $\mathbf{A}^{-1}$

\begin{itemize}
\item The cofactors are given by the formula $c_{ij}=\left( -1\right)
^{i+j}\det \left( \mathbf{M}_{ij}\right) $ where $\mathbf{M}_{ij}$ is the $%
1\times 1$ matrix obtained from $\mathbf{A}$ by deleting its $i^{th}$ row
and $j^{th}$ column.%
\begin{eqnarray*}
c_{11} &=&\left( -1\right) ^{1+1}\det \left( \left[ 4\right] \right) =4 \\
c_{12} &=&\left( -1\right) ^{1+2}\det \left( \left[ 3\right] \right) =-3 \\
c_{21} &=&\left( -1\right) ^{2+1}\det \left( \left[ 3\right] \right) =-2 \\
c_{22} &=&\left( -1\right) ^{2+2}\det \left( \left[ 2\right] \right) =2
\end{eqnarray*}%
So the cofactor matrix is%
\begin{equation*}
\mathbf{C}=\left[ 
\begin{array}{cc}
4 & -3 \\ 
-2 & 2%
\end{array}%
\right]
\end{equation*}%
We can now apply the cofactor formula for $\mathbf{A}^{-1}$ 
\begin{equation*}
\mathbf{A}^{-1}=\frac{1}{\det \left( \mathbf{A}\right) }\mathbf{C}^{T}=\frac{%
1}{\left( 8-6\right) }\left[ 
\begin{array}{cc}
4 & -2 \\ 
-3 & 2%
\end{array}%
\right] =\left[ 
\begin{array}{cc}
2 & -1 \\ 
-\frac{3}{2} & 1%
\end{array}%
\right]
\end{equation*}
\end{itemize}

\end{document}
