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\begin{document}

\begin{center}
\textbf{Math 3013}\\[0pt]
ANSWERS TO SECOND EXAM \\[0pt]
November 3, 2020
\end{center}

1. Consider the vectors $\left\{ [1,2,1,1,0] \; , \; [0,1,2,1,1]\; ,
\;[-1,-1,1,0,1] \; , \; [-1,0,3,1,2]\right\} \in \mathbb{R}^{5}$

\begin{eqnarray}
\left[ 
\begin{array}{ccccc}
1 & 0 & -3 & -1 & -2 \\ 
0 & 1 & 2 & 1 & 1 \\ 
0 & 0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 & 0%
\end{array}%
\right]  &=&R.R.E.F.\left( \mathbf{A}\right)   \notag \\
\quad  &\Rightarrow &\quad \text{not linearly independent}  \TCItag{a} \\
\dim RowSp\left( \mathbf{A}\right)  &=&2  \TCItag{b}
\end{eqnarray}

\begin{itemize}
\item 
2. Definitions
\end{itemize}

3. Given that the matrix $\displaystyle\mathbf{A}=\left[ 
\begin{array}{cccc}
2 & 4 & 0 & 2 \\ 
-1 & -2 & 1 & -3 \\ 
0 & 0 & 1 & 2%
\end{array}%
\right] $ \quad row reduces to \quad $\left[ 
\begin{array}{cccc}
1 & 2 & 0 & 1 \\ 
0 & 0 & 1 & -2 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right] $

\begin{itemize}
\item 
\begin{eqnarray}
\text{basis for row space } &\text{= }&\left\{ \left[ 1,2,0,1\right] \ ,\ %
\left[ 0,0,1,-2\right] \right\}   \TCItag{a} \\
\text{basis for column space } &\text{= }&\left\{ \left[ 
\begin{array}{c}
2 \\ 
-1 \\ 
0%
\end{array}%
\right] \ ,\ \left[ 
\begin{array}{c}
0 \\ 
1 \\ 
1%
\end{array}%
\right] \right\}   \TCItag{b} \\
\text{basis for }NullSp\left( \mathbf{A}\right)  &=&\left\{ \left[ -2,1,0,0%
\right] \ ,\ \left[ -1,0,2,1\right] \right\}   \TCItag{c} \\
rank &=&2  \TCItag{d}
\end{eqnarray}
\end{itemize}

4.

\begin{itemize}
\item 
\begin{eqnarray}
\mathbf{A}_{T} &=&=\left[ 
\begin{array}{ccc}
1 & 2 & 1 \\ 
1 & 1 & 0 \\ 
1 & 0 & -1%
\end{array}%
\right]   \TCItag{a} \\
\text{basis for Range}\left( T\right)  &=&\left\{ \left[ 1,1,1\right] \ ,\ %
\left[ 2,1,0\right] \right\}   \TCItag{b} \\
\text{basis for }\ker \left( T\right)  &=&\left\{ \left[ 1,-1,1\right]
\right\}   \TCItag{c}
\end{eqnarray}
\end{itemize}

5. (15 pts) cofactor expansions $\quad \Rightarrow \quad $ $\det \left( 
\mathbf{A}\right) =-2$

6. (15 pts) row reduction method $\quad \Rightarrow \quad \det \left( 
\mathbf{A}\right) =2$

\end{document}
