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\begin{document}

\begin{center}
\textbf{Math 3013} \\[0pt]
SOLUTIONS TO FIRST EXAM \\[0pt]
4:30 - 6:00pm, September 22, 2020
\end{center}

1. Let 
\begin{equation*}
\quad \mathbf{A}=\left[ 
\begin{array}{cc}
1 & -1%
\end{array}%
\right] \quad ,\quad \mathbf{B}=\left[ 
\begin{array}{ccc}
2 & 0 & 1 \\ 
0 & -1 & 0%
\end{array}%
\right]
\end{equation*}%
Compute the matrix products $\mathbf{AB}$ and $\mathbf{BA}$ (if they exist)%
\begin{equation*}
\mathbf{AB}=\left[ 
\begin{array}{cc}
1 & -1%
\end{array}%
\right] \left[ 
\begin{array}{ccc}
2 & 0 & 1 \\ 
0 & -1 & 0%
\end{array}%
\right] =\left[ 
\begin{array}{ccc}
2+0 & 0+1 & 1+0%
\end{array}%
\right] =\left[ 
\begin{array}{ccc}
2 & 1 & 1%
\end{array}%
\right]
\end{equation*}%
\begin{equation*}
\mathbf{BA=}\left[ 
\begin{array}{ccc}
2 & 0 & 1 \\ 
0 & -1 & 0%
\end{array}%
\right] \left[ 
\begin{array}{cc}
1 & -1%
\end{array}%
\right] \text{ is undefined}
\end{equation*}%
The reason $\mathbf{BA}$ is undefined is because the number of columns of
the first factor (3) doesn't match the number of rows (1) of the second
factor.

2. For each of the following augmented matrices, indicate

\begin{itemize}
\item the number of equations and the number of variables in the
corresponding linear system

\item whether or not the corresponding linear system has a solution

\item if the corresponding linear system does have a solution, the number of
free variables in the solution.
\end{itemize}

Hint: note that these augmented matrices are already in row echelon form and
you do not have to actually solve the systems.

(a) $\left[ 
\begin{array}{cccc|c}
1 & 1 & -1 & 2 & 1 \\ 
0 & 1 & 0 & 1 & 1 \\ 
0 & 0 & 0 & 1 & 2 \\ 
0 & 0 & 0 & 0 & 0%
\end{array}%
\right] $

\begin{itemize}
\item 4 equations in 4 unknowns

\item there is a solution

\item Column 3 is the only column on the left without a pivot. \ Therefore,
there will be 1 free parameter $\left( x_{3}\right) $ in the solution.
\end{itemize}

(b) $\left[ 
\begin{array}{cccc|c}
1 & 0 & 0 & 2 & 1 \\ 
0 & 0 & 1 & 1 & 2 \\ 
0 & 0 & 0 & 0 & 0%
\end{array}%
\right] $

\begin{itemize}
\item 3 equations in 4 unknowns

\item there is a solution

\item Columns 2 and 4 are the only columns on the left without pivots. \
Therefore, there will be 2 free parameters ($x_{2}$ and $x_{3}$)\ in the
solution.
\end{itemize}

(c) $\left[ 
\begin{array}{ccc|c}
1 & 1 & 0 & 3 \\ 
0 & 1 & 2 & 2 \\ 
0 & 0 & 1 & 1 \\ 
0 & 0 & 0 & 1%
\end{array}%
\right] $

\begin{itemize}
\item 4 equations in 3 unknowns

\item There is no solution since the equation corresponding to the last row
says $0=1$ (which implies there was a contradiction amongst the original
equations).
\end{itemize}

\newpage

3. Consider the following linear system 
\begin{equation*}
\begin{array}{rcr}
x_{1}-x_{2}+2x_{3}+x_{4} & = & 1 \\ 
2x_{1}+x_{2}+x_{3}+2x_{4} & = & -1 \\ 
-x_{1}+x_{2}+2x_{3}-x_{4} & = & 3%
\end{array}%
\end{equation*}%
Write down the corresponding augmented matrix and row reduce it to
row-echelon form.

\begin{itemize}
\item 
\begin{eqnarray*}
\left[ \mathbf{A}\mid \mathbf{b}\right]  &=&\left[ 
\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 1 \\ 
2 & 1 & 1 & 2 & -1 \\ 
-1 & 1 & 2 & -1 & 3%
\end{array}%
\right]  \\
&&\underrightarrow{%
\begin{array}{c}
R_{2}\rightarrow R_{2}-2R_{1} \\ 
R_{3}\rightarrow R_{3}+R_{1}%
\end{array}%
}\left[ 
\begin{array}{cccc|c}
1 & -1 & 2 & 1 & 1 \\ 
0 & 3 & -3 & -3 & 3 \\ 
0 & 0 & 4 & 0 & 4%
\end{array}%
\right] \qquad (R.E.F.)
\end{eqnarray*}
\end{itemize}

4. Row reduce the following augmented matrix to \textbf{reduced} row-echelon
form. 
\begin{equation*}
\left[ \mathbf{A}\mid \mathbf{b}\right] =\left[ 
\begin{array}{ccccc|c}
1 & 2 & 1 & 2 & 2 & 2 \\ 
0 & 0 & 2 & 4 & 4 & 4 \\ 
0 & 0 & 0 & 1 & -2 & 2 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right] 
\end{equation*}

\begin{itemize}
\item 
\begin{eqnarray*}
&&\underrightarrow{R_{2}\rightarrow \frac{1}{2}R_{2}}\left[ 
\begin{array}{ccccc|c}
1 & 2 & 1 & 2 & 2 & 2 \\ 
0 & 0 & 1 & 2 & 2 & 2 \\ 
0 & 0 & 0 & 1 & -2 & 2 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right]  \\
&&\underrightarrow{R_{1}\rightarrow R_{1}-R_{2}}\left[ 
\begin{array}{ccccc|c}
1 & 2 & 0 & 0 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 6 & -2 \\ 
0 & 0 & 0 & 1 & -2 & 2 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right] \qquad (R.R.E.F.)
\end{eqnarray*}
\end{itemize}

\newpage

5. Suppose the augmented matrix below is the Reduced Row Echelon Form of an
augmented matrix of a linear system. Display the solution of the linear
system as a hyperplane (within the space of variables). 
\begin{equation*}
\left[ 
\begin{array}{ccccc|c}
1 & 0 & 0 & -2 & 0 & 1 \\ 
0 & 0 & 1 & 1 & 0 & 2 \\ 
0 & 0 & 0 & 0 & 1 & 0%
\end{array}%
\right] 
\end{equation*}

\begin{itemize}
\item The augmented matrix in R.R.E.F. has columns 2 and 4 on the left
without pivots. \ This means that $x_{2}$ and $x_{4}$ are to be free
variables in the solution. \ Converting back to equations, and moving the
free variables to the right hand side yields%
\begin{equation*}
\left. 
\begin{array}{r}
x_{1}-2x_{4}=1 \\ 
x_{3}+x_{4}=2 \\ 
x_{5}=0%
\end{array}%
\right\} \quad \Rightarrow \quad \left\{ 
\begin{array}{l}
x_{1}=1+2x_{4} \\ 
x_{3}=2-x_{4} \\ 
x_{5}=0%
\end{array}%
\right. 
\end{equation*}%
Next we write down a typical solution vector, substituting where we can for
the fixed variables $x_{1},x_{3}$, and $x_{5}:$%
\begin{equation*}
\mathbf{x}=\left[ 
\begin{array}{l}
1+2x_{4} \\ 
x_{2} \\ 
2-x_{4} \\ 
x_{4} \\ 
0%
\end{array}%
\right] =\left[ 
\begin{array}{l}
1 \\ 
0 \\ 
2 \\ 
0 \\ 
0%
\end{array}%
\right] +x_{2}\left[ 
\begin{array}{l}
0 \\ 
1 \\ 
0 \\ 
0 \\ 
0%
\end{array}%
\right] +x_{4}\left[ 
\begin{array}{l}
2 \\ 
0 \\ 
-1 \\ 
1 \\ 
0%
\end{array}%
\right] 
\end{equation*}%
which displays the solution set as a $2$-dimensional hyperplane in $\mathbb{R%
}^{5}$.
\end{itemize}

6. Determine if the matrix below has an inverse and, if so, determine its
inverse. (Hint: a single row reduction process will answer both parts of to
this problem.) 
\begin{equation*}
\mathbf{A}=\left[ 
\begin{array}{ccc}
0 & 0 & 1 \\ 
1 & 1 & 1 \\ 
2 & 1 & 0%
\end{array}%
\right] 
\end{equation*}%
, inverse: $\left[ 
\begin{array}{ccc}
1 & -1 & 1 \\ 
-2 & 2 & -1 \\ 
1 & 0 & 0%
\end{array}%
\right] $

\begin{itemize}
\item We form the adjoined matrix $\left[ \mathbf{A}\mid \mathbf{I}\right] $
and row reduce it to its Reduced Row Echelon Form:%
\begin{eqnarray*}
\left[ \mathbf{A}\mid \mathbf{I}\right]  &=&\left[ 
\begin{array}{ccc|ccc}
0 & 0 & 1 & 1 & 0 & 0 \\ 
1 & 1 & 1 & 0 & 1 & 0 \\ 
2 & 1 & 0 & 0 & 0 & 1%
\end{array}%
\right]  \\
&&\underrightarrow{R_{1}\longleftrightarrow R_{2}}\left[ 
\begin{array}{ccc|ccc}
1 & 1 & 1 & 0 & 1 & 0 \\ 
0 & 0 & 1 & 1 & 0 & 0 \\ 
2 & 1 & 0 & 0 & 0 & 1%
\end{array}%
\right]  \\
&&\underrightarrow{R_{3}\rightarrow R_{3}-2R_{1}}\left[ 
\begin{array}{ccc|ccc}
1 & 1 & 1 & 0 & 1 & 0 \\ 
0 & 0 & 1 & 1 & 0 & 0 \\ 
0 & -1 & -2 & 0 & -2 & 1%
\end{array}%
\right]  \\
&&\underrightarrow{R_{3}\longleftrightarrow R_{2}}\left[ 
\begin{array}{ccc|ccc}
1 & 1 & 1 & 0 & 1 & 0 \\ 
0 & -1 & -2 & 0 & -2 & 1 \\ 
0 & 0 & 1 & 1 & 0 & 0%
\end{array}%
\right]  \\
&&\underrightarrow{R_{2}\rightarrow -R_{2}}\left[ 
\begin{array}{ccc|ccc}
1 & 1 & 0 & -1 & 1 & 0 \\ 
0 & 1 & 2 & 0 & 2 & -1 \\ 
0 & 0 & 1 & 1 & 0 & 0%
\end{array}%
\right]  \\
&&\underrightarrow{R_{2}\rightarrow R_{2}-2R_{3}}\left[ 
\begin{array}{ccc|ccc}
1 & 1 & 0 & -1 & 1 & 0 \\ 
0 & 1 & 0 & -2 & 2 & -1 \\ 
0 & 0 & 1 & 1 & 0 & 0%
\end{array}%
\right]  \\
&&\underrightarrow{R_{1}\rightarrow R_{1}-R_{2}}\left[ 
\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1 & 1 \\ 
0 & 1 & 0 & -2 & 2 & -1 \\ 
0 & 0 & 1 & 1 & 0 & 0%
\end{array}%
\right] 
\end{eqnarray*}%
Since the $\mathbf{A}$-part of the reduced row echelon form of $\left[ 
\mathbf{A}\mid \mathbf{I}\right] $ is the identity matrix the right hand
side (the $\mathbf{I}$-part) of the RREF is the inverse of $\mathbf{A}$.
Thus,%
\begin{equation*}
\mathbf{A}=\left[ 
\begin{array}{|ccc}
1 & -1 