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\begin{document}

\begin{center}
\textbf{Math 3013} \\[0pt]
SOLUTIONS TO FIRST EXAM \\[0pt]
11:30 - 12:20, February 18, 2022
\end{center}

1. Let 
\begin{equation*}
\quad \mathbf{A}=\left[ 
\begin{array}{cc}
1 & -1 \\ 
0 & 3%
\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{eqnarray*}
\mathbf{AB} &=&\left[ 
\begin{array}{cc}
1 & -1 \\ 
0 & 3%
\end{array}%
\right] \left[ 
\begin{array}{ccc}
2 & 0 & 1 \\ 
0 & -1 & 0%
\end{array}%
\right] =\left[ 
\begin{array}{ccc}
\left( 1\right) \left( 2\right) +\left( -1\right) \left( 0\right) & \left(
1\right) \left( 0\right) +\left( -1\right) \left( -1\right) & \left(
1\right) \left( 1\right) +\left( -1\right) \left( 0\right) \\ 
\left( 0\right) \left( 2\right) +\left( 3\right) \left( 0\right) & \left(
0\right) \left( 0\right) +\left( 3\right) \left( -1\right) & \left( 0\right)
\left( 1\right) +\left( 3\right) \left( 0\right)%
\end{array}%
\right] \\
&=&\left[ 
\begin{array}{ccc}
2 & 1 & 1 \\ 
0 & -3 & 0%
\end{array}%
\right]
\end{eqnarray*}
: 
\begin{equation*}
\mathbf{BA=}\left[ 
\begin{array}{ccc}
2 & 0 & 1 \\ 
0 & -1 & 0%
\end{array}%
\right] \left[ 
\begin{array}{cc}
1 & -1 \\ 
0 & 3%
\end{array}%
\right] \text{ is undefined since }\#Columns\left( \mathbf{B}\right) \neq
\#Rows\left( \mathbf{A}\right)
\end{equation*}

2. For each of the following augmented matrices, indicate

\begin{itemize}
\item[(i)] the number of equations and the number of variables in the
corresponding linear system

\item[(ii)] whether or not the corresponding linear system has a solution

\item[(iii)] 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
that you do not have to do any calculations to answer the questions.

(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[(i)] 4 equations in 4 unknowns

\item[(ii)] there is a solution

\item[(iii)] 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[(i)] 3 equations in 4 unknowns

\item[(ii)] there is a solution

\item[(iii)] 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[(i)] 4 equations in 3 unknowns

\item[(ii)] 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 & 0 & -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}
1 & 0 & 1 \\ 
0 & 0 & 1 \\ 
2 & 1 & 0%
\end{array}%
\right] 
\end{equation*}

\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}
1 & 0 & 1 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 1 & 0 \\ 
2 & 1 & 0 & 0 & 0 & 1%
\end{array}%
\right]  \\
&& \\
&&\underrightarrow{R_{3}\longleftrightarrow R_{3}-2R_{1}}\quad \left[ 
\begin{array}{ccc|ccc}
1 & 0 & 1 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 0 & 1 & 0 \\ 
0 & 1 & -2 & -2 & 0 & 1%
\end{array}%
\right]  \\
&&\underrightarrow{R_{2}\longleftrightarrow R_{3}\ }\quad \left[ 
\begin{array}{ccc|ccc}
1 & 0 & 1 & 1 & 0 & 0 \\ 
0 & 1 & -2 & -2 & 0 & 1 \\ 
0 & 0 & 1 & 0 & 1 & 0%
\end{array}%
\right]  \\
&&\underrightarrow{%
\begin{array}{c}
R_{1}\rightarrow R_{1}-R_{3} \\ 
R_{2}\rightarrow R_{2}+2R_{3}\ 
\end{array}%
}\quad \left[ 
\begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1 & 0 \\ 
0 & 1 & 0 & -2 & 2 & 1 \\ 
0 & 0 & 1 & 0 & 1 & 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 & 0 \\ 
-2 & 2 & 1 \\ 
0 & 1 & 0%
\end{array}%
\right] 
\end{equation*}
\end{itemize}

\end{document}
