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\begin{document}

\begin{center}
\textbf{Math 3013} \\[0pt]
SOLUTIONS TO FIRST EXAM \\[0pt]
12:30 - 1:20pm, September 15, 2021
\end{center}

1. Let 
\begin{equation*}
\quad \mathbf{A}=\left[ 
\begin{array}{ccc}
1 & 0 & -1 \\ 
0 & 1 & 0 \\ 
2 & 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{itemize}
\item 
\begin{equation*}
\mathbf{AB:}\text{ product is undefined since \#columns of the first factor }%
\neq \text{\#rows of the second factor }
\end{equation*}%
\begin{eqnarray*}
\mathbf{BA} &\mathbf{=}&\left[ 
\begin{array}{ccc}
2 & 0 & 1 \\ 
0 & -1 & 0%
\end{array}%
\right] \left[ 
\begin{array}{ccc}
1 & 0 & -1 \\ 
0 & 1 & 0 \\ 
2 & 1 & 1%
\end{array}%
\right] =\left[ 
\begin{array}{ccc}
2+0+2 & 0+0+1 & -2+0+1 \\ 
0+0+0 & 0-1+0 & 0+0+0%
\end{array}%
\right] \\
&=&\left[ 
\begin{array}{ccc}
4 & 1 & -1 \\ 
0 & -1 & 0%
\end{array}%
\right]
\end{eqnarray*}
:
\end{itemize}

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}{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 and 3 unknowns

no solution (since the last row corresponds to the equation $0=1$ which is a
contradiction)
\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 variables

solutions exist

2 free variables, $x_{2}$ and $x_{4}$, since columns 2 and 4 on the left
hand side of the R.E.F. don't have pivots
\end{itemize}

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

\begin{itemize}
\item 3 equation in 3 variables

there is a solution

no free variables in solution, since the left hand side of the R.E.F. does
not have any columns without pivots
\end{itemize}

\newpage

3. Consider the following linear system 
\begin{equation*}
\begin{array}{rcr}
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}|\mathbf{b}\right]  &=&\left[ 
\begin{array}{cccc|c}
0 & 1 & 2 & 1 & 1 \\ 
2 & 1 & 1 & 2 & -1 \\ 
1 & 1 & 2 & -1 & 3%
\end{array}%
\right] \quad \underrightarrow{\quad R_{1}\longleftrightarrow R_{3}\quad }%
\quad \left[ 
\begin{array}{cccc|c}
1 & 1 & 2 & -1 & 3 \\ 
2 & 1 & 1 & 2 & -1 \\ 
0 & 1 & 2 & 1 & 1%
\end{array}%
\right]  \\
\underrightarrow{\quad R_{2}\rightarrow R_{2}-2R_{1}\quad } &&\left[ 
\begin{array}{cccc|c}
1 & 1 & 2 & -1 & 3 \\ 
0 & -1 & -3 & 4 & -7 \\ 
0 & 1 & 2 & 1 & 1%
\end{array}%
\right] \quad \underrightarrow{\quad R_{3}\rightarrow R_{3}+R_{2}\quad }%
\quad \left[ 
\begin{array}{cccc|c}
1 & 1 & 2 & -1 & 3 \\ 
0 & -1 & -3 & 4 & -7 \\ 
0 & 0 & -1 & 5 & -6%
\end{array}%
\right]  \\
&=&R.R.E.F.\left( \left[ \mathbf{A}|\mathbf{b}\right] \right) 
\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%
\end{array}%
\right]
\end{equation*}

\begin{itemize}
\item 
\begin{eqnarray*}
&&\left[ 
\begin{array}{ccccc|c}
1 & 2 & 1 & 2 & 2 & 2 \\ 
0 & 0 & 2 & 4 & 4 & 4 \\ 
0 & 0 & 0 & 1 & -2 & 2%
\end{array}%
\right] \quad \underrightarrow{\quad R_{2}\rightarrow \frac{1}{2}R_{2}\quad }%
\quad \left[ 
\begin{array}{ccccc|c}
1 & 2 & 1 & 2 & 2 & 2 \\ 
0 & 0 & 1 & 2 & 2 & 2 \\ 
0 & 0 & 0 & 1 & -2 & 2%
\end{array}%
\right] \\
&&\underrightarrow{%
\begin{array}{c}
R_{1}\rightarrow R_{1}-2R_{3} \\ 
R_{2}\rightarrow R_{2}-2R_{3}\quad%
\end{array}%
}\left[ 
\begin{array}{ccccc|c}
1 & 2 & 1 & 0 & 6 & -2 \\ 
0 & 0 & 1 & 0 & 6 & -2 \\ 
0 & 0 & 0 & 1 & -2 & 2%
\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%
\end{array}%
\right] \\
&=&R.R.E.F.\left( \left[ \mathbf{A}|\mathbf{b}\right] \right)
\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 & 1%
\end{array}%
\right]
\end{equation*}

\begin{itemize}
\item Columns 2 and 4 of $R.R.E.F.\left( \left[ \mathbf{A}|\mathbf{b}\right]
\right) $ do not contain pivots, and so $x_{2}$ and $x_{4}$ will be free
variables in the solution. \ Converting the augmented matrix back to
equations, and moving the free variables to the right hand side we get%
\begin{equation*}
\left. 
\begin{array}{r}
x_{1}-2x_{4}=1 \\ 
x_{3}+x_{4}=2 \\ 
x_{5}=1%
\end{array}%
\right\} \quad \Rightarrow \quad \left\{ 
\begin{array}{l}
x_{1}=1+2x_{4} \\ 
x_{3}=2-x_{4} \\ 
x_{5}=1%
\end{array}%
\right.
\end{equation*}%
Thus, a typical solution vector will have the form%
\begin{equation*}
\mathbf{x}=\left[ 
\begin{array}{c}
x_{1} \\ 
x_{2} \\ 
x_{3} \\ 
x_{4} \\ 
x_{5}%
\end{array}%
\right] =\left[ 
\begin{array}{c}
1+2x_{4} \\ 
x_{2} \\ 
2-x_{4} \\ 
x_{4} \\ 
1%
\end{array}%
\right] =\left[ 
\begin{array}{c}
1 \\ 
0 \\ 
2 \\ 
0 \\ 
1%
\end{array}%
\right] +x_{2}\left[ 
\begin{array}{c}
0 \\ 
1 \\ 
0 \\ 
0 \\ 
0%
\end{array}%
\right] +x_{4}\left[ 
\begin{array}{c}
2 \\ 
0 \\ 
-1 \\ 
1 \\ 
0%
\end{array}%
\right]
\end{equation*}%
which displays the solution vectors as living on a $2$-dimensional plane.
\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 & -1 & 1 \\ 
-2 & 2 & -1 \\ 
1 & 0 & 0%
\end{array}%
\right]
\end{equation*}

\begin{itemize}
\item 
\begin{eqnarray*}
\left[ \mathbf{A}|\mathbf{I}\right] &=&\left[ 
\begin{array}{ccc|ccc}
1 & -1 & 1 & 1 & 0 & 0 \\ 
-2 & 2 & -1 & 0 & 1 & 0 \\ 
1 & 0 & 0 & 0 & 0 & 1%
\end{array}%
\right] \quad \underrightarrow{%
\begin{array}{c}
\quad R_{2}\rightarrow R_{2}+2R_{1} \\ 
R_{3}\rightarrow R_{3}-R_{1}%
\end{array}%
}\quad \left[ 
\begin{array}{ccc|ccc}
1 & -1 & 1 & 1 & 0 & 0 \\ 
0 & 0 & 1 & 2 & 1 & 0 \\ 
0 & 1 & -1 & -1 & 0 & 1%
\end{array}%
\right] \\
\underrightarrow{\quad R_{2}\longleftrightarrow R_{3}\quad } && \left[ 
\begin{array}{ccc|ccc}
1 & -1 & 1 & 1 & 0 & 0 \\ 
0 & 1 & -1 & -1 & 0 & 1 \\ 
0 & 0 & 1 & 2 & 1 & 0%
\end{array}%
\right] \underrightarrow{%
\begin{array}{c}
\quad R_{1}\rightarrow R_{1}-R_{3} \\ 
R_{2}\rightarrow R_{2}+R_{3}\quad%
\end{array}%
}\left[ 
\begin{array}{ccc|ccc}
1 & -1 & 0 & -1 & -1 & 0 \\ 
0 & 1 & 0 & 1 & 1 & 1 \\ 
0 & 0 & 1 & 2 & 1 & 0%
\end{array}%
\right] \\
\underrightarrow{\quad R_{1}\rightarrow R_{1}+R_{2}} && \left[ 
\begin{array}{ccc|ccc}
1 & 0 & 0 & 0 & 0 & 1 \\ 
0 & 1 & 0 & 1 & 1 & 1 \\ 
0 & 0 & 1 & 2 & 1 & 0%
\end{array}%
\right]
\end{eqnarray*}%
Since $R.R.E.F.\left( \left[ \mathbf{A}|\mathbf{I}\right] \right) $ has the
identity matrix on its left hand side, its right hand side must be $\mathbf{A%
}^{-1}$. \ Thus%
\begin{equation*}
\mathbf{A}^{-1}=\left[ 
\begin{array}{|ccc}
0 & 0 & 1 \\ 
1 & 1 & 1 \\ 
2 & 1 & 0%
\end{array}%
\right]
\end{equation*}
\end{itemize}

\end{document}
