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\begin{document}

\begin{center}
\textbf{Math 3013} \\[0pt]
SOLUSTIONS TO FIRST EXAM \\[0pt]
9:30 - 10:20, February 18, 2022
\end{center}

1. Let 
\begin{equation*}
\quad \mathbf{A}=\left[ 
\begin{array}{cc}
2 & 0 \\ 
1 & -1%
\end{array}%
\right] \quad \mathbf{B}=\left[ 
\begin{array}{cc}
1 & 0 \\ 
1 & -1 \\ 
2 & 1%
\end{array}%
\right] 
\end{equation*}%
Compute the matrix products $\mathbf{AB}$ and $\mathbf{BA}$ (if they exist).

\begin{itemize}
\item 
\begin{equation*}
\mathbf{AB}=\left[ 
\begin{array}{cc}
2 & 0 \\ 
1 & -1%
\end{array}%
\right] \left[ 
\begin{array}{cc}
1 & 0 \\ 
1 & -1 \\ 
2 & 1%
\end{array}%
\right] \text{ is undefined since }\#Columns\left( \mathbf{A}\right) \neq
\#Rows\left( \mathbf{B}\right) 
\end{equation*}%
\begin{equation*}
\mathbf{BA}=\left[ 
\begin{array}{cc}
1 & 0 \\ 
1 & -1 \\ 
2 & 1%
\end{array}%
\right] \left[ 
\begin{array}{cc}
2 & 0 \\ 
1 & -1%
\end{array}%
\right] =\left[ 
\begin{array}{cc}
2+0 & 0+0 \\ 
2-1 & 0+1 \\ 
4+1 & 0-1%
\end{array}%
\right] =\left[ 
\begin{array}{cc}
2 & 0 \\ 
1 & 1 \\ 
5 & -1%
\end{array}%
\right] 
\end{equation*}
\end{itemize}

2. For each of the following augmented matrices, indicate

\begin{itemize}
\item[(i)] the number of equations and the number of variables in the original
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 & 0 & 0 & 1 & 2 \\ 
0 & 0 & 0 & 0 & 1%
\end{array}%
\right] $

\begin{itemize}
\item[(i)] 3 equations in 4 unknowns

\item[(ii)] no solution (last row implies $0=1$ which indicates there was a
contradiction amongst the original equations)
\end{itemize}

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

\begin{itemize}
\item[(i)] 3 equations in 4 unknowns
\item[(ii)] there is a solution
\item[(iii)] Column 1 of the REF lacks a pivot while the other columns on the left
have pivots. This implies there is 1 free variable ($x_{1}$) in the solution.
\end{itemize}

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

\begin{itemize}
\item[(i)] 4 equations in 5 unknowns
\item[(ii)] there is a solution
\item[(iii)] Columns 3 and 5 on the left lack pivots. \ This implies there will be
two free variables ($x_{3}$ and $x_{5}$) in the general solution.
\end{itemize}

\newpage

3. Consider the following linear system 
\begin{equation*}
\begin{array}{rcr}
-x_{2}-3x_{3} & = & -5 \\ 
2x_{1}+x_{2}+x_{3} & = & 1 \\ 
x_{1}+x_{2}+2x_{3} & = & 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}{ccc|c}
0 & -1 & -3 & -5 \\ 
2 & 1 & 1 & 1 \\ 
1 & 1 & 2 & 3%
\end{array}%
\right] \\
&&\underrightarrow{R_{1}\longleftrightarrow R_{3}}\left[ 
\begin{array}{ccc|c}
1 & 1 & 2 & 3 \\ 
2 & 1 & 1 & 1 \\ 
0 & -1 & -3 & -5%
\end{array}%
\right] \\
&&\underrightarrow{R_{2}\longleftrightarrow R_{2}-2R_{1}}\left[ 
\begin{array}{ccc|c}
1 & 1 & 2 & 3 \\ 
0 & -1 & -3 & -5 \\ 
0 & -1 & -3 & -5%
\end{array}%
\right] \\
&&\underrightarrow{R_{3}\longleftrightarrow R_{3}-R_{2}}\left[ 
\begin{array}{ccc|c}
1 & 1 & 2 & 3 \\ 
0 & -1 & -3 & -5 \\ 
0 & 0 & 0 & 0%
\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 & 4 & 6 & 2 & 2 \\ 
0 & 0 & -1 & 2 & 2 & 3 \\ 
0 & 0 & 0 & 0 & -2 & 2 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right]
\end{equation*}%
, row echelon form: $\left[ 
\begin{array}{cccccc}
1 & 2 & 0 & 14 & 0 & 24 \\ 
0 & 0 & 1 & -2 & 0 & -5 \\ 
0 & 0 & 0 & 0 & 1 & -1 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right] $

\begin{itemize}
\item 
\begin{eqnarray*}
&&\underrightarrow{%
\begin{array}{c}
R_{2}\rightarrow -R_{2} \\ 
R_{3}\rightarrow -\frac{1}{2}R_{3}%
\end{array}%
}\left[ 
\begin{array}{ccccc|c}
1 & 2 & 4 & 6 & 2 & 2 \\ 
0 & 0 & 1 & -2 & -2 & -3 \\ 
0 & 0 & 0 & 0 & 1 & -1 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right] \\
&&\underrightarrow{%
\begin{array}{c}
R_{1}\rightarrow R_{1}-2R_{3} \\ 
R_{2}\rightarrow R_{2}+2R_{3}%
\end{array}%
}\left[ 
\begin{array}{ccccc|c}
1 & 2 & 4 & 6 & 0 & 4 \\ 
0 & 0 & 1 & -2 & 0 & -5 \\ 
0 & 0 & 0 & 0 & 1 & -1 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right] \\
&&\underrightarrow{R_{1}\rightarrow R_{1}-4R_{2}}\left[ 
\begin{array}{ccccc|c}
1 & 2 & 0 & 14 & 0 & 24 \\ 
0 & 0 & 1 & -2 & 0 & -5 \\ 
0 & 0 & 0 & 0 & 1 & -1 \\ 
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 & 1 & 0 & -2 & 1 & 1 \\ 
0 & 0 & 1 & -1 & -1 & 2 \\ 
0 & 0 & 0 & 0 & 0 & 0%
\end{array}%
\right]
\end{equation*}

\begin{itemize}
\item There are no pivots in columns 2, 4 or 5. This means that $x_{2}$, $%
x_{4}$, and $x_{5}$ will be free variables in the solution. \ Writing down
the equations corresponding to this augmented matrix in R.R.E.F., and then
solving for the \textquotedblleft fixed variables\textquotedblright\ $x_{1}$
and $x_{3}$ yields%
\begin{equation*}
\left. 
\begin{array}{r}
x_{1}+x_{2}-2x_{4}+x_{5}=1 \\ 
x_{3}-x_{4}-x_{5}=2 \\ 
0=0%
\end{array}%
\right\} \quad \Rightarrow \quad \left\{ 
\begin{array}{c}
x_{1}=1-x_{2}+2x_{4}-x_{5} \\ 
x_{3}=2+x_{4}+x_{5}%
\end{array}%
\right.
\end{equation*}%
Next, we try writing down a typical solution vector, substituting where we
can for the fixed variables, and then expanding the result in terms of the
free variables yields:%
\begin{equation*}
\mathbf{x}=\left[ 
\begin{array}{l}
1-x_{2}+2x_{4}-x_{5} \\ 
x_{2} \\ 
2+x_{4}+x_{5} \\ 
x_{4} \\ 
x_{5}%
\end{array}%
\right] =\left[ 
\begin{array}{l}
1 \\ 
0 \\ 
2 \\ 
0 \\ 
0%
\end{array}%
\right] +x_{2}\left[ 
\begin{array}{l}
-1 \\ 
1 \\ 
0 \\ 
0 \\ 
0%
\end{array}%
\right] +x_{4}\left[ 
\begin{array}{l}
2 \\ 
0 \\ 
1 \\ 
1 \\ 
0%
\end{array}%
\right] +x_{5}\left[ 
\begin{array}{l}
-1 \\ 
0 \\ 
1 \\ 
0 \\ 
1%
\end{array}%
\right]
\end{equation*}%
The last expression on the right exhibits the solution as a $3$-dimensional
hyperplane in $\mathbb{R}^{5}$.
\end{itemize}

6. Determine if the following matrix has an inverse. \ If the matrix does
have an inverse, determine its inverse. (Hint: a single row reduction
process can determine the answer to both parts of this problem.) 
\begin{equation*}
\mathbf{A}=\left[ 
\begin{array}{ccc}
0 & 1 & 1 \\ 
1 & 2 & 1 \\ 
2 & 3 & 1%
\end{array}%
\right]
\end{equation*}

\begin{itemize}
\item We will form the adjoined matrix $\left[ \mathbf{A}\mid \mathbf{I}%
\right] $ and then row reduce it to its Reduced Row Echelon Form:%
\begin{eqnarray*}
\left[ \mathbf{A}\mid \mathbf{I}\right] &=&\left[ 
\begin{array}{ccc|ccc}
0 & 1 & 1 & 1 & 0 & 0 \\ 
1 & 2 & 1 & 0 & 1 & 0 \\ 
2 & 3 & 1 & 0 & 0 & 1%
\end{array}%
\right] \\
&&\underrightarrow{R_{1}\longleftrightarrow R_{2}}\left[ 
\begin{array}{ccc|ccc}
1 & 2 & 1 & 0 & 1 & 0 \\ 
0 & 1 & 1 & 1 & 0 & 0 \\ 
2 & 3 & 1 & 0 & 0 & 1%
\end{array}%
\right] \\
&&\underrightarrow{R_{3}\longleftrightarrow R_{3}-2R_{1}}\left[ 
\begin{array}{ccc|ccc}
1 & 2 & 1 & 0 & 1 & 0 \\ 
0 & 1 & 1 & 1 & 0 & 0 \\ 
0 & -1 & -1 & 0 & -2 & 1%
\end{array}%
\right] \\
&&\underrightarrow{R_{3}\longleftrightarrow R_{3}+R_{2}}\left[ 
\begin{array}{ccc|ccc}
1 & 2 & 1 & 0 & 1 & 0 \\ 
0 & 1 & 1 & 1 & 0 & 0 \\ 
0 & 0 & 0 & 1 & -2 & 1%
\end{array}%
\right]
\end{eqnarray*}%
The zero row on the left hand side (the $\mathbf{A}$-part) of the adjoined
matrix shows that $\mathbf{A}$ cannot be row-reduced to the identity matrix.
\ Hence, the matrix $\mathbf{A}$ has no inverse.
\end{itemize}

\end{document}
