###### Item a

Because the leading entry in the first row is not a \(1\text{,}\) this is not in reduced row echelon form. If we scale the first row by \(1/2\text{,}\) however, we have a matrix in reduced row echelon form.

\begin{equation*}
\left[
\begin{array}{rrr|r}
2 \amp 0 \amp 4 \amp -8 \\
0 \amp 1 \amp 3 \amp 2 \\
\end{array}
\right]
\sim
\left[
\begin{array}{rrr|r}
1 \amp 0 \amp 2 \amp -4 \\
0 \amp 1 \amp 3 \amp 2 \\
\end{array}
\right].
\end{equation*}

We may write the corresponding linear system as

\begin{equation*}
\begin{alignedat}{4}
x_1 \amp \amp \amp {}+{} \amp 2x_3 \amp {}={} \amp -4 \\
\amp \amp x_2 \amp {}+{} \amp 3x_3 \amp {}={} \amp 2, \\
\end{alignedat}
\end{equation*}

which may be rewritten as

\begin{equation*}
\begin{alignedat}{2}
x_1 \amp {}={} -8 - 4x_3 \\
x_2 \amp {}={} 2 - 3x_3. \\
\end{alignedat}
\end{equation*}

Since \(x_3\) may take on any value, this shows that there are infinitely many solutions.