###### Item b

Once again, we begin with a sequence of replacement and scaling operations that lead to the triangular system

\begin{equation*}
\begin{alignedat}{4}
-x \amp {}-{} \amp 2y \amp {}+{} \amp 2z \amp
{}={} \amp -1 \\
\amp \amp \amp \amp z \amp {}={} \amp 1 \\
\amp \amp \amp \amp 0 \amp {}={} \amp 0 \\
\end{alignedat}
\end{equation*}

Back substituion gives us the system

\begin{equation*}
\begin{alignedat}{4}
x \amp {}+{} \amp 2y \amp \amp \amp
{}={} \amp 3 \\
\amp \amp \amp \amp z \amp {}={} \amp 1 \\
\amp \amp \amp \amp 0 \amp {}={} \amp 0 \\
\end{alignedat}
\end{equation*}

The third equation does not impose a restriction on the solutions since it is satisfied for any \((x,y,z)\text{.}\) The second equation tells us that \(z\) must equal \(1\text{;}\) however, there are infinitely many solutions to the first equation that have \(z=1\text{.}\) Therefore, this system has infinitely many solutions.