###### Item 2

Consider the matrix

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

The last equation gives

\begin{equation*}
0x +0y+0z = 0\text{,}
\end{equation*}

which is true for any \((x,y,z)\text{.}\) We may safely ignore this equation since it does not provide a restriction on the choice of \((x,y,z)\text{.}\) We then see that there is a unique solution \((x,y,z) = (4,-3,1)\text{.}\)