Item2

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 \((x,y,z) = (4,-3,1)\text{.}\)

in-context