Item 1.2.5.2.c

This is not in reduced row echelon form since the leading entry in the last row is not \(1\) and is not the only nonzero entry in its column.

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

This linear system has a single solution.

in-context