##### Item1.2.5.2.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.