###### Itema

Because the leading entry in the first row is not a $$1\text{,}$$ this is not in reduced row echelon form. If we scale the first row by $$1/2\text{,}$$ however, we have a matrix in reduced row echelon form.

\begin{equation*} \left[ \begin{array}{rrr|r} 2 \amp 0 \amp 4 \amp -8 \\ 0 \amp 1 \amp 3 \amp 2 \\ \end{array} \right] \sim \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp -4 \\ 0 \amp 1 \amp 3 \amp 2 \\ \end{array} \right]. \end{equation*}

We may write the corresponding linear system as

\begin{equation*} \begin{alignedat}{4} x_1 \amp \amp \amp {}+{} \amp 2x_3 \amp {}={} \amp -4 \\ \amp \amp x_2 \amp {}+{} \amp 3x_3 \amp {}={} \amp 2, \\ \end{alignedat} \end{equation*}

which may be rewritten as

\begin{equation*} \begin{alignedat}{2} x_1 \amp {}={} -8 - 4x_3 \\ x_2 \amp {}={} 2 - 3x_3. \\ \end{alignedat} \end{equation*}

Since $$x_3$$ may take on any value, this shows that there are infinitely many solutions.

in-context