Item a

Our aim is to apply a sequence of scaling, interchange, and replacement operations to first put the system into a triangular form. We begin by multiplying the first equation by \(-2\) and adding it to the second equation. Next, we add the first equation to the third. This leads us to:

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp y \amp {}+{} \amp 2z \amp {}={} \amp 1 \\ \amp \amp -3y \amp {}-{} \amp 6z \amp {}={} \amp 0 \\ \amp \amp 2y \amp {}+{} \amp 3z \amp {}={} \amp 1 \\ \end{alignedat} \end{equation*}

We will now apply a scaling operation to make the coefficients of \(y\) equal \(1\) in the second equation.

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp y \amp {}+{} \amp 2z \amp {}={} \amp 1 \\ \amp \amp y \amp {}+{} \amp 2z \amp {}={} \amp 0 \\ \amp \amp 2y \amp {}+{} \amp 3z \amp {}={} \amp 1 \\ \end{alignedat} \end{equation*}

Another replacement operation brings the system into a triangular form.

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp y \amp {}+{} \amp 2z \amp {}={} \amp 1 \\ \amp \amp y \amp {}+{} \amp 2z \amp {}={} \amp 0 \\ \amp \amp \amp \amp -z \amp {}={} \amp 1 \\ \end{alignedat} \end{equation*}

From here, we begin the process of back substitution seeking a decoupled system.

\begin{equation*} \begin{alignedat}{4} x \amp {}+{} \amp y \amp \amp \amp {}={} \amp 3 \\ \amp \amp y \amp \amp \amp {}={} \amp 2 \\ \amp \amp \amp \amp z \amp {}={} \amp -1 \\ \end{alignedat} \end{equation*}

Finally, we have the decopled system

\begin{equation*} \begin{alignedat}{4} x \amp \amp \amp \amp \amp {}={} \amp 1 \\ \amp \amp y \amp \amp \amp {}={} \amp 2 \\ \amp \amp \amp \amp z \amp {}={} \amp -1 \\ \end{alignedat} \end{equation*}

From this, we see that the solution space consists of the single solution \((1,2,-1)\text{.}\)