Answer
The complete solution is $\left\{ 2,\ \frac{z}{2}-1,\ z \right\}$.
Work Step by Step
We follow the following steps.
$\begin{align}
& 3x+2y-z=5 \\
& x+2y-z=1
\end{align}$
Now converting it into an augmented matrix, we get,
$\left[ \left. \begin{align}
& \begin{matrix}
3 & 2 & -1 \\
\end{matrix} \\
& \begin{matrix}
1 & 2 & -1 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
5 \\
1 \\
\end{matrix} \right]$
$\left[ \left. \begin{align}
& \begin{matrix}
3 & 2 & -1 \\
\end{matrix} \\
& \begin{matrix}
0 & 4 & -2 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
5 \\
-2 \\
\end{matrix} \right]$ $3{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$
Now we will express this matrix into equations as below:.
$\begin{align}
& 3x+2y-z=5 \\
& 4y-2z=-2
\end{align}$
Further we will solve the equations for values of $ x,y $ and $ z $.
$\begin{align}
& x=2 \\
& y=\frac{z}{2}-1 \\
& z=z
\end{align}$
Hence, the solution is:
$\left\{ 2,\ \frac{z}{2}-1,\ z \right\}$
