Answer
The solution of the system is $\left\{ \frac{z}{3}+1,\ \frac{z}{3},z \right\}$.
Work Step by Step
In this given linear equations, we have to find out the solution; for that we have to follow the following steps.
$\begin{align}
2x+y-z=2 & \\
3x+3y-2z=3 & \\
\end{align}$
Now, converting it into an augmented matrix, we get,
$\left[ \left. \begin{align}
& \begin{matrix}
2 & 1 & -1 \\
\end{matrix} \\
& \begin{matrix}
3 & 3 & -2 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
2 \\
3 \\
\end{matrix} \right]$
Use $2{{R}_{2}}-3{{R}_{1}}\to {{R}_{2}}$
Therefore, $\left[ \left. \begin{align}
& \begin{matrix}
6 & 3 & -3 \\
\end{matrix} \\
& \begin{matrix}
0 & 3 & -1 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
6 \\
0 \\
\end{matrix} \right]$
Now we will express this matrix into equations as below:.
$\begin{align}
6x+3y-3z=6 & \\
3y-z=0 & \\
\end{align}$
Further we will solve the equations for values of $ x, y $ and $ z $.
$\begin{align}
& x=\frac{z}{3}+1 \\
& y=\frac{z}{3} \\
& z=z
\end{align}$
Hence, the solution is, $\left\{ \frac{z}{3}+1,\ \frac{z}{3},z \right\}$.
