Answer
The complete solution is $\left\{ 2z-\frac{5}{4},\ \ \frac{13}{4},\ z \right\}$.
Work Step by Step
We start with:
$\begin{align}
& x+y-2z=2 \\
& 3x-y-6z=-7
\end{align}$
Now converting it into an augmented matrix, we get,
$\left[ \left. \begin{matrix}
\begin{matrix}
1 & 1 & -2 \\
\end{matrix} \\
\begin{matrix}
3 & -1 & -6 \\
\end{matrix} \\
\end{matrix} \right|\begin{matrix}
2 \\
-7 \\
\end{matrix} \right]$
$\left[ \left. \begin{matrix}
\begin{matrix}
1 & 1 & -2 \\
\end{matrix} \\
\begin{matrix}
0 & -4 & 0 \\
\end{matrix} \\
\end{matrix} \right|\begin{matrix}
2 \\
-13 \\
\end{matrix} \right]$ ${{R}_{2}}-3{{R}_{1}}\to {{R}_{2}}$
Now we will express this matrix into equations as below:.
$\begin{align}
& x+y-2z=5 \\
& -4y=-13
\end{align}$
Further we will solve the equations for values of $ x,y $ and $ z $.
$\begin{align}
& x=2z-\frac{5}{4} \\
& y=\frac{13}{4} \\
& z=z
\end{align}$
Hence, the complete solution is $\left\{ 2z-\frac{5}{4},\ \ \frac{13}{4},\ z \right\}$.
