Answer
Solution set: $\{(2$ , $-1$ , $1 )\}$
Work Step by Step
Rewrite the system in standard form.
$\left\{\begin{array}{l}
x+2y-z=-1\\
x-y+z=4\\
x+y-3z=-2
\end{array}\right.$
Augmented matrix:
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & -1\\
1 & -1 & -1 & | & 4\\
1 & 1 & -3 & | & -2
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftarrow R_{2}-R_{1}\\
\leftarrow R_{3}-R_{1}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & -1\\
0 & 1 & 2 & | & 1\\
0 & -3 & 2 & | & 5
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\leftarrow R_{3}+3R_{2}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & -1\\
0 & 1 & 2 & | & 1\\
0 & 0 & 8 & | & 8
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\div 8
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & -1 & | & -1\\
0 & 1 & 2 & | & 1\\
0 & 0 & 1 & | & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow z=1
\end{array}\right.$
Back substitute into row/equation 2,
$y+2(1)=1$
$y=1-2=-1$
Back substitute into row/equation 1:
$x+2(-1)-(1)=-1$
$x=-1+3=2$
Solution set: $\{(2$ , $-1$ , $1 )\}$
