Answer
Solution set: $\{( 1$ , $3$ , $-4 )\}$
Work Step by Step
Begin with the system's augmented matrix.
Use matrix row operations to get ls down the main diagonal from upper left to lower right, and 0s below the $1\mathrm{s}$.
$\left[\begin{array}{lllll}
1 & 2 & 3 & | & -5\\
2 & 1 & 1 & | & 1\\
1 & 1 & -1 & | & 8
\end{array}\right]\left\{\begin{array}{l}
.\\
-2R_{1}+R_{2}\\
-R_{1}+R_{3}
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & 3 & | & -5\\
0 & -3 & -5 & | & 11\\
0 & -1 & -4 & | & 13
\end{array}\right]\left\{\begin{array}{l}
.\\
\leftrightarrow R_{3}\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & 3 & | & -5\\
0 & -1 & -4 & | & 13\\
0 & -3 & -5 & | & 11
\end{array}\right]\left\{\begin{array}{l}
.\\
(-1)R_{2}\\
-3R_{2}+R_{3}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & 3 & | & -5\\
0 & 1 & 4 & | & -13\\
0 & 0 & 7 & | & -28
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow z=-4.
\end{array}\right.$
Back substitute z into row/equation 2
$y+4z=-13$
$y-16=-13$
$y=-13+16$
$y=3$
Back substitute y and z into row/equation 1
$x+2y+3z=-5$
$x+6-12=-5$
$x=-5+6$
$x=1$
Solution set: $\{( 1$ , $3$ , $-4 )\}$
