Answer
Solution set: $\{(2,-1,1)\}$
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 Os below the $1\mathrm{s}$ (row-echelon form).
$\left[\begin{array}{lllll}
2 & -1 & -1 & | & 4\\
1 & 1 & -5 & | & -4\\
1 & -2 & 0 & | & 4
\end{array}\right]\left\{\begin{array}{l}
R1\leftrightarrow R2.\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -5 & | & -4\\
2 & -1 & -1 & | & 4\\
1 & -2 & 0 & | & 4
\end{array}\right]\left\{\begin{array}{l}
.\\
-2R1.+R2\rightarrow R2\\
-R1+R3\rightarrow R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -5 & | & -4\\
0 & -3 & 9 & | & 12\\
0 & -3 & 5 & | & 8
\end{array}\right]\left\{\begin{array}{l}
.\\
\div(-3)\\
-R2+R3.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 1 & -5 & | & -4\\
0 & 1 & -3 & | & -4\\
0 & 0 & -4 & | & -4
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow z=1
\end{array}\right.$
Back substitute $z=1$ into row/equation 2,
$y-3(1)=-4$
$y=-4+3=-1$
Back substitute into row/equation 1:
$x+(-1)-5(1)=-4$
$x=-4+6=2$
Solution set: $\{(2,-1,1)\}$