Answer
Solution set: $\{(1,-1,2)\}$
Work Step by Step
Begin with the system's augmented matrix.
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & -2\\
2 & -1 & 1 & | & 5\\
-1 & 2 & 2 & | & 1
\end{array}\right]$
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).
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & -2\\
2 & -1 & 1 & | & 5\\
-1 & 2 & 2 & | & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
-2R1+R2\rightarrow R2\\
R1+R3\rightarrow R3
\end{array}\right\}$
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & -2\\
0 & -3 & 3 & | & 9\\
0 & 3 & 1 & | & -1
\end{array}\right]\left\{\begin{array}{l}
.\\
\times(-\frac{1}{3})\\
R2+R3\rightarrow R3
\end{array}\right\}$
$\left[\begin{array}{lllll}
1 & 1 & -1 & | & -2\\
0 & 1 & -1 & | & -3\\
0 & 0 & 4 & | & 8
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\Rightarrow 4z=8
\end{array}\right\}$
$z=2$
Back substitute $z=2$ into row 2,
$y-z=-3$
$y=-3+2=-1$
Back substitute into equation 1:
$x+(-1)-(2)=-2$
$x=-2+3=1$
Solution set: $\{(1,-1,2)\}$
