Answer
$\{(1,-1,2)\}$.
Work Step by Step
The given system of equations is
$x+y-z=-2$
$2x-y+z=5$
$-x+2y+2z=1$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & -1& -2\\
2 & -1 & 1& 5 \\
-1&2&2&1
\end{array}\right]$
Perform $R_2\rightarrow R_2-2\times R_1$ and $R_3\rightarrow R_3+ R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & -1& -2\\
2-2\times (1) & -1-2\times (1) & 1-2\times (-1)& 5-2\times (-2) \\
-1+1&2+1&2-1&1-2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & -1& -2\\
0 & -3 & 3& 9 \\
0&3&1&-1
\end{array}\right]$
Perform $R_2\rightarrow \frac{R_2}{-3}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & -1& -2\\
0 & -3/(-3) & 3/(-3)& 9/(-3) \\
0&3&1&-1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & -1& -2\\
0 & 1 & -1& -3 \\
0&3&1&-1
\end{array}\right]$
Perform $R_3\rightarrow R_3-3\times R_2$ and $R_1\rightarrow R_1- R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-0 & 1-1 & -1-(-1)& -2-(-3)\\
0 & 1 & -1& -3 \\
0-3\times (0)&3-3\times (1)&1-3\times (-1)&-1-3\times (-3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0 & 1 & -1& -3 \\
0&0&4&8
\end{array}\right]$
Perform $R_3\rightarrow \frac{R_3}{4}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0 & 1 & -1& -3 \\
0/(4)&0/(4)&4/(4)&8/(4)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0 & 1 & -1& -3 \\
0&0&1&2
\end{array}\right]$
Perform $R_2\rightarrow R_2+R_3$
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0+0 & 1+0 & -1+1& -3+2 \\
0&0&1&2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0 & 1 & 0& -1 \\
0&0&1&2
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=1$
and
$\Rightarrow y=-1$.
and
$\Rightarrow z=2$.
The solution set is $\{(x,y,z)\}=\{(1,-1,2)\}$.
