Answer
$\{(-1,2,-2)\}$.
Work Step by Step
The given system of equations is
$x+y=1$
$y+2z=-2$
$2x-z=0$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 0& 1\\
0 & 1 & 2& -2 \\
2&0&-1&0
\end{array}\right]$
Perform $R_3\rightarrow R_3-2 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 0& 1\\
0 & 1 & 2& -2 \\
2-2(1)&0-2(1)&-1-2(0)&0-2(1)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 0& 1\\
0 & 1 & 2& -2 \\
0&-2&-1&-2
\end{array}\right]$
Perform $R_1\rightarrow R_1- R_2$ and $R_3\rightarrow R_3+2 R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-0 & 1-1 & 0-2& 1-(-2)\\
0 & 1 & 2& -2 \\
0+2(0)&-2+2(1)&-1+2(2)&-2+2(-2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -2& 3\\
0 & 1 & 2& -2 \\
0&0&3&-6
\end{array}\right]$
Perform $R_3\rightarrow \frac{R_3}{3}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -2& 3\\
0 & 1 & 2& -2 \\
0/3&0/3&3/3&-6/3
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -2& 3\\
0 & 1 & 2& -2 \\
0&0&1&-2
\end{array}\right]$
Perform $R_1\rightarrow R_1+2 R_3$ and $R_2\rightarrow R_2-2 R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+2(0) & 0+2(0) & -2+2(1) & 3+2(-2) \\
0 -2(0)& 1-2(0) & 2-2(1)& -2-2(-2) \\
0&0&1&-2
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0 & -1 \\
0 & 1 & 0& 2 \\
0&0&1&-2
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-1$
and
$\Rightarrow y=2$.
and
$\Rightarrow z=-2$.
The solution set is $\{(x,y,z)\}=\{(-1,2,-2)\}$.
