Answer
$\{(-2,-1,0)\}$.
Work Step by Step
The given system of equations is
$x-2y+z=0$
$y-3z=-1$
$2y+5z=-2$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
1 & -2 & 1 &0\\
0 & 1 & -3 &-1 \\
0 & 2 & 5 &-2
\end{array}\right]$
Perform $R_1\rightarrow R_1+2 R_2$ and $R_3\rightarrow R_3-2 R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+2(0) & -2+2(1) & 1+2(-3) &0+2(-1)\\
0 & 1 & -3 &-1 \\
0-2(0) & 2-2(1) & 5-2(-3) &-2-2(-1)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -5 &-2\\
0 & 1 & -3 &-1 \\
0 & 0 & 11 &0
\end{array}\right]$
Perform $R_3\rightarrow R_3/(11)$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -5 &-2\\
0 & 1 & -3 &-1 \\
0/(11) & 0/(11) & 11/(11) &0/(11)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -5 &-2\\
0 & 1 & -3 &-1 \\
0 & 0 & 1 &0
\end{array}\right]$
Perform $R_1\rightarrow R_1+5 R_3$ and $R_2\rightarrow R_2+ 3R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+5(0) & 0+5(0) & -5+5(1) &-2+5(0)\\
0+3(0) & 1+3(0) & -3+3(1) &-1+3(0) \\
0 & 0 & 1 &0
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0 &-2\\
0 & 1 & 0 &-1 \\
0 & 0 & 1 &0
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=-2$
and
$\Rightarrow y=-1$.
and
$\Rightarrow z=0$.
The solution set is $\{(x,y,z)\}=\{(-2,-1,0)\}$.
