Answer
$(-1,5,0)$
Work Step by Step
Start with the augmented matrix, row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left[\begin{array}{llll}
1 & 2 & -1 & 9\\
2 & 0 & -1 & -2\\
3 & 5 & 2 & 22
\end{array}\right]\rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftarrow R_{2}-R_{1}.\\
R_{3}\leftarrow R_{3}-3R_{1}
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 2 & -1 & 9 \\
0 & -4 & 1 & -20 \\
0 & -1 & 5 & -5
\end{array}\right] \rightarrow\left(\begin{array}{l}
.\\
R_{2}\leftrightarrow R_{3}.\\
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 2 & -1 & 9 \\
0 & -1 & 5 & -5 \\
0 & -4 & 1 & -20
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}+2R_{2}.\\
R_{2}\leftarrow-R_{2}.\\
R_{3}\leftarrow R_{3}-4R_{2}
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & 9 & -1 \\
0 & 1 & -5 & 5 \\
0 & 0 & -19 & 0 \end{array}\right] \rightarrow
\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow R_{3}/(-19)
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & 9 & -1 \\
0 & 1 & -5 & 5 \\
0 & 0 & 1 & 0
\end{array}\right] \rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-9R_{3}.\\
R_{2}\leftarrow R_{2}+5R_{3}.\\
.
\end{array}\right)$
$\rightarrow \left[\begin{array}{cccc}
1 & 0 & 0 & -1 \\
0 & 1 & 0 & 5 \\
0 & 0 & 1 & 0
\end{array}\right]$
The solution is $(-1,5,0)$
