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