#### Answer

The solution to the system is
$$x_1=\frac{1}{3}\qquad x_2=1\qquad x_3=-\frac{1}{3}.$$

#### Work Step by Step

The system associated to this matrix is:
\begin{align*}
2x_1+x_2+x_3=&0\\
x_1-2x_2+x_3=&-2\\
x_1\qquad+x_3=&0
\end{align*}
To solve it follow the steps below:
Step 1: Subtract from the first equation the second one multiplied by $2$ to eliminate $x_1$:
$$2x_1-2x_1+x_2-2(-2x_2) + x_3-2x_3 = 0-2(-2)$$
which becomes
$$5x_2+3x_3=4$$
Now our system reads
\begin{align*}
5x_2+3x_3=&4\\
x_1-2x_2+x_3=&-2\\
x_1\qquad+x_3=0
\end{align*}
Step 2: Subtract from the second equation the third one to eliminate $x_1$ (and $x_3$ for free):
$$x_1-x_1-2x_2+x_3-x_3 = -2-0$$
which becomes $$-2x_2=-2$$ and this gives $$x_2=1.$$
Step 3: Put this into the first equation to find $x_3$:
$$5\times 1+ 3x_3=4\Rightarrow 3x_3 = -1$$
and this gives $$x_3=-\frac{1}{3}$$
Step 4: Put $x_2=1$ and $x_3=-\frac{1}{3}$ into the second equation and find $x_1$:
$$x_1-2\times1-\frac{1}{3} = -2\Rightarrow x_1-\frac{7}{3}=-2$$ and this will give
$$x_1 = \frac{1}{3}.$$