#### Answer

The solution is
$$x_1=1\qquad x_2=1\qquad x_3=0$$

#### Work Step by Step

The system associated with this matrix is
\begin{align*}
2x_1+x_2-x_3=&3\\
x_1-x_2+x_3=&0\\
x_2+2x_3=&1
\end{align*}
To solve it follow the steps below:
Step 1: Subtract from the first equation the second one multiplied by $2$:
$$2x_1-2x_1+x_2-2(-x_2)-x_3-2x_3=3-2\times 0$$ which becomes
$$3x_2-3x_3=3.$$
Divide this by $3$ for better looks
$$x_2-x_3=1.$$
Now we have a system
\begin{align*}
x_2-x_3=&1\\
x_1-x_2+x_3=&0\\
x_2+2x_3=&1.
\end{align*}
Step 2: Subtract from the third equation the first one:
$$x_2-x_2+2x_3-(-x_3) = 1-1$$
which becomes
$$3x_3=0$$ and this gives $$x_3 =0$$
Step 3: Put this into the first equation and find $x_1$:
$$x_2-0=1$$ which gives $$x_2=1.$$
Step 4: Put $x_3=0$ and $x_2=1$ into the second equation to find $x_1$:
$$x_1-1+0=0$$ and this gives
$$x_1 =1.$$