#### Answer

There are infinitely many solutions to this system and they are
$$x_1=1-2t\qquad x_2 = t\qquad x_3=-1$$

#### Work Step by Step

This matrix is associated with the following system:
\begin{align*}
x_1+2x_2+x_3=&0\\
x_3=&-1\\
\end{align*}
There is no third equation since the third row only gives the identity $0=0$
This is a system of two equations with three variables so it will have infinitely many solutions. To find them we have to set one of them to be equal to parameter $t$ that can take any real value. Note that we cannot set $x_3 = t$ because we have a precise condition $x_3=-1$, so we will put $x_2 = t.$
Now we get
\begin{align*}
x_1+2x_2+x_3=&0\\
x_2\qquad=&t\\
x_3=&-1\\
\end{align*}
Putting all of this into the firs equation we have
$$x_1+2t-1=0$$
and this gives
$$x_1=1-2t.$$