Gauss elimination. Where did I go wrong? 
Gaussian elimination with back sub:
So my starting matrix:
\begin{bmatrix} 
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2 
\end{bmatrix}
multiply the 2nd and 3rd row by -1 * (first row):
\begin{bmatrix} 
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & 2 & 0 & 4 
\end{bmatrix}
then add -1(third row) to the 2nd row-> 
\begin{bmatrix} 
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 2 & 0 & 4 
\end{bmatrix}
add -2(2nd row) to the third row -> 
\begin{bmatrix} 
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 0 & 10 & 0 
\end{bmatrix}
But then this seems to have no solution because $10z = 0$.... ugh
EDIT 
As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?
 A: Yes, your answer is correct. 
Check that $(1,2,0)$ is a solution and also since the rank is $3$, there is a unique solution.
$10z=0 \implies z=0$, substitute that to other equations, we easily get $y=2$ and then $x-y+0 = -1 \implies x-2=-1 \implies x=1$.
A: I don't understand your way to obtain the RREF, we can proceed as follow
$$\begin{bmatrix} 
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2 
\end{bmatrix}\stackrel{R3-R2}\to \begin{bmatrix} 
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\0 & -1 & 5 & -2 
\end{bmatrix}\stackrel{R2-2\cdot R1}\to \begin{bmatrix} 
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & -1 & 5 & -2 
\end{bmatrix}\stackrel{3\cdot R3+R2}\to \begin{bmatrix} 
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & 0 & 10 & 0 
\end{bmatrix}$$
and since the matrix is full rank (we have three pivots) we have an unique solution that is


*

*from the third row: $z=0$

*from the second row: $y=2$

*from the first row: $x=1$

A: I'd use a more systematic method:
\begin{align}
\begin{bmatrix} 
1 & -1 & 1 & -1\\
2 & 1 & -3 & 4\\
2 & 0 & 2 & 2 
\end{bmatrix}
&\to
\begin{bmatrix} 
1 & -1 & 1 & -1\\
0 & 3 & -5 & 6\\
0 & 2 & 0 & 4 
\end{bmatrix}
&&\begin{aligned} R_2&\gets R_2-2R_1 \\ R_3&\gets R_3-2R_1 \end{aligned}
\\ &\to
\begin{bmatrix} 
1 & -1 & 1 & -1\\
0 & 1 & -5/3 & 2\\
0 & 2 & 0 & 4 
\end{bmatrix}
&& R_2\gets\tfrac{1}{3}R_2
\\ &\to
\begin{bmatrix} 
1 & -1 & 1 & -1\\
0 & 1 & -5/3 & 2\\
0 & 0 & 10/3 & 0 
\end{bmatrix}
&& R_3\gets R_3-2R_2
\\ &\to
\begin{bmatrix} 
1 & -1 & 1 & -1\\
0 & 1 & -5/3 & 2\\
0 & 0 & 1 & 0 
\end{bmatrix}
&& R_3\gets\tfrac{3}{10}R_3
\\ &\to
\begin{bmatrix} 
1 & -1 & 0 & -1\\
0 & 1 & 0 & 2\\
0 & 0 & 1 & 0 
\end{bmatrix}
&& \begin{aligned} R_2 &\gets R_2+\tfrac{5}{3}R_3 \\ R_1&\gets R_1-R_3\end{aligned}
\\ &\to
\begin{bmatrix} 
1 & 0 & 0 & 1\\
0 & 1 & 0 & 2\\
0 & 0 & 1 & 0 
\end{bmatrix}
&& R_1\gets R_1+R_2
\end{align}
The solution, which is explicit when the RREF is reached, is
\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}
