<s>[INST] 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?
 [/INST] 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: $x=1$

</s>