How would I reduce my matrix even further? How would I reduce my matrix even further? 

 A: I usually do also pivot reduction:
\begin{align}
\large
\begin{bmatrix}
2 & 2 & 1 & 2\\
-1 & 2 & -1 & -5\\
1 & -3 & 2 & 8
\end{bmatrix}
\xrightarrow{E_1(1/2)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
-1 & 2 & -1 & -5\\
1 & -3 & 2 & 8
\end{bmatrix}
\\
\large
\xrightarrow{E_{21}(1)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 3 & -1/2 & -4\\
1 & -3 & 2 & 8
\end{bmatrix}
\\
\large
\xrightarrow{E_{31}(-1)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 3 & -1/2 & -4\\
0 & -4 & 3/2 & 7
\end{bmatrix}
\\
\large
\xrightarrow{E_{2}(1/3)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & -1/6 & -4/3\\
0 & -4 & 3/2 & 7
\end{bmatrix}
\\
\large
\xrightarrow{E_{32}(4)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & -1/6 & -4/3\\
0 & 0 & 5/6 & 5/3
\end{bmatrix}
\\
\large
\xrightarrow{E_{3}(6/5)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & -1/6 & -4/3\\
0 & 0 & 1 & 2
\end{bmatrix}
\\
\text{Backwards elimination starts}\\
\large
\xrightarrow{E_{23}(1/6)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2
\end{bmatrix}
\\
\large
\xrightarrow{E_{13}(-1/2)}
&\large
\begin{bmatrix}
1 & 1 & 0 & 0\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2
\end{bmatrix}
\\
\large
\xrightarrow{E_{12}(-1)}
&\large
\begin{bmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2
\end{bmatrix}
\end{align}
In this way you can read directly the solution in the last column:
$$
x=1,\quad y=-1,\quad z=2.
$$
Notation.


*

*$E_{i}(c)$ (with $c\ne0$) is “multiply the $i$-th row by $c$”.

*$E_{ij}(d)$ (with $i\ne j$) is “sum to the $i$-th row the $j$-th row multiplied by $d$”.
A: You can't reduce it any further. It is already in RREF.
HINT: What are the conditions for RREF?
