<s>[INST] What's wrong with how I calculated the inverse of this matrix? $\left( \begin{array}{ccc}
-1 & -1 & 2 & | & 1 & 0 & 0\\
2 & 0 & 0 & | & 0 & 1 & 0\\
2 & 2 & 0 & | & 0 & 0 & 1\end{array} \right) $ is the original matrix. Now, add 2*R1 to R2 and R3.
$\left( \begin{array}{ccc}
-1 & -1 & 2 & | & 1 & 0 & 0\\
0 & -2 & 4 & | & 2 & 1 & 0\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, add -R3 to R2.
$\left( \begin{array}{ccc}
-1 & -1 & 2 & | & 1 & 0 & 0\\
0 & -2 & 0 & | & 0 & 1 & -1\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, add -.5*R3 to R1.
$\left( \begin{array}{ccc}
-1 & -1 & 0 & | & 0 & 0 & -.5\\
0 & -2 & 0 & | & 2 & 1 & 0\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, add -.5*R2 to R1.
$\left( \begin{array}{ccc}
-1 & 0 & 0 & | & -1 & -.5 & -.5\\
0 & -2 & 0 & | & 2 & 1 & 0\\
0 & 0 & 4 & | & 2 & 0 & 1\end{array} \right) $ Now, simple division yields
$\left( \begin{array}{ccc}
1 & 0 & 0 & | & 1 & .5 & .5\\
0 & 1 & 0 & | & -1 & -.5 & 0\\
0 & 0 & 1 & | & .5 & 0 & .25\end{array} \right) $ 
 [/INST] It seems you made a small computational error in rewriting the right hand inverse matrix after step 3. You gotta be really careful about that,it's like a complicated integration-it can really shaft you on an exam or important homework problem. If a row reduction to yield an inverse matrix looks wrong to you but you're not sure if you made a mistake, you can always multiply the matrix by the obtained inverse. If you made a mistake, you won't get the identity matrix and row where the mistake is is where you won't have a 1 or a 0 usually. Even better,if you've got access to a computer algebra system like Mathematica, that'll make checking really quick and simple. 
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