<s>[INST] Finding inverse of a $3\times3$ matrix I am given a $3 \times3$ matrix and am asked to find the inverse using elementary row operations. I know how they work, but have no idea of which steps to apply first, followed by which steps.
First, the matrices:
$$\begin{pmatrix}
1 & 1 & -3\\
2 & 1 & -3\\
2 & 2 & 1
\end{pmatrix}$$
All I know thus far is that, if there is a series of operations (pre-multipliers)
$E_nE_{n-1}...E_2E_1A$ that reduces to the identity matrix, the same sequence $ E_nE_{n-1}...E_2E_1I$ reduces to the inverse of $A$, $A^{-1}$.
Any help? If not, I will use another method already because this is not working thus far.
UPDATE
Thanks to the community, I got the final answer:
$$\begin{pmatrix}
-1 & 1 & 0\\
\frac8 7 & -1 & \frac 3 7\\
\frac{-2}{7} & 0 & \frac 1 7
\end{pmatrix}$$
 [/INST] I was taught to augment the matrix with the identity, then apply the row operations:
$$\begin{pmatrix}
1 & 1 & -3 & 1 & 0 & 0\\
2 & 1 & -3&0&1&0\\
2 & 2 & 1&0&0&1
\end{pmatrix}$$
Subtract twice row 1 from row 2 and twice row 1 from row 3 (yes, this is two operations)
$$\begin{pmatrix}
1 & 1 & -3 & 1 & 0 & 0\\
0 & -1 & 3&-2&1&0\\
0 & 0 & 7&-2&0&1
\end{pmatrix}$$
Multiply row 2 by -1 and row 3 by $\frac 17$
$$\begin{pmatrix}
1 & 1 & -3 & 1 & 0 & 0\\
0 & 1 & -3&2&-1&0\\
0 & 0 & 1&-2/7&0&1/7
\end{pmatrix}$$
Subtract row 2 from row 1, then add three times the third to the second and you are there.  The right three columns will be your inverse.
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