<s>[INST] How to solve for matrix $A$ in $AB = I$ Given $B$ = $\begin{bmatrix}
1 & 0 & 0\\ 
1 & 1 & 0\\ 
1 & 1 & 1
\end{bmatrix}$
I know that $B$ is equal to inverse of $A$, how can I go backwards to solve for $A$ in $AB = I$?
 [/INST] Start by writing $[B|I]$ and apply row operations until you end up with $[I|A]$. This works because row operations are multiplications with suitable matrices from the left, and if this suitable matrix turns $B$ into $I$ it must be $B^{-1}=A$, which of course turns $I$ into $A$:
$$\begin{align}\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
1 & 1 & 0 & | & 0 & 1 & 0 \\ 
1 & 1 & 1 & | & 0 & 0 & 1 \\ 
\end{bmatrix}&\text{subtract first from second}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
0 & 1 & 0 & | & -1 & 1 & 0 \\ 
1 & 1 & 1 & | & 0 & 0 & 1 \\ 
\end{bmatrix}&\text{subtract first from third}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
0 & 1 & 0 & | & -1 & 1 & 0 \\ 
0 & 1 & 1 & | & -1 & 0 & 1 \\ 
\end{bmatrix}&\text{subtract second from third}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
0 & 1 & 0 & | & -1 & 1 & 0 \\ 
0 & 0 & 1 & | & 0 & -1 & 1 \\ 
\end{bmatrix}&\text{done}\end{align}$$
</s>