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 $AB = I$?

Using Gauss-Jordan to invert the given matrix $B$ you get
$$\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
1&1&0&0&1&0\\
1&1&1&0&0&1
\end{array}\right)
\leadsto
\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&-1&1&0\\
0&1&1&-1&0&1
\end{array}\right)
\leadsto
\left(\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&-1&1&0\\
0&0&1&0&-1&1
\end{array}\right)
$$
Where $\left(\begin{array}{ccc}1&0&0\\-1&1&0\\0&-1&1\end{array}\right)$ is the inverse of $B$. Now you have
$$AB=I \Longleftrightarrow ABB^{-1} = IB^{-1} \Longleftrightarrow A = B^{-1}$$
Thus the previous found inverse of your matrix $B$ is the same as the matrix $A$ you are searching for.
