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 $AB = I$?
 A: Use the fact that $(A^{-1})^{-1} = A$                  
A: 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.
A: 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 $A$:
$$\begin{align}\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
1 & 1 & 0 & | & 0 & 1 & 0 \\ 
1 & 1 & 1 & | & 0 & 0 & 1 \\ 
\end{bmatrix}&\text{subtract first from second}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
0 & 1 & 0 & | & -1 & 1 & 0 \\ 
1 & 1 & 1 & | & 0 & 0 & 1 \\ 
\end{bmatrix}&\text{subtract first from third}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
0 & 1 & 0 & | & -1 & 1 & 0 \\ 
0 & 1 & 1 & | & -1 & 0 & 1 \\ 
\end{bmatrix}&\text{subtract second from third}\\
\begin{bmatrix}
1 & 0 & 0 & | & 1 & 0 & 0 \\ 
0 & 1 & 0 & | & -1 & 1 & 0 \\ 
0 & 0 & 1 & | & 0 & -1 & 1 \\ 
\end{bmatrix}&\text{done}\end{align}$$
A: Write
$$\pmatrix{ a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}} 
\pmatrix{ 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1}=
\pmatrix{ 1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}.
$$
Note that the top right element of the left hand side product equals $a_{13}$ which must hence be zero.  Likewise for $a_{23}$.
Now note that the top center element on the left hand side is $a_{12}+a_{13}=0$ and hence $a_{12}=0$.
The top left element is then $a_{11}+a_{12}+a_{13}=a_{11}$ which must equal one, etcetera.
