Example of finding the inverse of a 3-by-3 matrix by row reducing the augmented matrix

Let $A=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}$. In $[A\ I]$, apply $R_2\leftarrow R_2-R_3$, then $R_1\leftarrow R_1-R_2$ using the updated second row. The left block becomes $I$ and the right block is $A^{-1}=\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}$.Copy