The inverse of a matrix (if it exists) can be found by row reducing the matrix augmented by the identity matrix.

To compute $A^{-1}$, row-reduce $[A\ I]$. If the left block becomes $I$, the right block is $A^{-1}$: elementary operations give $[EA\ E]=[I\ A^{-1}]$. If the left block lacks a pivot, $A$ is singular. For solving $Ax=b$ numerically, factor and solve rather than explicitly forming an inverse.Copy