Matrix Inverse Algorithm

Algorithm
The matrix inverse algorithm is an algorithm which either:
 * $(1): \quad$ converts a matrix into its inverse, if it exists

or:
 * $(2): \quad$ determines that such an inverse does not exist.

Let $\mathbf A$ be the $n \times n$ square matrix in question.

Let $\mathbf I$ be the unit matrix of order $n$.


 * Step $0$: Create the augmented matrix $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$.


 * Step $1$: Perform elementary row operations until $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ is in reduced echelon form.

By Matrix is Row Equivalent to Reduced Echelon Matrix, this is possible.

By Reduced Echelon Matrix is Unique, this process is well-defined.

Call this new augmented matrix $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$.


 * Step $2$:


 * If $\mathbf H = \mathbf I$, then take $\mathbf C = \mathbf A^{-1}$.


 * If $\mathbf H \ne \mathbf I$, $\mathbf A$ is not invertible.

Proof
Suppose $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ can be transformed into an upper triangular matrix with no zeroes on its main diagonal.

Then from Identity Matrix from Upper Triangular Matrix, it can further be transformed into $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$ where $\mathbf H = \mathbf I$.

From Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices, the row operation to convert $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ to $\begin {bmatrix} \mathbf H & \mathbf C \end {bmatrix}$ is equivalent to the matrix product of the elementary row matrices corresponding to the sequence of elementary row operations that compose that row operation.

Let $\mathbf R$ be that matrix corresponding to that row operation.

Because $\mathbf H = \mathbf I$, it follows that:
 * $\mathbf R \mathbf A = \mathbf I$

and so $\mathbf R$ is the inverse of $\mathbf A$.

That is:
 * $\mathbf R = \mathbf A^{-1}$

We also have, from the action of $\mathbf R$ on the of the augmented matrix $\begin {bmatrix} \mathbf A & \mathbf I \end {bmatrix}$ that:
 * $\mathbf R \mathbf I = \mathbf C$

and so:
 * $\mathbf C = \mathbf R = \mathbf A^{-1}$

Now suppose that $\mathbf H \ne \mathbf I$.