# Left or Right Inverse of Matrix is Inverse

## Theorem

Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over a commutative ring with unity $\left({R, +, \circ}\right)$.

Suppose that:

$\mathbf A \mathbf B = \mathbf I_n$

where $\mathbf I_n$ is the unit matrix of order $n$.

Then $\mathbf A$ and $\mathbf B$ are invertible matrices, and furthermore:

$\mathbf B = \mathbf A^{-1}$

where $\mathbf A^{-1}$ is the inverse of $\mathbf A$.

## Proof

When $1_R$ denotes the unity of $R$, we have:

 $\ds 1_R$ $=$ $\ds \map \det {\mathbf I_n}$ Determinant of Unit Matrix $\ds$ $=$ $\ds \map \det {\mathbf A \mathbf B}$ by assumption $\ds$ $=$ $\ds \map \det {\mathbf A} \map \det {\mathbf B}$ Determinant of Matrix Product

From Matrix is Invertible iff Determinant has Multiplicative Inverse, it follows that $\mathbf A$ and $\mathbf B$ are invertible.

Then:

 $\ds \mathbf B$ $=$ $\ds \mathbf I_n \mathbf B$ Unit Matrix is Unity of Ring of Square Matrices $\ds$ $=$ $\ds \paren {\mathbf A^{-1} \mathbf A} \mathbf B$ Definition of Inverse Matrix $\ds$ $=$ $\ds \mathbf A^{-1} \paren {\mathbf A \mathbf B}$ Matrix Multiplication is Associative $\ds$ $=$ $\ds \mathbf A^{-1} \mathbf I_n$ by assumption $\ds$ $=$ $\ds \mathbf A^{-1}$ Unit Matrix is Unity of Ring of Square Matrices

$\blacksquare$