# Left or Right Inverse of Matrix is Inverse

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## 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 |

$\blacksquare$

## Sources

- 1994: Robert Messer:
*Linear Algebra: Gateway to Mathematics*: $\S 5.2$