# Product of Matrices is Invertible iff Matrices are Invertible

## Theorem

Let $\mathbf A$ and $\mathbf B$ be square matrices of order $n$.

Let $\mathbf A \mathbf B$ denote the matrix product of $\mathbf A$ and $\mathbf B$.

Let $\mathbf I$ be the $n \times n$ unit matrix.

Let $\mathbf A$ and $\mathbf B$ be invertible.

Then:

$\mathbf A \mathbf B$ is invertible
both $\mathbf A$ and $\mathbf B$ are invertible.

## Proof

### Necessary Condition

Let both $\mathbf A$ and $\mathbf B$ be invertible.

$\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$

where $\map \det {\mathbf A}$ denotes the determinant of $\mathbf A$.

$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$

Thus as both $\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$, it follows that:

$\map \det {\mathbf A \mathbf B} = \ne 0$

Hence by Matrix is Invertible iff Determinant has Multiplicative Inverse, $\map \det {\mathbf A \mathbf B}$ is invertible.

### Sufficient Condition

Let $\mathbf A \mathbf B$ be invertible.

Aiming for a contradiction, suppose it is not the case that both $\mathbf A$ and $\mathbf B$ are invertible.

Then by Matrix is Invertible iff Determinant has Multiplicative Inverse, either:

$\map \det {\mathbf A} = 0$

or:

$\map \det {\mathbf B} = 0$
$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$

and so:

$\map \det {\mathbf A \mathbf B} = 0$

Hence by Matrix is Invertible iff Determinant has Multiplicative Inverse, $\map \det {\mathbf A \mathbf B}$ is not invertible.

This contradicts the assumption that $\mathbf A \mathbf B$ is invertible.

Hence by Proof by Contradiction it follows that both $\mathbf A$ and $\mathbf B$ are invertible.

$\blacksquare$