# Matrix is Invertible iff Determinant has Multiplicative Inverse

## Theorem

Let $R$ be a commutative ring with unity.

Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.

Then $\mathbf A$ is invertible if and only if its determinant is invertible in $R$.

If $R$ is one of the standard number fields $\Q$, $\R$ or $\C$, this translates into:

$\mathbf A$ is invertible if and only if its determinant is non-zero.

## Proof

### Necessary Condition

Let $\mathbf A$ be invertible with $\mathbf B = \mathbf A^{-1}$.

Let $1_R$ denote the unity of $R$.

Let $\mathbf I_n$ denote the unit matrix of order $n$.

Then:

 $\ds 1_R$ $=$ $\ds \map \det {\mathbf I_n}$ Determinant of Unit Matrix $\ds$ $=$ $\ds \map \det {\mathbf A \mathbf B}$ Definition of Inverse Matrix $\ds$ $=$ $\ds \map \det {\mathbf A} \, \map \det {\mathbf B}$ Determinant of Matrix Product

This shows that:

$\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$

$\Box$

### Sufficient Condition

Let $\map \det {\mathbf A}$ be invertible in $R$.

 $\ds \mathbf A \cdot \adj {\mathbf A}$ $=$ $\ds \map \det {\mathbf A} \cdot \mathbf I_n$ $\ds \adj {\mathbf A} \cdot \mathbf A$ $=$ $\ds \map \det {\mathbf A} \cdot \mathbf I_n$

Thus:

 $\ds \mathbf A \cdot \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} }$ $=$ $\ds \mathbf I_n$ $\ds \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} } \cdot \mathbf A$ $=$ $\ds \mathbf I_n$

Thus $\mathbf A$ is invertible, and:

$\mathbf A^{-1} = \map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A}$

$\blacksquare$