Matrix is Invertible iff Determinant has Multiplicative Inverse/Necessary Condition

Theorem

Let $\struct {R, +, \circ}$ be a commutative ring with unity.

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

Let $\mathbf B = \mathbf A^{-1}$ be the inverse of $\mathbf A$.

Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.

Then:

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

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:

$\ds 1_R$

$=$

$\ds \map \det {\mathbf I_n}$

$\ds $

$=$

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

Definition of Inverse Matrix

$\ds $

$=$

$\ds \map \det {\mathbf A} \, \map \det {\mathbf B}$

This shows that:

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

$\Box$