Matrix is Invertible iff Determinant has Multiplicative Inverse/Necessary Condition

Theorem
Let $\mathbf A = \begin{bmatrix} a \end{bmatrix}_n$ be an invertible square matrix of order $n$ over a commutative ring with unity $\left({R, +, \circ}\right)$.

Let $\mathbf B = \begin{bmatrix} b \end{bmatrix}_n = \mathbf A^{-1}$ be the inverse of $\mathbf A$.

Let $\det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

Then:
 * $\det \left({\mathbf B}\right) = \dfrac 1 {\det \left({\mathbf A}\right)}$

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

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

Then:

This shows that:
 * $\det \left({\mathbf B}\right) = \dfrac 1 {\det \left({\mathbf A}\right)}$