Matrix is Invertible iff Determinant has Multiplicative Inverse

Corollary of Inverse of Matrix
A square matrix $A$ over a commutative ring with unity $\left({R, +, \circ}\right)$ is invertible iff its determinant has a multiplicative inverse in $R$.

Necessary Condition
Suppose that $\mathbf A$ is an invertible square matrix over $R$.

It follows from the first section of the main proof that $\det \left({\mathbf A}\right)$ has a multiplicative inverse in $R$.

Sufficient Condition
Suppose that $\det \left({\mathbf A}\right)$ has a multiplicative inverse in $R$.

It follows from the two last sections of the main proof that the matrix $\mathbf B = \begin{bmatrix} b \end{bmatrix}_n$, defined by:


 * $b_{ij} = \dfrac {1} {\det \left({\mathbf A}\right) } A_{ji}$

is the inverse matrix of $A$.

Here, $A_{ji}$ denotes the cofactor of $a_{ji} \in \mathbf A$.