Matrix is Invertible iff Determinant has Multiplicative Inverse

Theorem
Let $\mathbf A$ be a square matrix over a commutative ring with unity $\left({R, +, \circ}\right)$.

Then $\mathbf A$ is invertible its determinant has a multiplicative inverse in $R$.

If the underlying structure of $\mathbf A$ is one of the standard sets of numbers: integers, rational numbers, real numbers or complex numbers, this translates into:


 * $\mathbf A$ is invertible its determinant is non-zero.

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

From Inverse of Matrix: Determinant has Inverse, $\det \left({\mathbf A}\right)$ has a multiplicative inverse in $R$.

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

From Inverse of Matrix, 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 $a_{ji} \in \mathbf A$.