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 its determinant is invertible in $R$.

If this is the case, then:
 * $\mathbf A^{-1} = \map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A}$

where $\adj {\mathbf A}$ is the adjugate of $\mathbf A$.

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


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

Also see

 * Determinant of Inverse Matrix