Matrix is Invertible iff Determinant has Multiplicative Inverse/Sufficient Condition

Theorem
Let $R$ be a commutative ring with unity.

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

Let the determinant of $\mathbf A$ be invertible in $R$.

Then $\mathbf A$ is an invertible matrix.

Proof
Let $\map \det {\mathbf A}$ be invertible in $R$.

From Matrix Product with Adjugate Matrix:

Thus:

Thus $\mathbf A$ is invertible, and:
 * $\mathbf A^{-1} = \map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A}$

Also see

 * Determinant of Inverse Matrix