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 $\det \left({\mathbf A}\right)$ be invertible in $R$.

From Matrix Product with Adjugate Matrix:


 * $\mathbf A \cdot \operatorname{adj}(\mathbf A) = \det \left({\mathbf A}\right) \cdot \mathbf I_n$
 * $\operatorname{adj}(\mathbf A) \cdot \mathbf A = \det \left({\mathbf A}\right) \cdot \mathbf I_n$

Thus:


 * $\mathbf A \cdot \left( \det \left({\mathbf A}\right)^{-1} \cdot \operatorname{adj}(\mathbf A) \right) = \mathbf I_n$
 * $\left( \det \left({\mathbf A}\right)^{-1} \cdot \operatorname{adj}(\mathbf A) \right) \cdot \mathbf A = \mathbf I_n$

Thus $\mathbf A$ is invertible and $\mathbf A^{-1} = {\det \left({\mathbf A}\right)}^{-1} \cdot \operatorname{adj}(\mathbf A)$.

Also see

 * Determinant of Inverse Matrix