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

where $\operatorname{adj}(\mathbf A)$ 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.

Necessary Condition
Suppose that $\mathbf A$ is invertible.

Sufficient Condition
Suppose that $\det \left({\mathbf A}\right)$ is 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