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:

\(\ds \mathbf A \cdot \adj {\mathbf A}\)

\(=\)

\(\ds \map \det {\mathbf A} \cdot \mathbf I_n\)

\(\ds \adj {\mathbf A} \cdot \mathbf A\)

\(=\)

\(\ds \map \det {\mathbf A} \cdot \mathbf I_n\)

Thus:

\(\ds \mathbf A \cdot \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} }\)

\(=\)

\(\ds \mathbf I_n\)

\(\ds \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} } \cdot \mathbf A\)

\(=\)

\(\ds \mathbf I_n\)

Thus $\mathbf A$ is invertible, and:

$\mathbf A^{-1} = \map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A}$

$\blacksquare$

Also see

• Determinant of Inverse Matrix