Matrix is Invertible iff Determinant has Multiplicative Inverse

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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 if and only if 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 if and only if its determinant is non-zero.


Proof

Necessary Condition

Let $\mathbf A$ be invertible.

Let $1_R$ denote the unity of $R$.

Let $\mathbf I_n$ denote the unit matrix of order $n$.


Then:

\(\displaystyle 1_R\) \(=\) \(\displaystyle \map \det {\mathbf I_n}\) Determinant of Unit Matrix
\(\displaystyle \) \(=\) \(\displaystyle \map \det {\mathbf A \mathbf B}\) Definition of Inverse Matrix
\(\displaystyle \) \(=\) \(\displaystyle \map \det {\mathbf A} \, \map \det {\mathbf B}\) Determinant of Matrix Product

This shows that:

$\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$

$\Box$


Sufficient Condition

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

From Matrix Product with Adjugate Matrix:

\(\displaystyle \mathbf A \cdot \adj {\mathbf A}\) \(=\) \(\displaystyle \map \det {\mathbf A} \cdot \mathbf I_n\)
\(\displaystyle \adj {\mathbf A} \cdot \mathbf A\) \(=\) \(\displaystyle \map \det {\mathbf A} \cdot \mathbf I_n\)

Thus:

\(\displaystyle \mathbf A \cdot \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} }\) \(=\) \(\displaystyle \mathbf I_n\)
\(\displaystyle \paren {\map \det {\mathbf A}^{-1} \cdot \adj {\mathbf A} } \cdot \mathbf A\) \(=\) \(\displaystyle \mathbf I_n\)


Thus $\mathbf A$ is invertible, and:

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

$\blacksquare$


Also see


Sources