Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Necessary Condition
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Theorem
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be an nonsingular square matrix of order $n$.
Let $\mathbf B = \mathbf A^{-1}$ be the inverse of $\mathbf A$.
Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$.
Then:
- $\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$
Proof
Let $\mathbf A$ be nonsingular with $\mathbf B = \mathbf A^{-1}$.
Let $1_R$ denote the unity of $R$.
Let $\mathbf I_n$ denote the unit matrix of order $n$.
Then:
\(\ds 1_R\) | \(=\) | \(\ds \map \det {\mathbf I_n}\) | Determinant of Unit Matrix | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \det {\mathbf A \mathbf B}\) | Definition of Inverse Matrix | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \det {\mathbf A} \, \map \det {\mathbf B}\) | Determinant of Matrix Product |
This shows that:
- $\map \det {\mathbf B} = \dfrac 1 {\map \det {\mathbf A} }$
$\Box$
Sources
- 1994: Robert Messer: Linear Algebra: Gateway to Mathematics: $\S 7.4$