CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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An invertible square matrix over a commutative ring has unit determinant

Statement

Let $n \geq 1$ . If $A \in M_{n} (R)$ is invertible over a commutative ring, then $\det (A)$ is a unit of $R$ . More precisely, $\det (A^{- 1})$ is its inverse.

Facts & Assumptions

Given: An invertible matrix $A$ with inverse $A^{- 1}$ .

[L1]

Determinants multiply: $\det (X Y) = \det (X) \det (Y)$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ).

[L2]

A unit is an element with a two-sided multiplicative inverse, and its inverse is unique (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring).

[L3]

Identity matrices are multiplicative identities (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

[L4]

Matrix invertibility gives $A A^{- 1} = I_{n} = A^{- 1} A$ (Invertible square matrices and similarity over a commutative ring).

[L5]

The determinant is normalized: $\det (I_{n}) = 1$ (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

Proof

technique · direct

1.1L1L3L4L5

Apply [L1] to the two equations in [L4]: $\det (A) \det (A^{- 1}) = \det (I_{n}) = 1$ and $\det (A^{- 1}) \det (A) = 1$ .

2.1step 1.1L2∎

Thus $\det (A)$ is a unit with inverse $\det (A^{- 1})$ by [L2]. In the zero ring, $0 = 1$ and the same two-sided equation still gives the unit conclusion.

Depends on

Dependency tree · 2 levels

Sources

S. New, MATH 146 Linear Algebra 1 Lecture Notes, Theorem 4.24