Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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An invertible square matrix over a commutative ring has unit determinant

Statement

Let n1n\ge1. If AMn(R)A\in M_n(R) is invertible over a commutative ring, then det(A)\det(A) is a unit of RR. More precisely, det(A1)\det(A^{-1}) is its inverse.

Facts & Assumptions

Given: An invertible matrix AA with inverse A1A^{-1}.

[L1]

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

[L4]

Matrix invertibility gives AA1=In=A1AAA^{-1}=I_n=A^{-1}A (Invertible square matrices and similarity over a commutative ring).

[L5]

The determinant is normalized: det(In)=1\det(I_n)=1 (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

Proof

technique · direct
1.1

Apply [L1] to the two equations in [L4]: det(A)det(A1)=det(In)=1\det(A)\det(A^{-1})=\det(I_n)=1 and det(A1)det(A)=1\det(A^{-1})\det(A)=1.

L1L3L4L5
2.1

Thus det(A)\det(A) is a unit with inverse det(A1)\det(A^{-1}) by [L2]. In the zero ring, 0=10=1 and the same two-sided equation still gives the unit conclusion.

step 1.1L2

Depends on

Used by

Dependency tree · next 3 levels

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