Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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For same-sized finite square matrices over a commutative ring, det(AB)=det(A)det(B)\det(AB)=\det(A)\det(B)

Statement

For n1n\ge1 and A,BMn(R)A,B\in M_n(R) over a commutative ring, det(AB)=det(A)det(B).\det(AB)=\det(A)\det(B).

Facts & Assumptions

Given: A fixed matrix AMn(R)A\in M_n(R) and a variable matrix BMn(R)B\in M_n(R).

[L1]

Determinant is the unique normalized alternating column-multilinear matrix function (The determinant is the unique normalized alternating multilinear function on the columns).

Proof

technique · direct
1.1

Define F(B):=det(AB)F(B):=\det(AB). Each column of ABAB is AA times the corresponding column of BB, so distributivity makes FF column-multilinear; equal columns of BB give equal columns of ABAB, so FF is alternating.

L1L2L3L4
2.1

Since AIn=AAI_n=A, one has F(In)=det(A)F(I_n)=\det(A).

step 1.1L3L4
3.1

Apply [L2] to FF: det(AB)=F(B)=F(In)det(B)=det(A)det(B)\det(AB)=F(B)=F(I_n)\det(B)=\det(A)\det(B). Neither matrix was assumed invertible.

step 2.1L2algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources