Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11
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For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B)

Statement

For n≥1 and A,B∈Mn(R) over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B).

Facts & Assumptions

Given: A fixed matrix A∈Mn(R) and a variable matrix B∈Mn(R).

[L1]

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

[L2]

Every alternating column-multilinear F satisfies F(B)=F(In)det⁡(B) (Every alternating multilinear F satisfies F(A)=F(I)∑σ∈Snsgn⁡(σ)∏iaσ(i),i).

Proof

technique · direct
1.1

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

L1L2L3L4
2.1

Since AIn=A, one has F(In)=det⁡(A).

step 1.1L3L4
3.1

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

step 2.1L2algebra∎

Depends on

Used by

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources