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

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For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$

Statement

For $n \geq 1$ and $A, B \in M_{n} (R)$ over a commutative ring, $\det (A B) = \det (A) \det (B) .$

Facts & Assumptions

Given: A fixed matrix $A \in M_{n} (R)$ and a variable matrix $B \in M_{n} (R)$ .

[L1]

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

[L2]

Every alternating column-multilinear $F$ satisfies $F (B) = F (I_{n}) \det (B)$ (Every alternating multilinear $F$ satisfies $F (A) = F (I) \sum_{σ \in S_{n}} sgn (σ) \prod_{i} a_{σ (i), i}$ ).

[L3]

Matrix multiplication is associative, distributive and unital (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

[L4]

Determinant is defined for every square matrix over the commutative ring (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

Proof

technique · direct

1.1L1L2L3L4

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

2.1step 1.1L3L4

Since $A I_{n} = A$ , one has $F (I_{n}) = \det (A)$ .

3.1step 2.1L2algebra∎

Apply [L2] to $F$ : $\det (A B) = F (B) = F (I_{n}) \det (B) = \det (A) \det (B)$ . Neither matrix was assumed invertible.

For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B)det(AB)=det(A)det(B)

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$