TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-13

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Cramer's rule over a commutative ring: every solution satisfies $\det (A) x_{j} = \det (A_{j} (b))$ , and a unit determinant gives the unique quotient formula

Statement

Let $R$ be a commutative ring, $n \geq 1$ , $A \in M_{n} (R)$ , and $b \in R^{n}$ . Let $A_{j} (b)$ be obtained from $A$ by replacing column $j$ by $b$ . Every solution $x$ of $A x = b$ satisfies

$\det (A) x_{j} = \det (A_{j} (b)) (0 \leq j < n) .$

If $\det (A)$ is a unit, the solution is unique and

$x_{j} = \det (A)^{- 1} \det (A_{j} (b)) .$

Facts & Assumptions

Given: $R, n, A, b, j$ as in the statement, and a solution $x$ when the first claim is used.

[L1]

The determinant is alternating and multilinear in columns (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

[L2]

Laplace expansion along a column computes the determinant as the sum of its entries times their cofactors (Laplace expansion computes the determinant along every row and every column over a commutative ring).

[L3]

If $\det (A)$ is a unit, then $A$ is invertible (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit) and $A^{- 1} = \det (A)^{- 1} adj (A)$ (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[F1]

Matrix multiplication obeys the entry formula (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L4]

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

Proof

technique · direct

1.1

If $A x = b$ , then $b = \sum_{k} x_{k} A_{k}$ , where $A_{k}$ is column $k$ of $A$ .

F1given

1.2

Now suppose $u : = \det (A)$ is a unit. By [L3], $A$ is invertible, so $x = A^{- 1} b = u^{- 1} adj (A) b$ is the unique solution.

L3L4

2.1

Replace column $j$ by the sum from step 1.1 and use column multilinearity. Every term with $k \neq j$ has two equal columns and vanishes; the term $k = j$ is $x_{j} \det (A)$ . Thus $\det (A_{j} (b)) = x_{j} \det (A)$ .

step 1.1L1

2.2

Its $j$ th coordinate is $u^{- 1} \sum_{i} C_{i j} (A) b_{i}$ . Deleting column $j$ removes the replaced column, so $C_{i j} (A_{j} (b)) = C_{i j} (A)$ ; expanding $A_{j} (b)$ along column $j$ therefore identifies the sum as $\det (A_{j} (b))$ .

step 1.2F1L2

3.1

Step 2.1 proves the unconditional cross-multiplied identity, and step 2.2 proves the quotient formula and uniqueness under the unit hypothesis.

step 2.1step 2.2∎

Depends on

Used by

Over a field, for each right-hand side b, Ax=b has a unique solution exactly when det(A) is nonzero, and then Cramer's quotient formula holds Corollary

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

Interactive Linear Algebra (standard reference, not scraped)
András Pál, Introduction to Commutative Algebra (standard reference, not scraped)

Cramer's rule over a commutative ring: every solution satisfies det⁡(A)xj=det⁡(Aj(b))det(A)xj​=det(Aj​(b)), and a unit determinant gives the unique quotient formula