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

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For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$

Statement

For a commutative ring $R$ , $n \geq 1$ , and $A \in M_{n} (R)$ ,

$A adj (A) = adj (A) A = \det (A) I_{n} .$

Facts & Assumptions

Given: $R, n, A$ as in the statement.

[L1]

Laplace expansion along row $i$ is $\det (A) = \sum_{k} a_{i k} C_{i k} (A)$ , and expansion along column $j$ is $\det (A) = \sum_{k} a_{k j} C_{k j} (A)$ (Laplace expansion computes the determinant along every row and every column over a commutative ring).

[L2]

The determinant is alternating and multilinear in rows (The determinant is alternating and multilinear in the rows as well as in the columns).

[L3]

A matrix with two equal columns has determinant $0$ (A square matrix with a zero column or two equal columns has determinant zero).

[F1]

Matrix multiplication is given by $(B C)_{i j} = \sum_{k} b_{i k} c_{k j}$ (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L4]

Matrix multiplication is associative and distributive, and $I_{n}$ is its identity (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

Proof

technique · direct

1.1F1L1

The $(i, j)$ entry of $A adj (A)$ is $\sum_{k} a_{i k} C_{j k} (A)$ . If $i = j$ , [L1] makes this $\det (A)$ .

1.2L1L2

If $i \neq j$ , replace row $j$ of $A$ by row $i$ . Expanding the resulting determinant along row $j$ gives $\sum_{k} a_{i k} C_{j k} (A)$ , because the minors used in that row do not involve row $j$ . The matrix has two equal rows, so its determinant is $0$ by alternation.

1.3F1L1L3

The $(i, j)$ entry of $adj (A) A$ is $\sum_{k} C_{k i} (A) a_{k j}$ . It equals $\det (A)$ when $i = j$ by column expansion. When $i \neq j$ , it is the column- $i$ expansion of the matrix obtained by replacing column $i$ by column $j$ , whose determinant is $0$ because it has two equal columns.

2.1step 1.1step 1.2L4

Thus $A adj (A) = \det (A) I_{n}$ .

2.2step 1.3L4

Hence $adj (A) A = \det (A) I_{n}$ .

3.1step 2.1step 2.2∎

Combining steps 2.1 and 2.2 proves both identities.

Depends on

Dependency tree · 2 levels

Sources

András Pál, Introduction to Commutative Algebra, Proposition 7.2

For every positive-sized square matrix over a commutative ring, Aadj⁡(A)=adj⁡(A)A=det⁡(A)IAadj(A)=adj(A)A=det(A)I