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

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Laplace expansion computes the determinant along every row and every column over a commutative ring

Statement

Let $R$ be a commutative ring, $n \geq 1$ , and $A = (a_{i j}) \in M_{n} (R)$ . For every row $i$ and every column $j$ ,

$\det (A) = \sum_{k = 0}^{n - 1} a_{i k} C_{i k} (A) and \det (A) = \sum_{k = 0}^{n - 1} a_{k j} C_{k j} (A) .$

Facts & Assumptions

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

[F1]

$C_{i k} (A) = (- 1)^{i + k} \det (A^{(i, k)})$ , with the determinant of the $0 \times 0$ minor defined to be $1$ (Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring).

[L1]

For $n \geq 1$ , the determinant is the Leibniz sum over permutations (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix), and it is alternating, column-multilinear, and normalized (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

[L2]

$\det (A^{T}) = \det (A)$ (For every square matrix over a commutative ring, $\det (A^{T}) = \det (A)$ ).

Proof

technique · direct

1.1F1L1algebra

If $n = 1$ , both displayed sums have one term, $a_{00} (- 1)^{0} \cdot 1 = a_{00} = \det (A)$ .

1.2L1

Suppose $n > 1$ . In the Leibniz sum for $\det (A)$ , group the terms according to the unique column $k$ whose chosen entry lies in row $i$ .

2.1step 1.2algebra

After the factor $a_{i k}$ is removed, the remaining choices are exactly the permutations contributing to $\det (A^{(i, k)})$ . Moving row $i$ and column $k$ to the last positions takes $(n - 1 - i) + (n - 1 - k)$ transpositions, whose parity is the parity of $i + k$ ; hence the sign contribution is $(- 1)^{i + k}$ .

3.1step 2.1F1

Summing the groups in step 2.1 gives $\det (A) = \sum_{k} a_{i k} (- 1)^{i + k} \det (A^{(i, k)}) = \sum_{k} a_{i k} C_{i k} (A)$ .

4.1step 3.1L2F1

Apply the row formula from step 3.1 to $A^{T}$ . Its row- $j$ cofactors are the column- $j$ cofactors of $A$ , and [L2] identifies the two determinants, giving the column formula.

5.1step 1.1step 3.1step 4.1∎

Steps 1.1, 3.1, and 4.1 prove both formulas for every $n \geq 1$ .

Laplace expansion computes the determinant along every row and every column over a commutative ring

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources