Leibniz formula for determinants

Statement

Let $K$ be a commutative ring, let $n \ge 1$, and let $A = (a_{ij})$ be an $n \times n$ matrix over $K$. Let $S_n$ be the group of permutations of $\{1, \dots, n\}$, and let $\sgn(\sigma) \in \{1, -1\}$ be the sign of $\sigma$. Then the determinant of $A$ is $$\det A = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i=1}^{n} a_{i, \sigma(i)}.$$

Remarks

The sum has $n!$ terms. Each term takes exactly one entry from each row and from each column. For $n = 3$, the formula is the rule of Sarrus. The formula shows that the determinant is a polynomial in the entries, with integer coefficients.

Explanations

L1 human

To compute a determinant by this formula, choose one entry in each row so that no two lie in the same column. Multiply the chosen entries, and give the product the sign of the permutation. Add these signed products over all choices.