Leibniz formula for determinants

Set

 context $n\in \mathbb N$ context $R$ … abelian ring
 definiendum $\mathrm{det}_n:\mathrm{SquareMatrix}(n,R)\to R$ definiendum $\mathrm{det}_n(A):=\sum_{j_1,\dots,j_n}^n\varepsilon_{j_1,\dots,j_n}\cdot \prod_{k=1}^n A_{k,j_k}$

Discussion

This function concides with the implicitly defined determinant of Determinant, if the matrices are taken to be linear operators in the usual way.