Permutation_of_Determinant_Indices

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$ over a field. Let $\lambda: \N_{> 0} \to \N_{> 0}$ be any fixed permutation on $\N_{> 0}$. Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$. Let $\struct {S_n, \circ}$ be the symmetric group of $n$ letters. Then $\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} }$ $\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \mu k, \map \lambda k} }$ where the summation $\displaystyle \sum_{\mu \mathop \in S_n}$ goes over all the $n!$ permutations of $\set {1, 2, \ldots, n}$ $\map \sgn \mu$ is the sign of the permutation $\mu$.