From Schaum's Outline in Tensor Calculus
If $A = [a_{ij}]_{nn} $ is any square matrix, then define $\text{ det } A = \epsilon_{i_1i_2i_3…i_{n-1}i_n}a_{1 \, \cdot \, i_1}a_{2 \, \cdot \, i_2}…a_{(n – 1) \, \cdot \, i_{n – 1}}a_{n \, \cdot \, i_n} $.
I can check this by expanding the product and sum in full, but what's the derivation or motivation behind this formula? I tried to find something on the Internet about this.
Best Answer
This is the Leibniz formula $$ \det(A)=\sum_{\sigma\in\mathcal S_n}\text{sign}(σ)a_{1σ(1)}a_{2σ(2)}...a_{nσ(n)}, $$ the eps-tensor is the sign of the permutation.
It follows from repeated application of the Laplace formula, essentially the multilinear nature of the determinant, so that in the end it is represented as a linear combination of determinants of permutation matrices. $$ \det(A)=\det(a_1,...,a_n)=\sum_{i_1,...,i_n\in\{1,...,n\}}a_{1i_1}...a_{ni_n}\det(e_{i_1},...,e_{i_n}) $$ where $\det(e_{i_1},e_{i_1}...,e_{i_n})=ϵ_{i_1i_2...i_n}$ is only different from zero if $(i_1,i_2,...,i_n)$ is a permutation of $(1,2,...,n)$, so that $(e_{i_1},e_{i_1}...,e_{i_n})$ is a permutation matrix, and the determinant of a permutation matrix is the sign of the permutation, i.e., the value of the Levi-Civita symbol for this permutation.