Determinant of matrix in index notation The determinant of the $3\times 3$ square matrix $A=[a_{ij}]$   in index form  is given by
$$\mathrm{det}(A)=\epsilon_{ijk}a_{1i}a_{2j}a_{3k}$$
Wikipedia suggests that I can write it as
$$\mathrm{det}(A)=\frac{1}{3!}\epsilon_{ijk}\epsilon_{pqr}a_{ip}a_{jq}a_{kr}$$
using two epsilon symbols. But I don't understand How this supposed to happen. Can someone help me with this?
 A: This just averages $3!$ copies of the first formula. For example,$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\epsilon_{ijk}a_{2j}a_{1i}a_{3k}=\epsilon_{jik}a_{2i}a_{1j}a_{3k}=\sigma_{213}\epsilon_{ijk}a_{2i}a_{1j}a_{3k}.$$where $\sigma_{213}=-1=-(-1)^p,\,2\nmid p$ is a symbol I've invented to represent a coefficient caused by $213$ being an odd permutation. (The second $=$ above results from relabelling the indices.) More generally, if a permutation $\sigma$ introduces coefficient $\epsilon_\sigma$,$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\epsilon_\sigma\prod_{l=1}^3a_{(ijk)_l\sigma_l}.$$In fact, $\sigma_{213}$ is just $\epsilon_{213}$, but if I'd called the coefficient $\epsilon_{213}$ up front that would have been more confusing, so I've had to delay the usual notation $\epsilon_\sigma$. We can rewrite my last equation as$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\epsilon_{ijk}\epsilon_{pqr}a_{ip}a_{jq}a_{kr},$$provided this is interpreted as using one choice of $p,\,q,\,r$ rather than summing over them. Since that's not the usual reading of this notation, when we do use Einstein summation we instead have$$\epsilon_{ijk}a_{1i}a_{2j}a_{3k}=\frac{1}{3!}\epsilon_{ijk}\epsilon_{pqr}a_{ip}a_{jq}a_{kr}.$$Incidentally, this points us to another identity, $\epsilon_{ijk}\epsilon_{ijk}=3!$. This is obvious because there are $3!$ nonzero terms, each $(\pm1)^2=1$.
