Intuition/Proof behind the fact that delta system $\delta_{ijk}^{rst}$ can be represented as a determinant

Question

I am following Pavel Grinfield's Tensor Calculus book, in page-137, he shows this remarkable equation:

$$ \epsilon_{rst} \cdot \epsilon^{ijk}=d_{rst}^{ijk} = \begin{vmatrix} \delta_r^i & \delta_s^i & \delta_t^i \\ \delta_r^j & \delta_s^j& \delta_t^j\\ \delta_r^k &\delta_s^k & \delta_t^k \end{vmatrix}$$

This is the definition of determinant given in his book:

$$ |A| = \delta_{rst}^{ijk} \frac{a_i^r a_j^s a_k^t}{3!}$$

Note: $\epsilon_{ijk}$ and $\epsilon_{rst}$ are permutation symbols.

Here is my attempt at proving the fact:

Call: $$B=\begin{bmatrix} \delta_r^i & \delta_s^i & \delta_t^i \\ \delta_r^j & \delta_s^j& \delta_t^j\\ \delta_r^k &\delta_s^k & \delta_t^k \end{bmatrix}$$

Then,

$$|B|= \frac{1}{3!}\delta_{rst}^{ijk} (\delta_{p(i)}^{p(r)}\delta_{p(j)}^{p(s)}\delta_{p(k)}^{p(t)})$$

Where $p$ is a function which maps the index set in the following way : $p(1)=i, p(2)=j,p(3)=k$.. now I don't get how to simplfy..

Edit: Ok, everyother exercise in the proceeding section is based on this..

Related

Answer 1

An overview of how we can avoid brute force:

We can write $B^a_b=\delta^{g(a)}_{h(b)}$ with $g(1)=i$ etc., so$$\det B=\epsilon_{ace}\epsilon^{bdf}B^a_bB^c_dB^e_f=\epsilon_{ace}\epsilon^{bdf}\delta^{g(a)}_{h(b)}B^{g(c)}_{h(d)}B^{g(e)}_{h(f)}=\epsilon_{ace}\epsilon^{j(a)j(c)j(e)},\,j(k):=h^{-1}(g(k)).$$The antisymmetries of this expression, together with its behaviour in the case $a=1,\,c=2,\,e=3$, finishes the job.