Prove that
$$
det
\left[
\begin{matrix}
δ_{ak} & δ_{al} & δ_{am} \\
δ_{bk} & δ_{bl} & δ_{bm} \\
δ_{ck} & δ_{cl} & δ_{cm} \\
\end{matrix}
\right]
=ε_{abc}ε_{klm}
$$
with $a,b,c,k,l,m \in \{1,2,3\}$ where $δ_{ij}$ is Kronecker delta and $ε_{ijk}$ is Levi-Civita symbol in dimension 3
I found this problem in a very old mathematical-competition.I would apreciate if we can find a solution for this.
Best Answer
Write the Kronecker Delta as
$$\delta_{ij}=\hat x_i \cdot \hat x_j \tag 1$$
in terms of then inner product of Cartesian unit vectors. Write the Levi-Civita symbol as
$$\epsilon_{ijk}=\hat x_i\cdot(\hat x_j \times \hat x_k) \tag 2$$
in terms of the scalar triple product of Cartesian unit vectors. Using $(2)$, we have
$$\begin{align} \epsilon_{abc}\epsilon_{k\ell m}&=\left(\hat x_a\cdot(\hat x_b \times \hat x_c)\right)\left(\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\tag3\\\\ &=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot \hat x_k\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\tag4\\\\ &+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot \hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m\times \hat x_k)\right)\\\\ &+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot \hat x_{m}\hat x_{m}\cdot(\hat x_k\times \hat x_{\ell})\right)\\\\ &=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot (\hat x_{\ell} \times \hat x_m)\right)\\\\ &+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot (\hat x_m\times \hat x_k)\right)\\\\ &+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot (\hat x_k\times \hat x_{\ell})\right)\\\\ &=\delta_{ak}\left(\delta_{b\ell}\delta_{cm}-\delta_{bm}\delta_{c\ell}\right)\\\\ &+\delta_{a\ell}\left(\delta_{bm}\delta_{ck}-\delta_{bk}\delta_{cm}\right)\\\\ &+\delta_{am}\left(\delta_{bk}\delta_{c\ell}-\delta_{b\ell}\delta_{ck}\right)\\\\ &= det \left[ \begin{matrix} δ_{ak} & δ_{al} & δ_{am} \\ δ_{bk} & δ_{bl} & δ_{bm} \\ δ_{ck} & δ_{cl} & δ_{cm} \\ \end{matrix} \right] \end{align}$$
as was to be shown!
NOTE: In going from $(3)$ to $(4)$, we used the result in This Answer.