I cannot wrap my head around this identity for a 3 by 3 Matrix:
$\sum_{l,m,n = 1}^{3} \varepsilon^{lmn} a_{li}a_{mj}a_{nk} = \varepsilon^{ijk}$det$M$
I have tried proving this using the Leibniz formula $\varepsilon^{ijk}\det(A)=\varepsilon^{ijk}\sum_{\sigma\in\mathcal S_n}\text{sign}(σ)a_{1σ(1)}a_{2σ(2)}a_{3σ(3)} = \sum_{\sigma\in\mathcal S_n}\varepsilon^{ijk}\varepsilon^{\sigma(1)\sigma(2)\sigma(3)}a_{1σ(1)}a_{2σ(2)}a_{3σ(3)} = \sum_{l,m,n = 1}^{3} \varepsilon^{ijk}\varepsilon^{lmn} a_{1l}a_{2m}a_{3n}$
where in the last step I tried to make use of the fact that $\varepsilon^{lmn}$ is $0$ whenever indices coincide. I am not sure where to go from here, or if this is even the right path to tackle this identity. Any help is greatly appreciated!
Recall that the determinant can also be written as
$$ \det M = \sum_{\sigma} \epsilon^\sigma \prod_s a_{\sigma(s)s}.$$
Then $$ \delta^{ijk} := \sum_{l,m,n = 1}^3 \epsilon^{lmn}a_{li}a_{mj}a_{nk}.$$ is simply the determinant of a matrix whose
It follows that $\delta^{ijk} = \epsilon^{ijk} \det M.$
Another way to see it is that $$ \epsilon^{ijk} = \det(M_{ijk})$$ where $M_{ijk} = (e_i',e_j',e_k')$ is the matrix with the canonical basis vectors as columns. With this notation the equation your trying to prove is just
$$ \det(MM_{ijk}) = \det(M)\det(M_{ijk})$$