Contraction of two Levi-Civita symbols

Contracting one index gives

For a $3\times 3$ matrix $A$, contraction of ${ϵ}_{lmn}{A}_{il}{A}_{jm}{A}_{kn}=(\det A){ϵ}_{ijk}$ with ${ϵ}_{ijk}$ proves the title identity, because ${ϵ}_{ijk}{ϵ}_{ijk}=6$. DefineExpansion shows that $C$ is the cofactor matrix. Equivalently, the determinant identity gives $A_{ib}{C}_{ab}=(\det A){\delta }_{ia}$. Thus for invertible $A$,Differentiating the cubic determinant formula gives three equal terms, so $\partial (\det A)/\partial A_{ab}=C_{ab}$. This is the entrywise derivative of the determinant and remains valid for singular $A$ through the cofactor expression. The inverse expression requires $\det A\ne 0$.