Proof with the Levi-Civita tensor

Question

This may seem like an amateur question at best, but I've just started to get my hands dirty with index notation and tensor calculus and am having some trouble (or overthinking) trying to prove the following identity:

$\epsilon_{ijk}a_{ip}a_{jq}a_{kr}=\epsilon_{pqr}det(A)$

What I've done is expand out the left side and used the following identity to replace the $det(A)$ in the right side:

$det(A)=\epsilon_{ijk}a_{i1}a_{j2}a_{k3}$

which gives me:

$\epsilon_{ijk}a_{ip}a_{jq}a_{kr}=\epsilon_{pqr}\epsilon_{ijk}a_{i1}a_{j2}a_{k3}$

I'm unsure how to proceed from here. Can anyone provide some suggestion(s)?

Answer 1

• Both of $\epsilon_{ijk}A_{ip}A_{jq}A_{kr}$ and $\epsilon_{pqr}\epsilon_{ijk}A_{i1}A_{j2}A_{k3}$ are antisymmetric in $p$, $q$ and $r$.
• In three dimensions, any tensors antisymmetric in three indices are proportional to each other*.
• Setting $p=1$, $q=2$ and $r=3$ we find that the two tensors are equal.

*Consider any tensor $a_{pqr}$ with three indices such that swapping any two negates it, i.e. $a_{pqr}=−a_{qpr}=−a_{prq}=−a_{rqp}$. Then the $123$, $231$ and $312$ components are equal, the $132$, $213$ and $321$ components are equal to the negation of this, and the remaining components are zero (because swapping two equal indices shows that that component is equal to its own negation). So we specify the whole tensor just by specifying $a_{123}$.