I want to prove the following identity: if $A$ is a $3\times 3$ matrix, then $$\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=\det(A)\varepsilon_{pqr},$$ where $\varepsilon$ is the Levi-Civita symbol.
I know that (Leibniz formula) $$\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}a_{1i}a_{2j}a_{3k}=\det(A).$$ However I cant see why $$\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{ijk}a_{pi}a_{qj}a_{rk}=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3\varepsilon_{pqr}\varepsilon_{ijk}a_{1i}a_{2j}a_{3k}.$$
It's easy to see that it's true for $p=1, q=2, r=3$. From this, we can see that it's true when $\epsilon_{pqr}=1$. Can you see why? If $\epsilon_{pqr}=-1$, we will get $-\det(A)$. To see why, let $p=2, q=1, r=3$. The lhs is now $\epsilon_{ijk}a_{2i}a_{1j}a_{3k}=\epsilon{ijk}a_{1j}a_{2i}a_{3k}=-\epsilon_{jik}a_{1j}a_{2i}a_{3k}$. What happens when $\epsilon_{pqr}=0$?