How to prove that determinant with permutation symbols How to prove that 
$$\varepsilon_{ijk}a_{i\ell}a_{jm}a_{kn} = \det[a]\epsilon_{\ell mn}$$
I'm trying to solve this problem with permutation symbol but i can't solve it 
Help me,please. Thank you 
(Original screenshot of formula)
 A: Here's is the typical physicist's (and maybe even a mathematician's) way to do this. If we exchange any two of the $l,m,n$ indices in $\epsilon_{ijk}a_{il}a_{jm}a_{kn}$, we get the following for, say, $l \leftrightarrow m$:
$$\epsilon_{ijk}a_{im}a_{jl}a_{kn}$$
$$=\epsilon_{jik}a_{jm}a_{il}a_{kn}$$
$$=-\epsilon_{ijk}a_{jm}a_{il}a_{kn}$$
$$=-\epsilon_{ijk}a_{il}a_{jm}a_{kn}$$
where in going to the second line we have renamed the indices $i\rightarrow j$, $j\rightarrow i$ independently/simultaneously, in going to the third line we used the anti-symmetry of the Levi-Civita symbol and in going to the last line we just exchanged $a_{jm}$ with $a_{il}$. For any two indices being the same, like $l=m$ we get
$$\epsilon_{ijk}a_{im}a_{jm}a_{kn}$$
$$=-\epsilon_{jik}a_{im}a_{jm}a_{kn}$$
$$=-\epsilon_{ijk}a_{jm}a_{im}a_{kn}$$
$$=-\epsilon_{ijk}a_{im}a_{jm}a_{kn}$$
$$=0$$
where in going to the second line we used the anti-symmetry of $\epsilon_{ijk}$, in going to the third line we renamed the indices $i\rightarrow j$, $j\rightarrow i$ independently/simultaneously, in going to the fourth line we exchanged $a_{jm}$ with $a_{im}$. The end result comes from the fact that the the first line is equal to minus itself.
These results are true for any exchange between the indices $l,m,n$. Hence, we can write
$$\epsilon_{ijk}a_{il}a_{jm}a_{kn}=A\ \epsilon_{lmn}$$
for some constant $A$. To determine the constant, we can set $l=1,m=2,n=3$. Doing so gives
$$A=\epsilon_{ijk}a_{i1}a_{j2}a_{k3}$$
For clarity, let's use the following notation: $a_{\mu\nu}=(\vec{a}^{(\nu)})_\mu$, which implies that it is the $\mu-$th component of the vector $\vec{a}^{(\nu)}$. With this notation, we can re-write the above result as:
$$A=\epsilon_{ijk}(\vec{a}^{(1)})_i (\vec{a}^{(2)})_j (\vec{a}^{(3)})_k=\det[a]$$
where $a$ is defined as
$$a=\left[\vec{a}^{(1)}\ \vec{a}^{(2)}\ \vec{a}^{(3)} \right]$$
i.e. a matrix whose rows are the vectors $\vec{a}^{\nu}$. So, the final result is what we wanted to reach, namely
$$\epsilon_{ijk}a_{il}a_{jm}a_{kn}=\det[a]\ \epsilon_{lmn}$$
Appendix:
We have used $\epsilon_{ijk}(\vec{a}^{(1)})_i (\vec{a}^{(2)})_j (\vec{a}^{(3)})_k=\det[a]$. To make it easier to see why this is true, in case that a reader might not know why this is so, we use yet another notation for the case of $\max(i)=3$:  $\vec{a}^{(1)}=\vec{u}$, $\vec{a}^{(2)}=\vec{v}$, $\vec{a}^{(3)}=\vec{w}$.
A very elementary result for determinants is that $\det[a]=\det[\vec{u}\ \vec{v}\ \vec{w}]=\vec{u}\cdot(\vec{v}\times\vec{w})$. Using index notation, we write the last part as
$$\vec{u}\cdot(\vec{v}\times\vec{w})=u_i(\vec{v}\times\vec{w})_i=u_i\epsilon_{ijk}v_jw_k=\epsilon_{ijk}u_iv_jw_k$$
which is exactly what we have.
A: Hint: Look at the left-hand side. What happens if you swap two of $l,m,n$ around? What does this tell you about the final answer?
