Prove that the third invariant is equal to the determinant

1. Sep 8, 2009

blalien

1. The problem statement, all variables and given/known data
This is all in summation notation.
Given a 3x3 matrix $$A_{ij}$$, show that $$det[A]=1/6(A_{ii}A_{jj}A_{kk}+2A_{ij}A_{jk}A_{ki}-3A_{ij}A_{ji}A_{kk})$$

2. Relevant equations
I've been told that we're supposed to begin with
$$det[A]=1/6\epsilon_{ijk}\epsilon_{pqr}A_{ip}A_{jq}A_{kr}$$

3. The attempt at a solution
I hate to say this, but I have no idea where to start. I would really appreciate if somebody could just give me the first step and push me in the right direction.

2. Sep 9, 2009

Hurkyl

Staff Emeritus
Definitions are often a good place to start.

3. Sep 9, 2009

blalien

Never mind, I got it. You set $$A_{ij} = \delta_{ij}$$ and get the epsilons in terms of the deltas.

4. Sep 9, 2009

Hurkyl

Staff Emeritus
Why can you do that?

5. Sep 9, 2009

blalien

Well, that's not exactly right. But you can work a little index magic and come up with
$$\epsilon_{ijk}\epsilon_{pqr}=\left| \begin{matrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \end{matrix} \right|$$