# 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
$$\epsilon_{ijk}\epsilon_{pqr}=\left| \begin{matrix} \delta_{ip} & \delta_{iq} & \delta_{ir} \\ \delta_{jp} & \delta_{jq} & \delta_{jr} \\ \delta_{kp} & \delta_{kq} & \delta_{kr} \end{matrix} \right|$$