Prove Determinant Using the Triple Scalar Product

1. Aug 30, 2015

ThirdEyeBlind

1. The problem statement, all variables and given/known data
I'm supposed to prove $det A = \frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{kr}$ using the triple scalar product.

2. Relevant equations
$\frac{1}{6} \epsilon_{ijk} \epsilon_{pqr} A_{ip} A_{jq} A_{ kr}$
$(\vec u \times \vec v) \cdot \vec w = u_i v_j w_k \epsilon_{ijk}$

3. The attempt at a solution
I have written out and understand that $det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k}$ but I don't understand where the triple scalar product comes into play.

I see the similarity between the shorthand notations of the triple scalar product and the determinant but don't see how I can relate the two. I figure I must be missing some geometric/mathematical interpretation that can help me.

I've used Einstein summation notation in a couple previous problems but this is really the first involved problem I have to do and am a bit lost as to what to do.

2. Aug 30, 2015

techmologist

Both the scalar triple product and the determinant (in 3 dimensions) have the geometrical interpretation of volume. $(\vec u \times \vec v) \cdot \vec w$ is the volume of the prism formed by vectors u, v, and w. They are probably expecting you to make the identification

$$\det A \quad = (\vec{A}_1 \times \vec{A}_2) \cdot \vec{A}_3 ,$$

where, for example, $\vec{A}_1$ is the first row in matrix A. The determinant of A is the volume spanned by its three row vectors.

That gets you up to the point where you are, namely $\det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k}$. To get the rest of it, note that

$$\det A = \epsilon_{ijk} A_{1i} A_{2j} A_{3k} = -\epsilon_{ijk} A_{2i} A_{1j} A_{3k} = \epsilon_{213}\epsilon_{ijk} A_{2i} A_{1j} A_{3k}$$

for example.