# Tensor determinant using box product

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## Homework Statement

Using index notation only (i.e. don't expand any sums) show that:
\begin{align*}
&\text{(a) } \epsilon_{ijk} \det \underline{\bf{A}} = \epsilon_{mnp} A_{mi} A_{nj} A_{pk} \\
& \text{(b) } \det \underline{\bf{A}} = \epsilon_{mnp} A_{m1} A_{n2} A_{p3}
\end{align*}

## Homework Equations

\begin{align*}
&\text{(1) }\epsilon_{ijk} \det \underline{\bf{A}} = [\underline{\bf{A}} \underline{e}_i, \underline{\bf{A}} \underline{e}_j, \underline{\bf{A}} \underline{e}_k] \\
& \text{(2) } \underline{\bf{A}} = A_{ij} \underline{e}_i \otimes \underline{e}_j \\
& \text{(3) } \underline{\bf{A}} \underline{e}_j= (A_{pq} \underline{e}_p \otimes \underline{e}_q)\underline{e}_j
= (\underline{e}_q \cdot \underline{e}_j) A_{pq} \underline{e}_p = \delta_{qj} A_{pq} \underline{e}_p = A_{pj} \underline{e}_p
\end{align*}

## The Attempt at a Solution

I got part (a) by using (3) for each term of the box product, substituting each of those expressions into the box product, and evaluating.

What baffles me is (b). If I multiply $\epsilon_{ijk}$ times (b) and compare to (a), the net result is:
\begin{align*}
&\epsilon_{mnp} (\epsilon_{ijk} A_{m1} A_{n2} A_{p3}) = \epsilon_{mnp} (A_{mi} A_{nj} A_{pk}) \\
& \epsilon_{ijk} A_{m1} A_{n2} A_{p3} = A_{mi} A_{nj} A_{pk}
\end{align*}

But I have no idea what to do with that (the indices look really goofy). Any tips?

Last edited:

\begin{align*}
&\epsilon_{mnp} (\epsilon_{ijk} A_{m1} A_{n2} A_{p3}) = \epsilon_{mnp} (A_{mi} A_{nj} A_{pk}) \\
& \epsilon_{ijk} A_{m1} A_{n2} A_{p3} = A_{mi} A_{nj} A_{pk}
\end{align*}
You can't go from the first line to the second line by dividing by ##\epsilon_{mnp}## throughout - because there is an summation involved.

To arrive at (b) from (a), consider the value of ##\epsilon_{123}##

Homework Helper
Ah, thanks. I was applying standard algebra techniques, which I now see is a mistake. Gosh, the answer is soooooo simple! Thanks.

P.S. I'd fiddled with this thing for probably 2 hours. At one point I was on the right track by looking at how to eliminate $\epsilon_{ijk}$ from (a) to isolate the determinant. I thought about multiplying both sides by $\epsilon_{ijk}$ to get $6 \det(A) = xxx$ but that didn't look promising. Drat!

Last edited:
Haha, I guess the trick is to realise that the only way you can have explicit labels such as 1,2 and 3 appearing in the indices is by setting them to be so - multiplying by other stuff with arbitrary indices can never give you specific indices.