- #1

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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 [itex] \epsilon_{ijk} [/itex] 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?

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