This question relates to this continuum mechanics online course. So far it has excellent lectures and problem sets but no solutions.
Problem 3.
Show that the permutation symbol, $\epsilon_{ijk}$ can be expressed as:
$$ \epsilon_{ijk} = \begin{vmatrix}
\delta_{i1} & \delta_{i2} & \delta_{i3} \\
\delta_{j1} & \delta_{j2} & \delta_{j3} \\
\delta_{k1} & \delta_{k2} & \delta_{k3} \\
\end{vmatrix}$$
I am also posting my answer. I am working on my rigorous mathematical writing as well as the continuum mechanics, so I would appreciate feedback of my proof-writing style as well.
Best Answer
Starting with the conjecture:
$$ \epsilon_{ijk} = \begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}$$
It should be true that:
$$ \epsilon_{ijk}A_iB_jC_k = \begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}A_iB_jC_k$$
We can expand the left hand side of the statement to be the following, keeping in mind that even though there are 27 permutations of $ijk$, only the following six of them are non-zero.
$$\epsilon_{ijk}A_iB_jC_k = A_1B_2C_3 + A_2B_3C_1 + A_3B_1C_2 - A_3B_2C_1 - A_2B_1C_3 - A_1B_3C_2$$
Likewise expanding the right hand side of statement:
$$\begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}A_iB_jC_k = (\delta_{i1}\delta_{j2}\delta_{k3} + \delta_{i2}\delta_{j3}\delta_{k1} + \delta_{i3}\delta_{j1}\delta_{k2} - \delta_{i3}\delta_{j2}\delta_{k1} - \delta_{i2}\delta_{j1}\delta_{k3} - \delta_{i1}\delta_{j3}\delta_{k2})A_iB_jC_k$$
$$\begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}A_iB_jC_k = A_1B_2C_3 + A_2B_3C_1 + A_3B_1C_2 - A_3B_2C_1 - A_2B_1C_3 - A_1B_3C_2$$
$$\therefore$$
$$\epsilon_{ijk}A_iB_jC_k = \begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}A_iB_jC_k$$
$$\epsilon_{ijk} = \begin{vmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \\ \delta_{j1} & \delta_{j2} & \delta_{j3} \\ \delta_{k1} & \delta_{k2} & \delta_{k3} \\ \end{vmatrix}$$
QED