Show that the permutation symbol, $\epsilon_{ijk}$ can be expressed as $\det\{\delta_{k,l}\}$ 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.
 A: 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
