Show that
$\epsilon_{ijk} A_{il} A_{jm} A_{kn} = \det(A) \epsilon_{lmn}$
where $\epsilon$ epsilon is the standard Levi-Civita symbol and A is a three dimensional matrix.
I found the above property in a book used for my course about turbulence, where it is used to prove that the $\epsilon$ is isotropic. The proof for the property however is not provided, and I'm struggling to prove it myself.
Thanks.
Best Answer
Consider the following matrix $$ \mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$ Then the $\det(\mathbf{A})$ can be written as \begin{align} \det(\mathbf{A}) &= (-1)^0a_{11}(a_{22}a_{33}-a_{23}a_{32})+(-1)^1a_{12}(a_{21}a_{33}-a_{23}a_{31})\\ &+(-1)^2a_{13}(a_{21}a_{32}-a_{22}a_{31}) \end{align} Now, the minor of a determinant is denoted as $A_{nm}$ where $n,m$ correspond to the row and column of the deleted entries. Therefore, $a_{22}a_{33}-a_{23}a_{32} = A_{11}$. What are the other minors? Recall that $$ \epsilon_{ijk} = \begin{cases} 0, & \text{when } i = j\text{ or } i = k\text{ or } j = k\\ 1, & \text{if we have an even permutation}\\ -1, & \text{if we have an odd permutation} \end{cases} $$ Can you figure the rest out now?