Levi-Civita symbol with matrix

Question

What is the matrix form of $\epsilon_{ijk} A_{jk}$?

Answer 1

$\epsilon_{ijk} A_{jk}$ has only one free index (the $i$), assuming that repeated indices are summed. Hence, it is a vector. I assume you are working in 3D.

Define the vector $v_i$

$$v_i = \epsilon_{ifg} A_{fg}$$

and calculate its components by using the fact that, for example, $\epsilon_{112}=\epsilon_{133}=0$ (i.e. the $\epsilon$ is zero when you have the same index appearing twice):

$$ v_1 = \epsilon_{1fg} A_{fg} = \epsilon_{123} A_{23} + \epsilon_{132} A_{32} $$

You know that $\epsilon_{123}=1$ and $\epsilon_{132}=-1$. Therefore,

$$ v_1 = \epsilon_{1fg} A_{fg} = A_{23} - A_{32} $$

Similarly for the other two components $v_2$ and $v_1$.