I'm solving problem 3.D in H. Georgi *Lie Algebra etc* for fun where one is to compute the matrix elements of the direct product $\sigma_2\otimes\eta_1$ where $[\sigma_2]_{ij}\text{ and }[\eta_1]_{xy}$ are two different Pauli matrices in two different two dimensional spaces. 

Defining the basis in our four dimensional tensor product space 
$$\tag{1}\left|1\right\rangle = \left|i=1\right\rangle\left|x=1\right\rangle\\
\left|2\right\rangle = \left|i=1\right\rangle\left|x=2\right\rangle\\
\left|3\right\rangle = \left|i=2\right\rangle\left|x=1\right\rangle\\
\left|4\right\rangle = \left|i=2\right\rangle\left|x=2\right\rangle$$

Now we know that when we multiply representations, the generators add in the sense of 

$$\tag{2}[J_a^{1\otimes2}(g)]_{jyix} = [J_a^1]_{ji}\delta_{yx} +\delta_{ji}[J_a^2]_{yx}, $$ where the $J$s are the generators corresponding to the different representations $D_1$ and $D_2$ ($g$ stands for the group elements). 

Using all of this I find that in the basis of $(1)$ the matrix representation of the tensor product is given by 

$$\tag{3}\sigma_2\otimes\eta_1 = \begin{pmatrix} 
0 & 1 & -i & 0 \\
1 & 0 & 0 & -i \\ 
i & 0 & 0 & 1 \\
0 & i & 1 & 0 \end{pmatrix}$$

I am not asking you to redo the calculations$^*$ for me but does $(3)$ make sense? 

Appendix: 
The Pauli matrices
$$\begin{align}
  \sigma_1 &=
    \begin{pmatrix}
      0&1\\
      1&0
    \end{pmatrix} \\
  \sigma_2  &=
    \begin{pmatrix}
      0&-i\\
      i&0
    \end{pmatrix} \\
  \sigma_3 &=
    \begin{pmatrix}
      1&0\\
      0&-1
    \end{pmatrix} \,.
\end{align}
$$ 
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$*$ Or perhaps you know the answer already.