The magic square game is a two-player pseudo-telepathy game that was presented by Padmanabhan Aravind, who built on work by Mermin. In the magic square we have ones in columns (odd number) and rows (even number).
According to https://arxiv.org/abs/quant-ph/0407221v3
$$ A2= \frac12\left[ {\begin{array}{ccccc} i & 1 & 1 & i \\-i & 1 & -1 & i\\i & 1 & -1 & -i\\-i & 1 & 1 & -i \end{array} } \right] $$
$$ B3= \frac1{\sqrt{2}}\left[ {\begin{array}{ccccc} 1 & 0 & 0 & 1 \\-1 & 0 & 0 & 1\\0 & 1 & 1 & 0\\0 & 1 & -1 & 0 \end{array} } \right] $$
We have an input the entangled state shared by Alice and Bob
$ \mid \psi \rangle = \frac{1}{2}\mid0011 \rangle -\frac{1}{2}\mid0110 \rangle -\frac{1}{2}\mid1001 \rangle +\frac{1}{2}\mid1100 \rangle$
Consider for example inputs x =2 and y =3. After Alice and Bob apply A2 and B3 respectively, the state evolves to $$ A2 \otimes B3 \mid \psi\rangle = \frac{1}{2\sqrt{2}} \left[\mid0000\rangle -\mid0010\rangle -\mid0101\rangle +\mid 0111\rangle +\mid 1001\rangle +\mid 1011\rangle -\mid 1100\rangle -\mid 1110\rangle \right] $$
Question is how to obtain this result. I did multiplication of the matrices
$ +(A2 \otimes B3) \mid 00 \rangle \otimes \mid 11 \rangle $
$ -(A2 \otimes B3) \mid 01 \rangle \otimes \mid 10 \rangle $
$ -(A2 \otimes B3) \mid 10 \rangle \otimes \mid 01 \rangle $
$ +(A2 \otimes B3) \mid 11 \rangle \otimes \mid 00 \rangle $
Calculate the tensor product
$ +(A2 \mid 00 \rangle \otimes B3\mid 11 \rangle $ part 1
$ -(A2 \mid 01 \rangle \otimes B3\mid 10 \rangle $ part 2
$ -(A2 \mid 10 \rangle \otimes B3\mid 01 \rangle $ part 3
$ +(A2 \mid 11 \rangle \otimes B3\mid 00 \rangle $ part 4
Let's calculate part 2 with step 1 and step 2
$$ step 1 = A2 | 01 \rangle = \left[ {\begin{array}{ccccc} i & 1 & 1 & 1 \\-i & 1 & -1 & 1\\i & 1 & -1 & -i\\-i & 1 & 1 & -i \end{array} } \right] \left[ {\begin{array}{c} 0 \\ 1 \\ 0 \\ 0\end{array}} \right] = \left[ {\begin{array}{c} 1 \\ 1 \\ 1 \\ 1\end{array}} \right] $$
$$ step 2 = B3 | 10 \rangle = \left[ {\begin{array}{ccccc} 1 & 0 & 0 & 1 \\-1 & 0 & 0 & 1\\0 & 1 & 1 & 0\\0 & 1 & -1 & 0 \end{array} } \right] \left[ {\begin{array}{c} 0 \\ 0 \\ 1 \\ 0\end{array}} \right] = \left[ {\begin{array}{c} 0 \\ 0 \\ 1 \\ -1\end{array}} \right] $$
$ step1 \otimes step2 = \left[ {\begin{array}{c} 0 \\ 0 \\ 1 \\ -1 \\ 0 \\ 0 \\ 1 \\ -1 \\ 0 \\ 0 \\ 1 \\ -1 \\0 \\ 0\\ 1 \\ -1 \end{array}} \right] $
Is this the way to go?