Shahn Majin and Xavier Gomez say in the beginig of their article (Noncommutative cohomology and electromagnetism on $\mathbb{C}_q [SL_2]$ at roots of unity) that tha action of left $\mathbb{C}_q [SL_2]$-crossed modules is given by: \begin{eqnarray} \nonumber a\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} q e_a +q\mu^2 e_d &e_b\\ e_c&q^{-1}e_d\\ \end{array} \right) \end{eqnarray}

\begin{eqnarray} \nonumber b\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} \mu e_c &\mu e_d\\ 0&0\\ \end{array} \right) \end{eqnarray}

\begin{eqnarray} \nonumber c\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} \mu e_b &0\\ q\mu e_d&0\\ \end{array} \right) \end{eqnarray} \begin{eqnarray} \nonumber d\triangleright \left(\begin{array}{cc} e_a&e_b\\ e_c&e_d\\ \end{array} \right)= \left(\begin{array}{cc} q^{-1} e_a &e_b\\ e_c&q e_d\\ \end{array} \right) \end{eqnarray}

My question is how (or where) can we find the details of this result ? Thank you