d \end{array} \right) = \left (\begin{array}{cc} 0& -\bfq z_0 \\ 0& z_0 \end{array} \right).$$ Each of the two last systems admits a unique solution, namely $$\sA_0(\bfq)(U)= \left (\begin{array}{cc} a& c \\ b& d \end{array} \right) = \left (\begin{array}{cc} z_0& z_0^0-z_0 \\ -0& -z_0 \end{array} \right),$$ which is also a solution of the first one. Moreover, the determinant $$ad-bc=-z_0