Spinning Matrix

We need to solve the equations $\begin{array}{rclcrcl} a^2 + bc &= & c & \hspace{2cm} & b(a+d) & = & a \\ c(a+d) & = & d & & bc + d^2 & = & b \end{array}$ Thus $a^2 - d^2 \,=\, c-b$ and $(b-c)(a+d) \,=\, a-d$ , so that $\begin{aligned} (a^2-d^2)(a+d) & = \; (c-b)(a+d) \; = \; d-a \\ (a-d)\big[(a+d)^2+ 1\big] & = \; 0 \end{aligned}$ and hence $a=d$ , and so $b = c = a^2+ bc$ . Thus we have $a=d$ and $b=c$ and the equations $a^2 \; = \; b(1-b) \hspace{2cm} 2ab = a$ Thus we can have $a=0$ , in which case $b=0,1$ , or we can have $b=\tfrac12$ , in which case $a = \pm\tfrac12$ . Thus there are $\boxed{4}$ posssible matrices: $\left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right) \hspace{1cm} \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \hspace{1cm} \left(\begin{array}{cc} \tfrac12 & \tfrac12 \\ \tfrac12 & \tfrac12 \end{array}\right) \hspace{1cm} \left(\begin{array}{cc}-\tfrac12 & \tfrac12 \\ \tfrac12 & -\tfrac12 \end{array}\right)$