Let  $A\text{ and }B$are two square idempotent matrices such that $AB\pm BA$ is a null matrix. Then value of $\det (A-B)$ can be equal

 $$\begin{gathered} {(A-B)}^2=A^2-AB-BA+B^2 \\ =A+B[\because AB+BA=0\text{ and }{A}^2=A,B^2=B] \\ {|A-B|}^2=|A+B| \end{gathered}$$
 And  ${(A+B)}^2=A^2+AB+BA+B^2$
 $$\begin{gathered} =A+B[\because AB+BA=0\text{ and }{A}^2=A,B^2=B] \\ {|A+B|}^2=|A+B| \\ \Rightarrow |A+B|(|A+B|-1)=0 \\ |A+B|=0,1 \end{gathered}$$
 
  From, eq (1)
 ${|A-B|}^2=0,1\Rightarrow |A-B|=0;\pm 1$
 $\det (A-B)=0,-1,1$