2a+c=1, 2b+d=0. C b=1, d=2. 又因为 AB BA 0 2 0-1 所以       − = − = + = + =  1, 0, 2 0, 2 1, b a b d a c        = = = − =  2. 1, 1, 0, d c b a 又因为       − 1 0 2 1       − 1 2 0 1       − 1 0 2 1 =       − 1 2 0 1 , 0 1 1 0       = 所以 . 1 2 0 1 1       − = − A AB BA