2a+c=1, 2b+d=0, b=-1 今 c=1, b=1, d=2. 又因为 AB BA 21/0 0-121 10八12 所以 4 125  − = − = + = + =  1, 0, 2 0, 2 1, ba b d a c  === −=  2. 1,1 , 0 , dcba 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