(後只人季 则(11)的解为 12 21 22 11 21 21则（1.1）的解为 , 2 1 2 2 1 1 1 2 2 2 2 1 1 2 1 1 a a a a b a b a A A x = = . 2 1 2 2 1 1 1 2 2 1 2 1 1 1 2 2 a a a a a b a b A A x = = (1.4)