0-2 -1-2 0 1 B= 2 0 20 -1 0 B 1 0 0 1 于是 AB= E 0 0 A B21 0 B23 (0 B12 B3 B21 421B12 A21B3+B23. 返回        =               − − − − = 2 1 2 3 1 2 1 3 0 0 1 0 0 1 2 0 0 1 0 1 2 1 0 2 1 2 B B B B B                 = 2 1 2 3 1 2 1 3 2 1 0 0 0 B B B B A E E AB 于是         + = 21 21 12 21 13 23 0 12 13 B A B A B B B B