0 0 6 0 2 2 0 3 0 解 4 7 3 X 2 Ix X 2 3 5 十 7 2 0 4 9 (2) 2 2 2 3 1) (一 2 2 2 (3 2 2 3 3 X 2 0 6 8 3 3 0 2 5 4 0 12a13x1 x3 12a22m 3 十 a12C1+a22X2+a23k3a13X1+a23x2+a3x3 xe1 十 2 2 2 x 2 0 0 0 2 2 0 0 0 3 3 0 5 A B 问 A B B 吗2 (6)               − − −               0 0 0 3 0 0 2 3 0 1 2 1 1 0 3 1 0 0 0 3 0 0 2 1 0 1 0 1 1 2 1 0 . 解 (1)                     − 1 2 7 5 7 0 1 2 3 4 3 1            +  +   + −  +   +  +  = 5 7 7 2 0 1 1 7 ( 2) 2 3 1 4 7 3 2 1 1           = 49 6 35 (2) ( )           1 2 3 1 2 3 = (1 3 + 2 2 + 3 1) = (10) (3) ( 1 2) 3 1 2 −                      −   −   −  = 3 ( 1) 3 2 1 ( 1) 1 2 2 ( 1) 2 2           − − − = 3 6 1 2 2 4 (4)               − − −         − 4 0 2 1 3 1 0 1 2 1 3 1 1 1 3 4 2 1 4 0         − − − = 20 5 6 6 7 8 (5) ( )                     3 2 1 13 23 33 12 22 23 11 12 13 1 2 3 x x x a a a a a a a a a x x x ( ) = a11 x1 + a12 x2 + a13 x3 a12 x1 + a22 x2 + a23 x3 a13 x1 + a23 x2 + a33 x3            3 2 1 x x x 12 1 2 13 1 3 23 2 3 2 33 3 2 22 2 2 = a11 x1 + a x + a x + 2a x x + 2a x x + 2a x x (6)               − − −               0 0 0 3 0 0 2 3 0 1 2 1 1 0 3 1 0 0 0 3 0 0 2 1 0 1 0 1 1 2 1 0               − − − = 0 0 0 9 0 0 4 3 0 1 2 4 1 2 5 2 5．设         = 1 3 1 2 A ,         = 1 2 1 0 B ，问： (1) AB = BA 吗?