(1)A= B 12 12 则AB: Ba= AB≠B4 46 38 (2)(A+B)2 22Y22)(814 25八25 1429 但A2+2AB+B 38 68 1016 1)(812)(3 1527 故(A+B)2≠A2+2AB+B2 2202)(06 (3)(A+B)(A-B)= 25八01 09 10 28 而 42-B2= 故 (A+B)(A-B)≠A2-B2 6.举反列说明下列命题是错误的: (1)若A2=0,则A=0; 2)若A2=A,则A=0或A=E; (3)若AX=AY,且A≠0,则X=Y. 解(1)取A= A2=0,但A≠0 (2)取A A2=A,但A≠0且A≠E 00 (3)取A= X 00 11 AX=AY且A≠0但X≠Y 7.设A=/0) ,求A2,A,,A 见1 1010 解A 元1人x1)(2x 1010 21八 324 (1)         = 1 3 1 2 A ,         = 1 2 1 0 B 则         = 4 6 3 4 AB         = 3 8 1 2 BA  AB  BA (2)                 + = 2 5 2 2 2 5 2 2 ( ) 2 A B         = 14 29 8 14 但 + + = 2 2 A 2AB B         +        +        3 4 1 0 8 12 6 8 4 11 3 8         = 15 27 10 16 故 2 2 2 (A+ B)  A + 2AB + B (3) (A+ B)(A− B) = =                0 1 0 2 2 5 2 2         0 9 0 6 而 − = 2 2 A B =        −        3 4 1 0 4 11 3 8         1 7 2 8 故 2 2 (A+ B)(A− B)  A − B 6．举反列说明下列命题是错误的: （１）若 0 2 A = ,则 A = 0 ; （２）若 A = A 2 ,则 A = 0 或 A = E ; （３）若 AX = AY ,且 A  0 ,则 X = Y . 解 (1) 取         = 0 0 0 1 A 0 2 A = ,但 A  0 (2) 取         = 0 0 1 1 A A = A 2 ,但 A  0 且 A  E (3) 取         = 0 0 1 0 A         − = 1 1 1 1 X         = 0 1 1 1 Y AX = AY 且 A  0 但 X Y 7．设         = 1 1 0  A ,求 k A ,A , ,A 2 3  . 解         =                = 2 1 1 0 1 1 0 1 1 0 2    A         =                = = 3 1 1 0 1 1 0 2 1 1 0 3 2    A A A