a12a1 (5)(x x2 x3)a2 a22 a23x2 q12a1 (x x, x3 22a2 a3a23a3人x M =(a1x+012X2+01x301X+0x+023X①13x1+02X2+03x3)x2 =a1x2+a2x2+a32x2+2a12x1x2+2a1x2+2a22x3 3.举反列说明下列命题是错误的 (1)若A2=0,则A=0; 解取A=0 0 则A2=0,但A≠0 2)若A2=A,则A=0或A=E; 解取4(0)0但0且ME (3)若AX=Ay,且A≠0,则XY(5)                 3 2 1 332313 232212 131211 321 )( x x x aaa aaa aaa xxx  解                 3 2 1 332313 232212 131211 321 )( x x x aaa aaa aaa xxx (a11x1a12x2a13x3 a12x1a22x2a23x3 a13x1a23x2a33x3)         3 2 1 x x x 322331132112 2 333 2 222 2 111      222 xxaxxaxxaxaxaxa  3 举反列说明下列命题是错误的 (1)若 A2 0 则 A0 解 取        00 10 A  则 A2 0 但 A0 (2)若 A2 A 则 A0 或 AE 解 取        00 11 A  则 A2 A 但 A0 且 AE (3)若 AXAY 且 A0 则 XY  解 取