解 4×7+3×2+1x1 232|=1×7+(-2)×2+3×1=6 5×7+7×2+0×1 (232|=(1×3+2x2+3×1)=(10) 2×(-1)2×2 24 1×(-1)1×2 12 3×(-1)3×2 36 21400-12 6-78 1-134月1-31 20-5-6 40-2 (5) 2 23 a3a23a3八x =(a1x1+a12x2+a1x3a2x+a2x2+a2x3a13x1+a2x2+a3x) x2=a1x2+a2x2+a3x3+2a12xx2+2a1x1x+22x2x 12101 01010 00210 0100 2-1012-4 00-43 0003人0 0 000-9 设A B 问 (1)AB=BA吗? (2)(A+B)2=A2+2AB+B2吗? (3)(4+B)(A-B)=A2-B吗? 解3 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? (2) 2 2 2 (A+ B) = A + 2AB + B 吗? (3) 2 2 (A+ B)(A− B) = A − B 吗? 解