例3A= B AB [注]AB≠BA;A≠O,B≠O,但是BA=O 算律:(1)(Am,B,n)Cm=A(BC (2)A(Bn+Cen=AB+Ac (A +B) Cen= AC +BC (3)k(A Bm)=(kA)B=A(kB) (4)Em A=A, AE=A 验证(1)设A=(an)m,B=(b)n,C=(cn)m,则 BCl=∑ab akber a, but t=1(k=1 C 4(BC刀=[n C ∑b,c ∑ akb,=AB)Cl Vi,J) y 应用:A=4a2 b2 y24 例 3       = 1 2 1 2 A ,       − − = 1 1 1 1 B       − − = 1 1 1 1 AB ,       = 0 0 0 0 BA [注] AB  BA ； A  O, B  O, 但是 BA = O． 算律：(1) (A B )C A(BC) ms sn nl = (2) Ams (Bsn +Csn ) = AB + AC (Ams + Bms )Csn = AC + BC (3) k(A B ) (kA)B A(kB) ms sn = = (4) Em Amn = A, AmnEn = A 验证(1) 设 A = aij ms ( ) , B = bij sn ( ) , ij n l C c =  ( ) , 则                 =   = = nj j s k i k kn s k ij i k k c c AB C a b  a b  1 1 1 1 [( ) ] tj n t s k ik kt  a b c = =       = 1 1                 =   = = n t st tj n t t tj ij i i s b c b c A BC a a 1 1 1 1 [ ( )]     = =       = s k n t ik kt tj a b c 1 1 tj n t s k ik kt  a b c = =       = 1 1 AB C ij = [( ) ] ( i, j) 应用：             = m m mn n n a a a a a a a a a A       1 2 21 22 2 11 12 1 ,             = n x x x x  2 1 ,             = mb b b b  2 1 ,             = m y y y y  2 1