矩阵乘积的运算法则的证明

1、矩阵乘积的运算法则的证明精品资料矩阵乘积的运算法则1 乘法结合律：若2乘法左分配律：若(A+B)C =AC +BC .A 亡 Cm 同，B<CnM , C忘 Cp 鸿，贝U A(BC) = (AB)C.A和B是两个 mxn矩阵，且C是一个nx p矩阵，则3"乘法右分配律：若A是一个 mxn矩阵，并且B和C是两个nx p矩阵，则A(B +C) =AC +BC .4若a是一个标量,并且 A和B是两个nm矩阵，则a (A + B)=aA+aB.证明1 先设 n 阶矩阵为 A=(aij),B =(bij), C = (q ), AB = (% ), BC =(列)ABC =(fij)

2、,A(BC) =(gij),有矩阵的乘法得:djeii= ai1b1j +ai2b2j + +3山匕可.i, j = 1,2n= bi1C1j +bi2C2j + +binCnj.i, j =1,2nfij= diiCij +di2C2ji +dinCnj.i,j =1,2ngj-ai1e1 ai2e2j中十 a inenj .i , j= 1,2 n故对任意i,j =1,2n 有:fij =di1C1j +di2C2j+ dinCnj= (ai1b11 +ai2b21+ainbn1)C1(ai1b12 中 ai2b22 中 十 ainbn2)C2j 中十佝4 +ai2b2n + +ainbn

3、n)Cnj=ai1 (biici j +bi2c2j + + binCnj ) +ai2 (b21C1j + b22C2+ b2ncnj ) +"*'ain (bn1C1j "*'bn2C2j*'bnnCnj )= aiieij +ai2e2j +中 ainenj= gij故(AB)C = A(BC)再看 A =(aik )mn , B = (bkj )n p , C = (Cjt )p q, AB = (d j ) m p , BC=(ekt)nqA(BC) =(git)mq,有矩阵的乘法得:dij=aiibi j+ ai2b2jiekt= bki

4、cit+ bk2C2tfit=d ii cit+ di2C2t+git=aii eit中 ai2e2t+故对任意的i =1,2m,fit= diiCit"*'di2C2t+-+ainbnj.i, j =i,2n+ bkpCpt.k =i,2n,t =i,2q+ dipCpt.i =1,2m,t =1,2q+ ain ent .i =1,2 m, t =1,2 qj =1,2p, k=1,2n, t=1,2q 有:+ di pCpt= (aiibii +ai2b2i +十玄禹勺心+(ailbi2 +印2&22 + ainbn2 )c2t +-+ (aiibi p + a

5、i2b2 p + ainbn p)Cptaii (biiCit +bi2C2t +bipCpt) +32(&21卬中 b22C2t 中*" b2 pCp t) +ain(bniCit +bn2C2t+bn pCpJaiieit +ai2e2tainent=gij故(AB)C = A(BC)证明2设Aij表示矩阵 A的第i行,第j列上的元素，则有(a + B)C j(Aik + Bik )CkjAik Ckj + 送 Bik Ckjk= (AC)ij +(BC)ij故证出矩阵乘法左分配律证明3同理矩阵乘法左分配律可得(AC)ij +(BC)ij=送 AikCkj BikCkjk

6、k=2( Aik 中 Bik )Ckjk(A + B)C故证出矩阵乘法左分配律证明4设 A =(aij)mnraiiai2 ainfbibi2bin "a21 a22 a2n，B=(bij) mn =b2ib22b2nhhhhh: ，Lam1a m2 amn_bmibm2bmn ”aii +biiai2+ bi2 ain +b,na21 +b2iba22+ b22h a2n +b2nb|_am1 + bmiam2+ bm2 amn + bmn .a(A+B)=ot(aii +bii) a(a2i +b2i)«(ai2Ct 22讥)+ b22)a (am + bin) Ct (a2n + b2n )otA =a (ami +bmi) giaa2iaA +aB =“2 gn 'fabiiaga bin 'aa22aa2nab2iab22ab2n,a Bhh«am2 amnLabmi«bm2 a bmnlii +bii)a(ai22) a(ain +bin) 1Si + b2i)Ct (a22 + b22) a(a2n + b2n )+ bm2)a (amn +