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矩阵乘积的运算法则的证明

矩阵乘积的运算法则

1? 乘法结合律：若A?Cm?n,B?Cn?p , C?Cp?q，则A(BC)?(AB)C.

2? 乘法左分配律：若A和B是两个m?n矩阵，且C是一个n?p矩阵，则

(A?B)C?AC?BC.

3? 乘法右分配律：若A是一个m?n矩阵，并且B和C是两个n?p矩阵，则A(B?C)?AC?BC.

4? 若?是一个标量，并且A和B是两个n?m矩阵，则?(A?B)??A??B.

证明1

①先设n阶矩阵为A?(aij),B?(bij), C?(cij),AB?(dij),BC?(eij)

?ABC?(fij),A(BC)?(gij),有矩阵的乘法得： dij?ai1b1j?ai2b2j???ainbnj.i,j?1,2?n eij?bi1c1j?bi2c2j???bincnj.i,j?1,2?n fij?di1c1j?di2c2j???dincnj.i,j?1,2?n gij?ai1e1j?ai2e2j???ainenj.i,j?1,2?n

故对任意i,j?1,2?n有：

fij?di1c1j?di2c2j???dincnj

?(ai1b11?ai2b21???ainbn1)c1j?

(ai1b12?ai2b22???ainbn2)c2j? ??(ai1b1n?ai2b2n???ainbnn)cnj ?ai1(b11c1j?b12c2j???b1ncnj)?

ai2(b21c1j?b22c2j???b2ncnj)?

??ain(bn1c1j?bn2c2j???bnncnj) ?ai1e1j?ai2e2j???ainenj

=gij

故(AB)C?A(BC)

②再看 A?(aik)mn ,B?(bkj)np,C?(cjt)pq, AB?(dij)mp , BC?(ekt)nq ,

A(BC)?(git)mq,

有矩阵的乘法得：

dij?ai1b1j?ai2b2j???ainbnj.i,j?1,2?n

ekt?bk1c1t?bk2c2t???bkpcpt.k?1,2?n,t?1,2?q fit?di1c1t?di2c2t???dipcpt.i?1,2?m,t?1,2?q git?ai1e1t?ai2e2t???ainent.i?1,2?m,t?1,2?q

故对任意的i?1,2?m, j?1,2?p, k?1,2?n, t?1,2?q有：

fit?di1c1t?di2c2t???dipcpt

?(ai1b11?ai2b21???ainbn1)c1t?

(ai1b12?ai2b22???ainbn2)c2t?

??(ai1b1p?ai2b2p???ainbnp)cpt ?ai1(b11c1t?b12c2t???b1pcpt)?

ai2(b21c1t?b22c2t???b2pcpt)?

??ain(bn1c1t?bn2c2t???bnpcpt)

6?ai1e1t?ai2e2t???ainent =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?ij??(Aik?Bik)Ckj

k ??AikCkj?k?BikCkj

k =(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??AikCkj?Ckj

k?Bikk ??(Aik?Bik)Ckj

k = ?(A?B)C?ij 故证出矩阵乘法左分配律. 证明4?

??a11a12?a1n?b12 设A?(a??a21a??b1122?a?2nij)mn??b22?????，B?(bbij)mn??21????am1am2?a??mn??bm1bm2??a11?b11a12?b12?a1n?b1n?可得A?B??a21?b21a22?b22?a2n?b?2n???????，

?am1?bm1am2?bm2?a?mn?bmn????(a11?b11)?(a12?b12)??(a1n?b1n)??(A?B)???(a?b21)?(a22?b22)??(a2n?b2n)?21?????????(a?m1?bm1)?(am2?bm2)??(amn?bmn)?b1n?b?2n???，b?mn???

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