余子式与代数余子式

,312213332112322311322113312312332211aaaaaaaaaaaaaaaaaa333231232221131211aaaaaaaaa例如3223332211aaaaa3321312312aaaaa3122322113aaaaa333123211333312321123332232211aaaaaaaaaaaaaaa一、余子式与代数余子式在阶行列式中，把元素所在的第行和第列划去后，留下来的阶行列式叫做元素的余子式，记作nijaij1nija.Mij，记ijjiijMA1叫做元素的代数余子式．ija例如44434241343332312423222114131211aaaaaaaaaaaaaaaaD44424134323114121123aaaaaaaaaM2332231MA.23M,44434241343332312423222114131211aaaaaaaaaaaaaaaaD,44434134333124232112aaaaaaaaaM1221121MA.12M,33323123222113121144aaaaaaaaaM.144444444MMA.行列式的每个元素分别对应着一个余子式和一个代数余子式引理一个阶行列式，如果其中第行所有元素除外都为零，那末这行列式等于与它的代数余子式的乘积，即．ijijAaDniijaija44434241332423222114131211000aaaaaaaaaaaaaD.14442412422211412113333aaaaaaaaaa例如证当位于第一行第一列时,ijannnnnaaaaaaaD21222211100即有.1111MaD又1111111MA,11M从而.1111AaD在证一般情形,此时nnnjnijnjaaaaaaaD1111100,1,2,1行对调第行第行行依次与第的第把iiiD得nnnjnnijiiijiaaaaaaaD1,1,11,11001ijaija1,2,1,Djjj再把的第列依次与第列第列第列对调得nnjnnjnijijiijjiaaaaaaaD1,,11,1,1110011ijannjnnjnijijiijjiaaaaaaa1,,11,1,12001nnjnnjnijijiijjiaaaaaaa1,,11,1,1001ijaijannnjnijnjaaaaaaaD1111100中的余子式.ijM在余子式仍然是中的在行列式元素ijnnjnnjnijijiijijaaaaaaaaa1,,11,1,100ijaija故得nnjnnjnijijiijjiaaaaaaaD1,,11,1,1001.1ijijjiMa于是有nnjnnjnijijiijaaaaaaa1,,11,1,100,ijijMaijaija定理３行列式等于它的任一行（列）的各元素与其对应的代数余子式乘积之和，即ininiiiiAaAaAaD2211ni,,2,1证nnnniniinaaaaaaaaaD212111211000000二、行列式按行（列）展开法则nnnninaaaaaaa2111121100nnnninaaaaaaa2121121100nnnninnaaaaaaa211121100ininiiiiAaAaAa2211ni,,2,1例13351110243152113D03550100131111115312cc34cc0551111115)1(330550261155526)1(315028.4012rr证用数学归纳法21211xxD12xx,)(12jijixx）式成立．时（当12n例2证明范德蒙德(Vandermonde)行列式1112112222121).(111jinjinnnnnnnxxxxxxxxxxxD)1(，阶范德蒙德行列式成立）对于假设（11n)()()(0)()()(0011111213231222113312211312xxxxxxxxxxxxxxxxxxxxxxxxDnnnnnnnnn就有提出，因子列展开，并把每列的公按第)(11xxi)()())((211312jjininnxxxxxxxxD).(1jjinixx223223211312111)())((nnnnnnxxxxxxxxxxxxn-1阶范德蒙德行列式推论行列式任一行（列）的元素与另一行（列）的对应元素的代数余子式乘积之和等于零，即.ji,AaAaAajninjiji02211,11111111nnnjnjininjnjnjjaaaaaaaaAaAa证行展开，有按第把行列式jaDij)det(,11111111nnniniininjninjiaaaaaaaaAaAa可得换成把),,,1(nkaaikjk行第j行第i,时当ji).(,02211jiAaAaAajninjiji同理).(,02211jiAaAaAanjnijiji相同关于代数余子式的重要性质;,0,,1jijiDDAaijnkkjki当当;,0,,1jijiDDAaijnkjkik当当.,0,1jijiij当，当其中例３计算行列式277010353D解27013D.27按第一行展开，得27005771030532004140013202527102135D例４计算行列式解0532004140013202527102135D660270132106627210.1080124220532414132525320414013202135215213rr122rr1.行列式按行（列）展开法则是把高阶行列式的计算化为低阶行列式计算的重要工具.;,0,,.21jijiDDAaijnkkjki当当;,0,,1jijiDDAaijnkjkik当当.,0,1jijiij当，当其中三、小结思考题阶行列式设nnnDn00103010021321求第一行各元素的代数余子式之和.11211nAAA思考题解答解第一行各元素的代数余子式之和可以表示成nAAA11211n001030100211111.11!2njjn