1-3行列式按行(列)展开

第一章行列式中南财经政法大学信息系,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.个代数余子式对应着一个余子式和一行列式的每个元素分别引理一个阶行列式，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再证一般情形,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ijaija,1,2,1对调列第列第列列依次与第的第再把jjjD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证二、行列式按行（列）展开法则项的和；相同，都是等式左右两边所含项数!n下证这些项一一对应；中的任意一项为右边ijijAaniiniinjjijijjjjjjjjiijaaaaaa112111211121)()1()1(级排列；一个的，，为其中1,,1,1211121nnjjjjjjjnii其符号为这一项，左边也含有niinjjijijjijaaaaaa11211121)()1112(1121)1(niijjjjjjniii)(1)1112(11121)1(niijjjjjjniii)(1121)1(niijjjjjji。边的项一一对应，得证故符号也相同，左右两例1计算行列式277011353D解27013D.58按第一行展开，得27015771132735D按第二行展开，得273377530.58例23351110243152113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3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阶范德蒙德行列式例４计算4324324324325555444433332222323232325551444133312221!5D333322225432543254321111!51)(!5jinjixx)45)(35)(34)(25)(24)(23(!5解：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jijiDAaAaAajninjiji02211jijiDAaAaAanjnijiji022110532004140013202527102135D例5计算行列式解0532004140013202527102135D660270132106627210.1080124220532414132525320414013202135215213rr122rr例6计算行列式xyyxyxyx000000000000解:按第一列展开xxyxx00000000原式＝nnnyx1)1(＝yxyxyyn0000000)1(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.三、小结jijiDAaAaAajninjiji02211jijiDAaAaAanjnijiji02211练习计算行列式aaaa0001000000001000.2nnaa结果提示：按第一列展开，