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§4n级行列式的性质§8Laplace定理行列式乘法法则§3n级行列式§2排列§1引言§5行列式的计算§7Cramer法则§6行列式按行(列)展开第二章行列式一、余子式、代数余子式二、行列式按行(列)展开法则§2.6行列式按一行(列)展开引入,312213332112322311322113312312332211aaaaaaaaaaaaaaaaaa333231232221131211aaaaaaaaa3223332211aaaaa3321312312aaaaa3122322113aaaaa2223113233aaaaa可见,三级行列式可通过二级行列式来表示.2123123133aaaaa2123133133aaaaa§2.6行列式按一行(列)展开一、余子式、代数余子式定义在n级行列式中将元素所在的ijadet()ija第i行与第j列划去,剩下个元素按原位置2(1)n次序构成一个级的行列式,1n111,11,111,11,11,11,1,11,11,11,1,1,1jjniijijiniijijinnnjnjnnaaaaaaaaaaaaaaaa称之为元素的余子式,记作.ijMija§2.6行列式按一行(列)展开(1)ijijijAM令称之为元素的代数余子式.ijaijA注:①行列式中每一个元素分别对应着一个余子式和代数余子式.无关,只与该元素的在行列式中的位置有关.②元素的余子式和代数余子式与的大小ijaija§2.6行列式按一行(列)展开元素除外都为0,则ija.ijijDaA1.引理二、行列式按行(列)展开法则若n级行列式D=的中第i行所有det()ija§2.6行列式按一行(列)展开证:先证的情形,即ijnnaa111,111,11,11,00nnnnnnnnnaaaDaaaa由行列式的定义,有112112()121,(1)nnnnjjjjnjnjjjjDaaaa1112111()121,(1)nnnjjnjjnjnnjjnaaaa111111()11,(1)nnnjjnnjnjjjaaa§2.6行列式按一行(列)展开111,11,11,1nnnnnnaaaaa.nnnnaAnnnnaM结论成立。一般情形:111,111,111,11,11,1,11,1,11,11,1,11,1,1,10000jjjniijijijinijiijijijinnnjnjnjnnaaaaaaaaaaaaaaaaaaaaa§2.6行列式按一行(列)展开111,11,1111,11,11,11,1,1,11,11,11,1,1,1,1(1)(1)0000jjnjiijijinijninjiijijinijnnjnjnnnjijaaaaaaaaaaaaaaaaaaaaa(1)ijijijaM2()(1)nijijijaM(1).ijijijijijaMaA结论成立。111,111,111,11,11,1,11,1,11,11,1,11,1,1,1(1)0000jjjniijijijinniiijijijinnnjnjnjnnijaaaaaaaaaaaaaaaaaaaaa§2.6行列式按一行(列)展开2.定理行列式D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1122jjjjnjnjDaAaAaA1122iiiiininDaAaAaA1nikikkaA1,2,,in1nkjkjkaA1,2,,jn或行列式按行(列)展开法则§2.6行列式按一行(列)展开证:111211212000000niiinnnnnaaaaaaDaaa1122iiiiininaAaAaA11121111211112112121212000000nnniiinnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaaani,,2,1§2.6行列式按一行(列)展开例1.计算行列式3112513420111533D解:11130153D51100051151111115505116205501362(1)5540§2.6行列式按一行(列)展开例2.证明范德蒙行列式1232222123111111231111()nnnijjinnnnnnaaaaaaaaDaaaaaa§2.6行列式按一行(列)展开证:用数学归纳法.时,211211.aaaa2n01假设对于级范德蒙行列式结论成立.即1n02结论成立.121222121111222121111()nnnijjinnnnnaaaaaaDaaaaa§2.6行列式按一行(列)展开把从第n行开始,后面一行减去前面一行的nD倍,得na1212221122111212121122111111000nnnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaDaaaaaaaaa下证对于n级范德蒙行列式结论也成立.nD§2.6行列式按一行(列)展开1212221112211121212112211(1)nnnnnnnnnnnnnnnnnnnnnaaaaaaaaaaaaaaaaaaaaaaaa1212221121121222121111(1)()()()nnnnnnnnnnnaaaaaaaaaaaaaaa§2.6行列式按一行(列)展开121222121121222121111()()()nnnnnnnnnnaaaaaaaaaaaaaaa1()ijjinaa范德蒙行列式中至少两个相等.120,nnDaaa注:12111()()()()nnnnijjinaaaaaaaa1nD§2.6行列式按一行(列)展开3.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11220,ijijninjaAaAaAij11220,ijijinjnaAaAaAij§2.6行列式按一行(列)展开证行展开,有按第把行列式jaDij)det(11111111,niinjjjnjnjjnnnnaaaaaAaAaaaa可得换成把),,,1(nkaaikjk§2.6行列式按一行(列)展开11111111,niinijinjniinnnnaaaaaAaAaaaa行第j行第i相同11220,ijijninjaAaAaAij11220.ijijinjnaAaAaA∴当时,ij同理可证,§2.6行列式按一行(列)展开10nikjkkDijaAij10nkikjkDijaAij综合定理及推论,有关于代数余子式的重要性质:§2.6行列式按一行(列)展开例3.设求35211105,13132413D解:11121314AAAA11111105131324134.和11213141.MMMM11121314AAAA§2.6行列式按一行(列)展开11213141MMMM11213141AAAA15211105131314130.§2.6行列式按一行(列)展开例4.证明:111111111111111111110000kkrkkkkrkkkrrrrrkrrraaaabbaaccbbaabbccbb§2.6行列式按一行(列)展开练习:1.计算行列式001000.000100naaDaa123123123123,aaafbbbfDcccfdddf11213141.AAAA2.设求答案:21.2.0nnaa
本文标题:高等代数【北大版】2.6
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