行列式按行按列展开VIP免费

1例3（n阶行列式的计算）axaaaaaxaaaaaxaaaaaxDLMMMMMLLLnccc21])2([anxaxaaaaxaaaaxaaa1111211312rrrrrrn])2([anxaxaxaxaaa20000200002011)2]()2([naxanx3例4nD001030100211111箭形行列式目标：把第一列化成三角形行列式ncnccc13121321nini00003000020111112)11(!2niin4例5baaaaabaaaaabaaaaabaDnnnn32132132132111312rrrrrrnbbbbbbaaaban000000321nccc21箭形行列式5bbbaaabaaann000000000)(3221121)]()[(nnbbaaa6例64321xaaaaxaaaaxaaaaxD)4,3,2,1,(iaxi（可以化为箭形行列式）213141rrrrrr1121314000000xaaaaxxaaxxaaxxa710010101001143211axaaxaaxaaxx4321cccc41)(iiax1000010000104324211axaaxaaxaaxaaxxii414211)(][iiiiaxaxaaxx41()iixa8aaaaaaaaa111213212223313233问题：一个n阶行列式是否可以转化为若干个n－1阶行列式来计算？（降阶的思想）§1.3行列式按行（列）展开定理一.按行列式某行(列)展开,312213332112322311322113312312332211aaaaaaaaaaaaaaaaaa9aaaaaaaaaaaaaaa222321232122111213323331333132可见一个三阶行列式可以转化成三个二阶行列式的计算。3223332211aaaaa3321312312aaaaa3122322113aaaaa问题：一个n阶行列式如何转化为n个n-1阶行列式来计算？首先介绍两个概念10定义1在n阶行列式中，把元素ija所在的第i行和第j列划去后，余下的n－1阶行列式叫做元素的余子式。记为ijM称1ijijijAM为元素的代数余子式。例如44434241343332312423222114131211aaaaaaaaaaaaaaaaD44424134323114121123aaaaaaaaaM2332231MA23Mijaija1144434241343332312423222114131211aaaaaaaaaaaaaaaaD44434134333124232112aaaaaaaaaM1221121MA12M33323123222113121144aaaaaaaaaM444444441MMA注意行列式的每个元素都分别对应着一个余子式和一个代数余子式。12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，即定理1证明（先特殊，再一般）分三种情况讨论，我们只对行来证明此定理。(1)假定行列式D的第一行除11a外都是0。1122...iiiiininDaAaAaA1,2,...,in1121222120...0..................nnnnnaaaaDaaa13由行列式定义，D中仅含下面形式的项其中恰是11M的一般项。所以，1111MaD111111)1(Ma1111Aa232323(1...)11231...(1)...nnnjjjjjnjjjjDaaaa2323(1...)1123(1)...nnjjjjjnjaaaa2323(...)23(1)...nnjjjjjnjaaa14(2)设D的第i行除了ija外都是0。把D转化为(1)的情形把D的第i行依次与第1i行，第2i行，······，第2行，第1行交换；再将第j列依次与第1j列，第2j列，······，第2列，第1列交换，这样共经过2)1()1(jiji次交换行与交换列的步骤。11111.....................0......0......................jnijnnjnnaaaaDaaa15由性质2，行列式互换两行（列）行列式变号，得，(1)ijijijijijaMaA21,1,11,,1...0...0.....................(1).....................ijijijijinnjnjnnaaaaDaaa16(3)一般情形111211212.................................niiinnnnnaaaDaaaaaa111211212...............0...00...0...0...0................niiinnnnnaaaaaaaaa17例如，行列式277010353D27013D按第一行展开，得27005771032711121112...............0...0...............ninnnnaaaaaaa11121212...............0...0...............ninnnnaaaaaaa1112112..................00..................ninnnnnaaaaaaa1122...iiiiininaA...