第一章行列式按行列展开ppt课件

1、,312213332112322311322113312312332211aaaaaaaaaaaaaaaaaa 333231232221131211aaaaaaaaa例如例如 3223332211aaaaa 3321312312aaaaa 3122322113aaaaa 333123211333312321123332232211aaaaaaaaaaaaaaa nijaij1 nija.Mij ，记记ijjiijMA 1ija44434241343332312423222114131211aaaaaaaaaaaaaaaaD 44424134323114121123aaaaaaaaaM 2332

2、231MA .23M ,44434241343332312423222114131211aaaaaaaaaaaaaaaaD ,44434134333124232112aaaaaaaaaM 1221121MA .12M ,33323123222113121144aaaaaaaaaM .144444444MMA .个个代代数数余余子子式式对对应应着着一一个个余余子子式式和和一一行行列列式式的的每每个个元元素素分分别别ininiiiiAaAaAaD 2211 ni, 2 , 1 111211111121110001nnnnnnnnaaaaDaaaa 111211212221111211nnnnnna

3、aaaaaaaa 121122( 1)( 1)( 1)iii niiiiininaMaMaMnnnjnijnjaaaaaaaD1111100 1 11111111111111111111111111110000jjjniijijijinn iiijijijinnijnjnjnjnnaaaaaaaaaaaaaaaDaaaaaa 1 1ijijM 111111111111111111111111111110001jjnjiijijinijn injiijijinijnnjnjnnnjaaaaaaaaaaDaaaaaaaaaa 1.ijA nijijAaD iijaija44434241332423

4、222114131211000aaaaaaaaaaaaaD .14442412422211412113333aaaaaaaaaa 例如例如nnnniniinaaaaaaaaaD212111211000000 nnnninaaaaaaa2111121100 nnnninaaaaaaa2121121100 nnnninnaaaaaaa211121100 ininiiiiAaAaAa 2211 ni, 2 , 1 例例13351110243152113 D03550100131111115 312 cc 34cc 0551111115)1(33 055026115 12rr 5526)1(31 50

5、28 .40 行列式任一行列的元素与另一行列的对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，即应元素的代数余子式乘积之和等于零，即. ji,AaAaAajninjiji 02211代数余子式的重要性质代数余子式的重要性质 ;,0,1jijiDDAaijnkkjki当当当当 ;,0,1jijiDDAaijnkjkik当当当当 .,0,1jijiij当当，当当其其中中 证证21211xxD 12xx , )(12 jijixx）式式成成立立时时（当当12 n例例 1112112222121).(111jinjinnnnnnnxxxxxxxxxxxD)1(，阶范德蒙德行列式成

6、立阶范德蒙德行列式成立）对于）对于假设（假设（11 n)()()(0)()()(0011111213231222113312211312xxxxxxxxxxxxxxxxxxxxxxxxDnnnnnnnnn 就就有有提提出出，因因子子列列展展开开，并并把把每每列列的的公公按按第第)(11xxi )()()(211312jjininnxxxxxxxxD ).(1jjinixx 223223211312111)()( nnnnnnxxxxxxxxxxxx n-1阶范德蒙德行列式阶范德蒙德行列式0532004140013202527102135 D例例 计算行列式计算行列式解解053200414001

7、3202527102135 D23110 072066 6627210 .1080124220 2312 5414235 53204140132021352152 13rr 122 rr 例例 计算计算 阶行列式阶行列式nabbbbabbbbabbbbaD 解解 abbbnababbnabbabnabbbbna1111 D将第将第 都加到第一列得都加到第一列得n, 3 , 2用化三角形行列式计算用化三角形行列式计算 abbbabbbabbbbna1111) 1( babababbbbna 1) 1(00 .)() 1(1 nbabna例例.43213213213211xaaaaaaxaaaaa

8、xaaaaaxDnnnn 解解列列都都加加到到第第一一列列，得得将将第第1, 3 , 2 nxaaaxaxaaxaaxaxaaaaxDniinniinniinniin32121212111 .1111)(32222111xaaaxaaaxaaaaxDnnnniin axaaaaaxaaaxaxDnniin 23122121111010010001)(. )()(11 niiniiaxax用降阶法计算用降阶法计算例计例计算算.4abcdbadccdabdcbaD 解解abcd ,1111)(4abcdbadccdabdcbaD 列列，得得列列都都减减去去第第、再再将将第第1432,0001)(4

9、dadbdcdcbcacdcbcbdbabdcbaD .)(4dadbdccbcacdbcbdbadcbaD 110()(),abcdabcddcacbccdbdadD)()( )(22cbdadcbadcba )()(dcbadcbadcbadcba 4100()(),abcdabcddcadbccdbcadDdacbcbdadcbadcbaD )(用递推法计算用递推法计算例例 计算计算.21xaaaaxaaaaxaDnn 解解.000121xaaaxaaaaxaaaaxann aaaaaxaaaaaxaaaaaxaDnn121 12110000,00000nnnxaxax Dxaa Dxx

10、xxxaxxxaxxxaxxxDnnnnnnn23142122121 ).(323112121xxxxxxxxxaxxxnnnn ).111(12121xxxaxxxDnnn 12,0nx xx 112212nnnnDx xxaxD 1211nnnnDx xxax D 12112212nnnnnnnDx xxax xxaxx xD 12nnaxaaaaxaDaaax 112100nnDaxaaxxxx 120nx xx 120nx xx 111222333222111nnnnnnnn 1112nnnnnnDDD 2112112112112112 nD例例nnDn00103010021321 .

11、11211nAAA nAAA11211 n001030100211111 .11!2 njjn,2122221112111aaaaaaaaaDnnnnnn ,221122222111112112abababaabababaaDnnnnnnnnnn .2DD 证证明明：证明证明由行列式的定义有由行列式的定义有.,)1( 2121121的逆序数的逆序数是排列是排列其中其中ppptaaaDnpnpptn .,)1( )()()1( 21)()21(212211221212211的的逆逆序序数数是是排排列列其其中中ppptbaaabababaDnpppnpnpptpnpnpppptnnnn ,212n

12、pppn 而而.)1(121221DaaaDpppnnt 所所以以利用范德蒙行列式计算利用范德蒙行列式计算.333222111222nnnDnnnn 解解.1333122211111!121212nnnnDnnnn !.1 !2)!2()!1( !)1()2()24)(23()1()13)(12( !)(!1 nnnnnnnnxxnDjinjin用数学归纳法用数学归纳法证明证明.coscos21000100000cos210001cos210001cos nDn 证证.,2, 1,2cos1cos22cos11cos,cos 221结结论论成成立立时时当当所所以以因因为为 nnDD .cos221DDDnnn ,)2cos( ,)1cos( ,21 nDnDnn由由归归纳纳假假设设;cos)2cos()2cos(cos)2cos()1cos(cos2 nnnnnnDn 评注评注小结小结12, ,ki