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

2、231MA .23M ,44434241343332312423222114131211aaaaaaaaaaaaaaaaD ,44434134333124232112aaaaaaaaaM 1221121MA .12M ,33323123222113121144aaaaaaaaaM .144444444MMA .個個代代數(shù)數(shù)余余子子式式對對應應著著一一個個余余子子式式和和一一行行列列式式的的每每個個元元素素分分別別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）的元素與另一行（列）的對行列式任一行（列）的元素與另一行（列）的對應元素的代數(shù)余子式乘積之和等于零，即應元素的代數(shù)余子式乘積之和等于零，即. ji,AaAaAajninjiji 02211代數(shù)余子式的重要性質代數(shù)余子式的重要性質 ;,0,1jijiDDAaijnkkjki當當當當 ;,0,1jijiDDAaijnkjkik當當當當 .,0,1jijiij當當，當當其中其中五、設行列式五、設行列式nnnDn00103010021321 求第一行各元素的代數(shù)余子式之和求第一行各元素的代數(shù)余子式之和.11211nAAA 例例3.設設解解41424344222AAAA

6、414243442220AAAA 112342151,31231111D 21211435 12 5216132D 31232151031232212 44142434421222iiAAAAB 與與42,jiAB42,jiab421iiB 1111415121216132 11141515212 1211435125216132B 42CC 后后 443424144, . 5AAAAcdbaacbdadbcdcbaD則則設四階行列式設四階行列式 證證21211xxD 12xx , )(12 jijixx）式成立）式成立時（時（當當12 n例例 1112112222121).(111jinji

7、nnnnnnnxxxxxxxxxxxD)1(，階范德蒙德行列式成立階范德蒙德行列式成立）對于）對于假設（假設（11 n)()()(0)()()(0011111213231222113312211312xxxxxxxxxxxxxxxxxxxxxxxxDnnnnnnnnn 就就有有提提出出，因因子子列列展展開開，并并把把每每列列的的公公按按第第)(11xxi )()()(211312jjininnxxxxxxxxD ).(1jjinixx 223223211312111)()( nnnnnnxxxxxxxxxxxx n-1階范德蒙德行列式階范德蒙德行列式0532004140013202527102

8、135 D例例 計算行列式計算行列式解解0532004140013202527102135 D23110 072066 6627210 .1080124220 2312 5414235 53204140132021352152 13rr 122 rr 例例 計算計算 階行列式階行列式nabbbbabbbbabbbbaD 解解 abbbnababbnabbabnabbbbna1111 D將第將第 都加到第一列得都加到第一列得n, 3 , 2用化三角形行列式計算用化三角形行列式計算 abbbabbbabbbbna1111) 1( babababbbbna 1) 1(00 .)() 1(1 nbab

9、na121121121121nnnnnnnnxxxxyxxxyxDxxyxxxyxxx 例例.43213213213211xaaaaaaxaaaaaxaaaaaxDnnnn 解解列都加到第一列，得列都加到第一列，得將第將第1, 3 , 2 nxaaaxaxaaxaaxaxaaaaxDniinniinniinniin32121212111 .1111)(32222111xaaaxaaaxaaaaxDnnnniin axaaaaaxaaaxaxDnniin 23122121111010010001)(. )()(11 niiniiaxax用降階法計算用降階法計算例例計計算算.4abcdbadccd

10、abdcbaD 解解abcd,1111)(4abcdbadccdabdcbaD 列，得列，得列都減去第列都減去第、再將第再將第1432,0001)(4dadbdcdcbcacdcbcbdbabdcbaD .)(4dadbdccbcacdbcbdbadcbaD 110()(),abcdabcddcacbccdbdadD)()( )(22cbdadcbadcba )()(dcbadcbadcbadcba 4100()(),abcdabcddcadbccdbcadDdacbcbdadcbadcbaD )(用遞推法計算用遞推法計算例例 計算計算.21xaaaaxaaaaxaDnn 解解.000121x

11、aaaxaaaaxaaaaxann aaaaaxaaaaaxaaaaaxaDnn121 12110000,00000nnnxaxax Dxaa DxxxxxaxxxaxxxaxxxDnnnnnnn23142122121 ).(323112121xxxxxxxxxaxxxnnnn ).111(12121xxxaxxxDnnn 12,0nx xx 112212nnnnDx xxaxD1211nnnnDx xxax D12112212nnnnnnnDx xxax xxaxx xD12nnaxaaaaxaDaaax 112100nnDaxaaxxxx 120nx xx 120nx xx 1112223

12、33222111nnnnnnnn 1112nnnnnnDDD2112112112112112 nD例例nnDn00103010021321 .11211nAAA nAAA11211 n001030100211111 .11!2 njjn,2122221112111aaaaaaaaaDnnnnnn ,221122222111112112abababaabababaaDnnnnnnnnnn .2DD 證明：證明：證明證明由行列式的定義有由行列式的定義有.,)1( 2121121的逆序數(shù)的逆序數(shù)是排列是排列其中其中ppptaaaDnpnpptn .,)1( )()()1( 21)()21(21221

13、1221212211的逆序數(shù)的逆序數(shù)是排列是排列其中其中ppptbaaabababaDnpppnpnpptpnpnpppptnnnn ,212npppn 而而.)1(121221DaaaDpppnnt 所以所以利用范德蒙行列式計算利用范德蒙行列式計算.333222111222nnnDnnnn 解解.1333122211111!121212nnnnDnnnn !.1 !2)!2()!1( !)1()2()24)(23()1()13)(12( !)(!1 nnnnnnnnxxnDjinjin用數(shù)學歸納法用數(shù)學歸納法證明證明.coscos21000100000cos210001cos210001cos nDn 證證.,2, 1,2cos1cos22cos11cos,cos 221結論成立結論成立時時當當所以所以因為因為 nnDD .cos221DDDnnn ,)2cos( ,)1cos( ,21 nDnDnn由歸納假設由歸納假設;cos)2cos()2cos(cos)2cos()1cos(cos2 nnnn