【正文】 中 , x3的系數(shù)是 [ ]. . 4. 設(shè) a, b為實(shí)數(shù) , 則當(dāng) a=[ ], b=[ ]時(shí) , 0 0 xxxxxxf21112)( ???? 2 010100???? abba 5． 的第四行各元素余子式之和的等于 [ ]. 2235007022220403??125305403?? M41+M42+M43+M44 . 所以，第四行各元素余子式之和等于 [ 28 ]. 2235007022220403?? =A41+A42A43+A44 1111007022220403????1112220437???20011104314? 28??三、解答題 1．設(shè) ，試求 A41+4A42+2A43的值 . 5314102302411312－－?D 解 A41+4A42+2A43 00241102302411312??－－ 2. 設(shè) , 已知代數(shù)余子式 A31=2, 求 A12. 5312421xxD ? 解 由于 A31=24x=2, 所以 , x=1. 于是 A12=9. 000)1()1(12,11,12,332311,211???????nnnnnnnnaaaaaaaa???????計(jì)算下列行列式 (1) D= 解 D= 00000012,11,11,2222111,11211?????????nnnnnnaaaaaaaaaa????000)1(12,11,11,2222111???????nnnnnnaaaaaaa?????12,11,2131 )1()1()1( nnnnnn aaaa ??? ???? ??12,11,212)1)(4()1( nnnnnnaaaa ?????? ? 12,11,212 )1()1( nnnnnn aaaa ???? ? 解 100020001)1(1n????nnD???????0000100002000010)2(nn ?????????!1)( 1 nn ??? 解 按第一列展開(kāi) , 有 1110 0 0 0 0 0 10 0 0 0 0 0= ( 1 ) 0 0 00 0 00 0 0 0 0 0nnnaaaa aaaa?????原 式naaaaa00010000000000001000)3(???????????1110 0 0 0 0 0 10 0 0 0 0 0= ( 1 ) 0 0 00 0 00 0 0 0 0 0nnnaaaa aaaa?????原 式120000= ( 1 ) ( 1 )00n n nnaaaa??? ? ?2= nnaa ??22= ( 1 )naa? ?22( 4)nnabababDcdcdcd? 解 按第一列展開(kāi) , 有 2 1 2 100=00nna b bababac c d a bcdcdd c d???2nD2 1 2 100=00nna b bababac c d a bcdcdd c d???2nD2 2 2 2=nna b a ba b a ba d b cc d c dc d c d???22= ( ) nad bc D ?? 2 24= ( ) na d b c D ?? 1 2= = ( ) na d b c D?? = ( ) na d b c? 解 n=1時(shí) 原式 =|1|=1 n=2時(shí) 13=732 ??原 式 n?3時(shí) , 讓各列都減去第三列 , 則有 nn333331333333333332333331)5(????????????2 0 3 0 00 1 3 0 00 0 3 0 0=0 0 3 1 00 0 3 0 3n???原 式2 0 0 0 00 1 0 0 00 0 3 0 0=0 0 0 1 00 0 0 0 3n??? =6(n?3)! 解 11111)6(321321121121121nnnnaaaaxaaaaxaaaaxaaaaxD???????????????1000010001001011313221322111???????????nnnnnnncacaxaaaxaaaaaxaaaaaaaxDnii????????????????)(1????niiax 解 12125431432321)7(??nnnD?????????1212)1(2542)1(1432)1(322)1(??????nnnnnnnnnnD?????????1211254114313212)1(???nnnnD?????????1110111011103212)1(?????????nnnnn????111111111111111112)1(???????nnnnnnn?????????111111111111111112)1(??????????nnnnnn?????????100010010010012)1(??????????nnnnnn?????????212)2)(1()1()1(2 )1( ???????? nnnnnnnD 2 )1()1(12)1( ????? nnDnnnnnnnnxaaaaxaaaaxaaaaxD?????????321321321321)8( ? 解 nnnnaxxaaxxaaxxaaaaxD????????????????0000001133112211321nnnnaxxaaxxaaxxaaaaxD????????????????0000001133112211321111331122113211)(000000)1(????????????nnnnnnnnDaxaxxaaxxaaxxaaaaxa?????????1112211 )()())(( ??? ?????? nnnnnn Daxaaxaxax ?111)()( ??????? ? nnnniiin Daxaxa21111111))(()()( ???????????????? ?? nnnnnnniiiinniiin Daxaxaxaaxa2313)()( Daxaxaniiinjiiiinjj ???????????2112111 )()( ??????? ???? ? nnnniiinn DaxaxaD???????????njiiiinjjniii axaax111)()(nnnnaxxaaxxaaxxaaaaxD????????????????0000001133112211321111331122113211)(000000)1(????????????nnnnnnnnDaxaxxaaxxaaxxaaaaxa?????????1112211 )()())(( ??? ?????? nnnnnn Daxaaxaxax ?11)()()( ??????? ? nnnniiinnn Daxaxaxa21111111))(()()()()( ????????????????? ?? nnnnnniiinnnniiinnn Daxaxaxaxaaxaxa?????????niiini nnn axaxa11)(])(1[?12211000000000100001)9(axaaaaxxxxnnn????????????????? 解 1 2 2 11 0 00 0 00 0 1nnnxxDxxa a a a x??????nn axD ?? ? 1 nnn axaDx ??? ?? 122 nnnn axaxax ?????? ? 11 ??11 0 0 01 0 0( 1 )0 0 1 00 0 1nnxax???????nnD??????????????????????10000000010001000)10(??????????? 解 ????????????????????????????????????????1000000001000100010000000010000000??????????????????????nnD???????????????????1000000001000100001???????????nnDD???????????????10000000010000100001???????????nnD??????100000000010000100001