线性代数习题

1第一章行列式4.计算下列各行列式：（1）71100251020214214；（2）2605232112131412；（3）efcfbfdecdbdaeacab；（4）dcba100110011001解(1)7110025102021421434327cccc0100142310202110214=34)1(143102211014=143102211014321132cccc1417172001099=0(2)260523211213141224cc260503212213041224rr041203212213041214rr0000032122130412=0(3)efcfbfdecdbdaeacab=ecbecbecbadf=111111111adfbce=abcdef4(4)dcba10011001100121arrdcbaab100110011010=12)1)(1(dcaab101101223dcc010111cdcadaab=23)1)(1(cdadab111=1adcdababcd5.证明:(1)1112222bbaababa=3)(ba;(2)bzaybyaxbxazbyaxbxazbzaybxazbzaybyax=yxzxzyzyxba)(33;(3)0)3()2()1()3()2()1()3()2()1()3()2()1(2222222222222222ddddccccbbbbaaaa;(4)444422221111dcbadcbadcba))()()()((dbcbdacaba))((dcbadc;(5)1221100000100001axaaaaxxxnnnnnnnaxaxax111.证明(1)00122222221312ababaabaabacccc左边abababaab22)1(2221321))((abaabab右边3)(ba(2)bzaybyaxzbyaxbxazybxazbzayxa分开按第一列左边bzaybyaxxbyaxbxazzbxazbzayyb002ybyaxzxbxazyzbzayxa分别再分bzayyxbyaxxzbxazzybzyxyxzxzybyxzxzyzyxa33分别再分3右边233)1(yxzxzyzyxbyxzxzyzyxa(3)2222222222222222)3()2()12()3()2()12()3()2()12()3()2()12(dddddcccccbbbbbaaaaa左边9644129644129644129644122222141312ddddccccbbbbaaaacccccc964496449644964422222ddddccccbbbbaaaa分成二项按第二列964419644196441964412222dddcccbbbaaa949494949464222224232423ddccbbaacccccccc第二项第一项06416416416412222dddcccbbbaaa(4)444444422222220001adacabaadacabaadacaba左边=)()()(222222222222222addaccabbadacabadacab=)()()(111))()((222addaccabbadacabadacab=))()((adacab)()()()()(00122222abbaddabbaccabbbdbcab=))()()()((bdbcadacab)()()()(112222bdabbddbcabbcc=))()()()((dbcbdacaba))((dcbadc(5)用数学归纳法证明.,1,2212122命题成立时当axaxaxaxDn假设对于)1(n阶行列式命题成立，即,122111nnnnnaxaxaxD4:1列展开按第则nD1110010001)1(11xxaxDDnnnn右边nnaxD1所以，对于n阶行列式命题成立.6.设n阶行列式)det(ijaD,把D上下翻转、或逆时针旋转90、或依副对角线翻转，依次得nnnnaaaaD11111，11112nnnnaaaaD，11113aaaaDnnnn，证明DDDDDnn32)1(21,)1(.证明)det(ijaDnnnnnnnnnnaaaaaaaaaaD2211111111111)1(nnnnnnnnaaaaaaaa331122111121)1()1(nnnnnnaaaa111121)1()1()1(DDnnnn2)1()1()2(21)1()1(同理可证nnnnnnaaaaD11112)1(2)1(DDnnTnn2)1(2)1()1()1(DDDDDnnnnnnnn)1(2)1(2)1(22)1(3)1()1()1()1(7.计算下列各行列式（阶行列式为kDk）：5(1)aaDn11,其中对角线上元素都是a，未写出的元素都是0；(2)xaaaxaaaxDn;(3)1111)()1()()1(1111naaanaaanaaaDnnnnnnn;提示：利用范德蒙德行列式的结果．(4)nnnnndcdcbabaD000011112;(5)jiaaDijijn其中),det(;(6)nnaaaD11111111121,021naaa其中.解(1)aaaaaDn00010000000000001000按最后一行展开)1()1(100000000000010000)1(nnnaaa)1)(1(2)1(nnnaaa6(再按第一行展开)nnnnnaaa)2)(2(1)1()1(2nnaa)1(22aan(2)将第一行乘)1(分别加到其余各行，得axxaaxxaaxxaaaaxDn0000000再将各列都加到第一列上，得axaxaxaaaanxDn0000000000)1()(])1([1axanxn(3)从第1n行开始，第1n行经过n次相邻对换，换到第1行，第n行经)1(n次对换换到第2行…，经2)1(1)1(nnnn次行交换，得nnnnnnnnnnaaanaaanaaaD)()1()()1(1111)1(1112)1(1此行列式为范德蒙德行列式112)1(1)]1()1[()1(jinnnnjaiaD111)1(2)1(112)1()][()1()1()]([)1(jinnnnnjinnnjiji711)(jinji(4)nnnnndcdcbabaD00011112nnnnnnddcdcbabaa0000000011111111展开按第一行000000)1(1111111112cdcdcbababnnnnnnn2222nnnnnnDcbDda都按最后一行展开由此得递推公式：222)(nnnnnnDcbdaD即niiiiinDcbdaD222)(而111111112cbdadcbaD得niiiiincbdaD12)((5)jiaij0432140123310122210113210)det(nnnnnnnnaDijn,3221rrrr0432111111111111111111111nnnn8,,141312cccccc1524232102221002210002100001nnnnn=212)1()1(nnn(6)nnaaDa11111111121,,433221ccccccnnnnaaaaaaaaaa10000100010000100010001000011433221展开（由下往上）按最后一列))(1(121nnaaaannnaaaaaaaaa00000000000000000000000000022433221nnnaaaaaaaa000000000000000001133221nnnaaaaaaaa000000000000000001143322nnnnnnaaaaaaaaaaaa322321121))(1()11)((121niinaaaa8.用克莱姆法则解下列方程组：;01123,2532,242,5)1(4321432143214321xxxxxxxxxxxxxxxx.15,065,065,065,165)2(5454343232121xxxxxxxxxxxxx9解(1)11213513241211111D81207350321011111450081300321011111421420005410032101111112105132412211151D1121051329050111511210233130905091512331309050112109151120230046100011210915114200038100112109151142112035122412111512D811507312032701151313900112300231011512842840001910023101151426110135232422115113D;14202132132212151114D1,3,2,144332211DDxDDxDDxDDx(2)5100065100065100065100065D展开按最后一行61000510065100655DDD65DDD6)65(5DD3019DD1146566551141965(,11的余子式中为行列式aDD,11的余子式中为aDD类推DD,)5100165100065