其次章 行列式

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习题

1.决定以下9级排列的逆序数，从而确定它们的奇偶性①.134782695；②.217986354；③.987654321.解.①.?(134782695)?0?0?0?0?0?4?2?0?4?10,偶排列；②.?(217986354)?0?1?0?0?1?3?4?4?5?18,偶排列；③.?(987654321)?0?1?2?3?4?5?6?7?8?36,偶排列。2.选择i,k使得：1).1274i56k9成偶排列;2).1i25k4897成奇排列.解.1).在排列中缺少3,8；取i?3,k?8,所得排列的逆序数为??5.2).取i?3,k?6，?(132564897)?5;满足条件。3.写出排列12435变成排列25341的那些对换。解.这种过程不唯一，如：12435?21435?25431?25341.(1,2)(1,5)(4,3)

由于交换改变排列的奇偶性，于是取i?8,k?3；所得排列为偶排列。

4.决定排列n(n?1)(n?2)解.逆序数为：21的逆序数，并探讨它的奇偶性。n(n?1)2?(n(n?1)(n?2)显然有21)?0?1?2??n?2?n?1??偶排列n?4k,4k?1?.奇排列n?4k?2,4k?3?5.假使排列x1x2xn?1xn的逆序数为k,那么排列xnxn?1x2x1的逆序数为多少？解.由于比xi大的数有n?xi个,所以在两个排列中由xi与比它大的各数构成的逆序数的和为:n?xi于是在两个排列中由xi构成的逆序数总数为:?(n?xi)?(n?1)?(n?2)?xi?1n?1

?2?1?n(n?1)2所以当x1x2xn?1xn的逆序数为k时,xnxn?1x2x1的逆序数为:n(n?1)?k.26.在6级行列式中a23a31a42a56a14a65,a32a43a14a51a66a25这两项应有什么符号？解.a23a31a42a56a14a65的符号为(?1)?(431265)?(?1)0?1?2?2?0?1?1;a32a43a14a51a66a25的符号为(?1)?(452316)?(?1)0?0?2?2?4?1.

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7.写出4级行列式中所有含有负号并且包含因子a23的项。解.设所求项为?a1ia23a3ja4k,则列标i,3,j,k为奇排列,且i,j,k为取自1,2,4中的不同值。令i?1,j?2,k?4,于是排列1324是奇排列;由于“对换改变排列的奇偶性〞，于是排列4312与2341是奇排列。那么所求为?a11a23a32a44，?a14a23a31a42，?a12a23a34a41.

8.按定义计算行列式：001).0n?1n000000001202).000n00n?10010002023).n?10?(?1)00000nn(n?1)2n(n?1)200010200.解.1).非零项仅有一项，其符号为(?1)?(n,n?1,n?2,,2,1),于是原行列式值为(?1)n!;

3，4，,n-1,1)2).非零项也仅有一项，其符号为(?1)?(2，?(?1)n?1,?原行列式值为(?1)n?1n!;3).非零项仅有一项，其符号为(?1)a1a2a3b3000b1b29.按行列式定义证明：c1c2d1d2e1e2?(n?1,n?2,,2,1,n)?(?1)(n?2)(n?1)2;于是行列式值为(?1)(n?2)(n?1)2n!.a4b4000a5b50?0.00

解.此行列式的一般项为a1j1a2j2a3j3a4j4a5j5,列标j3j4j5为取自1.2.3.4.5中的不同值，于是j3j4j5中至少有一项为0，故行列式的每一项均为0，所以行列式等于0.所证成立。2x10.由行列式定义计算f(x)?131同理x3的系数为?1.x12x1?1中x4与x3的系数。2x111x解由行列式定义知：.f(x)中仅有2x?x?x?x?2x4为含x4的项，故x4的系数为2；

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11.由111111111?0证明奇偶排列各半。证明：由题意行列式展开中的任一项(?1)?(j1j2jn)a1j1a2j2anjn的绝对值为1，而行列式的值为零，说明带正号的项与带负号的项数数目一致。而各项的符号为：(?1)?(j1j2于是奇偶排列各半。12.设jn)??1j1j2???1j1j2jn为奇排列jn为偶排列p(x)?11xa1x2a122an?1xn?1a1n?1n?1an?11an?1其中a1,a2,an?1是互不一致的数，1).由行列式定义说明p(x)是一个(n?1)次多项式；2).由行列式性质求行列式的根。解.1).将p(x)按第一行展开可知xn?1的系数为：11a1a2a122a2a1n?2n?2a21an?1这是一个范德蒙行列式，由于a1,a2,所以p(x)是一个(n?1)次多项式。2an?1n?2an?1an?1是互不一致的数，故此行列式不为零。,n?1)时p(x)?0，an?1.2).由范德蒙行列式的性质可知：当x取ai(i?1,2,

又由于p(x)是一个(n?1)次多项式，于是p(x)有n?1个不同的根a1,a2,-3-

13.计算以下行列式：2461).1014427327246100327543443l2?l31014100443?342100621l3?3l2l2?2l2468141002746127100143?1008141143?342721621l1?46l2l1?2l2x2).yx?y0100768yx?yxy?xyx?542100321?54213211076811610001161116??100??100??294?105.?5882940294?5881294x?y2x?2yyx?yxxx?yx?yxy1yx?yxyxl1?l2?l32x?2yy2x?2yxy?xx?y?2(x?y)yy6111?2(x?y)1x?y1xh1?h3h2?h302(x?y)01?2(x?y)(?(x?y)2?xy)??2(x3?y3).111131113).11111311631113110200??6?6?48.1131613111310020231312342341341241236113102341034110412101231113123413411412112300021021314?311?34).??10?10?102?2?202?2?2?1?1?10?1?1?10?1020?42?2?2?2??40?160.?1?1?1?1?1111xx001y?xy1100111?x111?y111001?x101001h1?h211?x11h3?h400y1?y1100按拉普拉斯定理11?x10011011?x5).111l2?l1l4?l311?y10111?yxyxy(?0?0?0?0?0)?xy?xy?(xy)2.

10展开，取定前两行1?x1?y1?y-4-

b?c14.证明：b1?c1b2?c2b?c证明：b1?c1b2?c2c?c1c21c?ac1?a1c2?a2c?ac1?a1c2?a2bb2a?ba2?b2a?bca1?b1?c1a2?b2c?ac1?a1c2?a2c2aa2aabb1b2cc1.c2a?bba1?b1?b1a2?b2aa1a2bb2bb2b1?b1b2cc1c2c?ac1?a1c2?a2aa2a?ba1?b1a2?b2aabb1b2cc1c2

a1?b1?2a1c?ac1?a1c2?a2cc2aa1a22a?ba2?b214a1?b1?b1a1?c1a1?2a115.写出以下行列式的全部代数余子式：1?120?1211),2)321.00210140003解.1)?1210210?110?12A11?021??6,A12?(?1)1?2023?0,A13?001?0,A14?(?1)1?4002?0;003003003000A21?(?1)1?2214114124121021??12,A22?021?6,A23?(?1)2?3001?0,A24?002?0;003003003000214114124121A31??121?15,A32?(?1)3?2023??6,A33?0?11??3,A34??0?12?0;003003003000214114124121

A41???121?7,A42?021?0,A43??0?11?1,A44?0?12??2.0210210010022).A11?A23??213132?1212?7,A12????12,A13??3;A21???6,A22??41?12121?1??1;A31???5,A32???5,A11??5.01213132-5-

1?a115）.1111?a1111111?a21111111?a31111111?an?111111?an?1110a30100an11111?an11111?anhi?h1i?2,3,,n?an?1?an0000an?100an1?a1?a?a31a2023a3100100?a1a2a3an(1??i?1n1).ai证明：1?a2111?a311n11a1ai1a2023?a1??i?2l1?a1liai000

nnnna1a111??ai(1?a1??)??ai(1???)??ai(1??).a1i?2aii?2aii?1aii?2i?1i?1nn19.用克拉默法则求解以下方程组：?2x1?x2?3x3?2x4?6?3x?3x?3x?2x?5?12341).?.3x?x?x?2x?34?123??3x1?x2?3x3?x4?42?1323?332解.D???70?0;D1?3?1?123?13?16534?1322632?3323532??70,D2???70,?1?1233?12?13?1343?1D1?x??1D?1??x?D2?1?2D??70;于是由克拉默法则可得?.D?x?3?1?3D?D?x4?4?1?D23D3?33?1?3?1?1622?13523?33??70,D4?323?1?14?13?136534-11-

?x1?2x2?3x3?2x4?6?2x?x?2x?3x?8解可得.D?324,D1?324,D2?648,D3??324,D4??648?12342).?;?x1?1,x2?2,x3??1,x4??2.?3x1?2x2?x3?2x4?4??2x1?3x2?2x3?x4??8?x1?2x2?2x3?4x4?x5??1?2x?x?3x?4x?2x?812345?解.D?24,D1?96,D2??336,D3??96,D4?168,D5?312;?3)?3x1?x2?x3?2x4?x5?3.于是x1?4,x2??14,x3??4,x4?7,x5?13.?4x?3x?4x?2x?2x??22345?1??x1?x2?x3?2x4?3x5??3?5x1?6x2?1?x?5x?6x?0解.D?665,D1?1507,D2??1145,D3?703,D4??395,D5?212;123??4).?x2?5x3?6x4?0.15072293779212于是x1?,x2??,x3?,x4??,x5?.?x?5x?6x?06651333513366545?3??x4?5x5?120.设a1,a2,,an是数域P中互不一致的数；b1,b2,,bn是数域P中的任意一组给定的数。使用克拉默法则证明：存在唯一的数域P中的多项式f(x)?c0xn?1?c1xn?2?使得f(ai)?bi(i?1,2,3,,n).,n)得：?cn?1?b1?cn?1?b2?cn?1?bn;证明：由f(ai)?bi(i?1,2,3,?cn?1,?f(a1)?c0a1n?1?c1a1n?2??n?1n?2?f(x)?c0a2?c1a2????f(x)?can?1?can?2?0n1n?这个线性方程组可以看做关于未知数ci(i?0,1,由于a1,a2,

,n?1)的方程组，它的系数矩阵是一个范得蒙行列式。,an是数域P中互不一致的数，因此这个范得蒙行列式的值不等于0，,n?1)使方程成立。得证。那么由克拉默法则可知有数域P中唯一确定的一组ci(i?0,1,21.设水银的密度与h与温度t之间的关系为h?a?a1t?a2t2?a3t3.由试验测得数据如下表；

求t?15℃与t?40℃时水银的温度(确切到两位小数).t/℃h013.601013.572013.553013.52-12-

解.将表中的数据代入给定的关系式可得：?a0?13.60?a?10a?100a?1000a?13.57?0123??a0?20a1?400a2?8000a3?13.55??a0?30a1?900a2?27000a3?13.52可以解得：?a0?13.60?a??0.0042?1??a2?0.00015??a3??0.0000033于是有h?13.60?0.0042t?0.00015t2?0.0000033t3.那么h(15)?13.56,h(40)?13.46即水银在15℃与40℃时，密度分别为13.56，13.46

补充题

a1j11.求j1j2

a1j2a2j2anj2a1jna2jnanjna11a12a22an2a1jna2jnanjn?(?1)?(j1j2jn)?a2j1jn,这里?是对所有的n级排列求和。anj1解.令a1na2nannD?a21an1那么a1j1a2j1anj1a1j2a2j2anj2D由于所有n级排列中奇偶排列各半，从而所求为0.

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a11(t)a12(t)da21(t)a22(t)2.证明：dtan1(t)an2(t)d?(i(?1)1?dti1ijini1ija1n(t)a2n(t)ann(t)a11(t)??j?1na21(t)da1j(t)a1n(t)dtda2j(t)a2n(t).dtdanj(t)dtainn(t)ainn(t)]ainn(t)ainn(t)]ann(t)an1(t)证明：右边???ijin)ai11(t)aijj(t)aijj(t)daij(t)dtj??(?1)in?(i1ijin)d[ai1(t)dt1(?1)in?(i1ijin)i1ijn?ai11(t)j?1in)n??[j?1i1ij?(?1)in?(i1ijai11(t)daijj(t)dta11(t)??j?1na21(t)da1j(t)a1n(t)dtda2j(t)a2n(t)?左边.得证。dtdanj(t)dtann(t)

an1(t)-14-

3.证明：a11?xa12?xa21?xa22?xan1?xan2?xa1n?xa2n?xann?xa11a21an1a12a22an2a1na2nannxa1n?xa2n?xann?xnn1).??x??Aij;i?1j?1其中Aij是aij的代数余子式.a11?xa12?xa?xa22?x证明：21an1?xan2?x1x?1a11??1a21?1an1na1n?xa2n?xann?xxa11a1na21a2nl1展开annaj?1,1aj?1,2aj?1,nnn1xx0a11?xa12?x加边0a21?xa22?x0an1?xan2?xa12a22an2a1na2nanna1na2nanna12a22an2a1na2nannnnxa12a22an2?1a11?1a21?1an1an1?x?(?1)j?1j?iaj?1,1aj?1,2aj?1,na11a21an1?a11a21an1a12a22an2a1na2nann?x??Aji?j?1i?1?x??Aij.i?1j?1

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a11?a12a12?a13a1,n?1?a1n1nn2).??Aa21?a22a22?a23a2,n?1?a2n1ij?.i?1j?1an1?an2an2?an3an,n?1?ann1证明：在1)中令x?1,有a11?1a12?1a1n?1a11a12a1na21?1a22?1a2n?1nn?a21a22a2n???Aij*；i?1j?1an1?1an2?1ann?1an1an2anna11?a12a12?a13a1,n?1?a1n1此外，a21?a22a22?a23a2,n?1?a2n1an1?an2an2?an3an,n?1?ann1ha12?a13a1,n?1?a1na1n?1n?1?ha11?a12na21?a22a22?a23a2,n?1?a2na2n?1h1?h2an1?an2an2?an3an,n?1?annann?1a11?a12a12?a13a1,n?1?a1na1na11?a12a12?a13?a21?a22a22?a23a2,n?1?a2na2n?a21?a22a22?a23an1?an2an2?an3an,n?1annan1?an2an2?an3a11a12a1na11?a12a12?a13a1,n?1?a1n1?a21a22a2n?a21?a22a22?a23a2,n?1?a2n1**an1an2annan1?an2an2?an3an,n?1?ann1*式与**式可知所证成立。-16-

a1,n?1?a1na2,n?1?a2nan,n?1?ann1

11比较4.计算：1232341).345n121h3?h2hn?h210210310123n1111hn?hn?11112h2?h1111?nn?111?n1n?11?n00(n?2)(n?1)2n?1n11?n1?n11121031011n?11?n00n(n?1)20000n11?nln?ln?11n0nnln?l100?n0?n01?2?00?n0?n0按第1列展开(?1)(?n)n?2n(n?1)n(n?1)n?1n?1?(?1)2n(注意符号！).22?aaaa?b?b?2).b?b?????????h2?hn???hn?1?hna0???0000aa00???00???a???????????b??(n?1)a000??

ln?ln?1ln?l2a0???0000aa00???00???b?????(n?2)???(???)n?2[??(n?2)?]?(?1)1?nb(?1)n?1(n?1)a(???)n?2?(???)n?2[???(n?2)???(n?1)ab].-17-

xaaaaxaaaaxaaa?axaaa?axaaa?axaa3).?a?axaa?a?axaaxa?a?a?axa拆n行?a?a?axa??a?a?a?a?ax?a?a?a?ax0000x?a?a?a?a?ax?a2a2a2aa0x?a2a2aa?(x?a)D0x?a2aan?1?0000x?aa?(x?a)Dn?1?a(x?a)n?1;0000a:xaaaaxaaa0xaaa?axaaa?axaa0?axaa?a?axaa?aa0a?a?a?axa拆n列?ax?a?a?ax0??a?ax?a?a?ax?a?a?a?ax?a?a?a?ax?a?a?a?a?ax?a0000?2ax?a000?(x?a)D?2a?2ax?a00n?1??2a?2a?2ax?a0?(x?a)Dn?1?a(x?a)n?1.?a?a?a?a?a于是???D?a)Dn?1n?(xn?1?a(x?a)(x?a)n?(x?a)n??D?Dn?1n?n?(x?a)Dn?1?a(x?a)2.xyyyyzxyyy当y?z时，解法如上题可得???Dz)Dx?y)n?1n?(x?n?1?z(Dn?(x?y)Dn?1?y(x?z)n4).zzxyy???1zzzxy;于是，D?y(x?z)n?z(x?y)nny?z;zzzzx当y?z时，易解得Dn?[x?(n?1)y](x?y)n?1.-18-

aaaaaaaaa?a同理

1x15).x12x1n?2x1n1x22x21x32x31xn2xnn?2x2nx2n?2x3nx3n?2xnnxn解.考虑这个范得蒙行列式：1x1f(y)?xn?21n?11n11x2xn?22n?12n21xnxn?2nn?1nnn1yyn?2n??(y?xi)i?1n?i?j?1?(xi?xj).xx显然f(y)中yn?1的系数为xxxxyn?1yn

可以知道所求行列式的值为f(y)中yn?1的系数的相反数。n??xii?1nn?i?j?1?(xi?xj)(xi?xj)于是D??xii?1n?i?j?1?-19-

111