其次章习题选讲

本文格式为Word版，下载可任意编辑——其次章习题选讲5.若排列x1x2?xn?1xn的逆序数为k,则排列xnxn?1?x2x1的逆序数是多少.

解:根据定义看,假设x1x2?xn?1xn中x1后比x1小的数有k1个,x2后比x2小的数有k2个,一直假设xn?1后比xn?1小的数有kn?1个,则k1?k2???kn?1?k.从而在排列xnxn?1?x2x1中,

x1前比x1小的数有k1个,比x1大的数有n?1?k1个,

x2前比x2小的数有k1个,比x2大的数有n?2?k2个,

xn?1前比xn?1小的数有k1个,比xn?1大的数有1?kn?1个,

则xnxn?1?x2x1的逆序数为n?1?k1?n?2?k2?1?kn?1?n(n?1)?k.200?01008.(1)??0n?1n0010?20n(n?1)?(n(n?1)?21)a1na2n?1?an?12an1?(?1)2n!.????(?1)???00000002?0?(23?n1)a12a23?an?1nan1?(?1)n?1n!.(2)??????(?1)000?n?1n000000(3)?n?103?010?200(n?1)(n?2?((n?1)(n?2)?21n)?(?1)a1n?1a2n?2?an?11ann?(?1)2n!.?????00000n10.由行列式的定义,要想出现4次方,只能是主对角线元素的乘积,故系数为2,同理x的系数的取法为a12a21a33a44,系数为?1.

12.设p(x)?11xa1x2a12?xn?1?a1n?1??1an?1???2n?1an?an?1?1,其中a1,a2,?,an?1互不一致,

1)由行列式定义说明p(x)是一个n?1此多项式.

1

2)由行列式性质,求p(x)的根.13.

2461).10144273271132701327543443?100?100021443?10511443??294?105.

?3427216211162101621xyx?yxx?yxy1yx?yxy1yx?y?y?x2)

yx?y?2(x?y)1x?y1x?2(x?y)0x0x?y?2(x?y)(?x2?xy?y2)??2(x3?y3).

xyx?yyx?yxx?yxy?3x2y?3xy2?(x?y)3?x3?y3??2(x3?y3).

1?x111xx00x00011?x1111?x111?x105)???x2y2.

111?y100yy00y01111?y1111?y101?ya2b26)2cd2a2b2?2cd2(a?1)2(b?1)2(c?1)2(d?1)2(a?2)2(b?2)2(c?2)2(d?2)2(a?3)2a22a?14a?42b?14b?42c?14c?46a?96b?9?

6c?9(b?3)2b2?22(c?3)c(d?3)2d22d?14d?46d?92a?1492b?149?0

2c?1492d?149b?cc?ac1?a1c2?a2a?babb1b2cc1c2a?bcc?ac1?a1c2?a2a?ba1?b1a2?b2a1?b1?2a1a2?b2a2b14.证明:b1?c1b2?c2b?c证明:b1?c1c?ac1?a1c2?a2ba?bc?ac1?a1c2?a2caa1a2b2?c2b?b1b2cc1c2a1?b1?b1a2?b2b2aa1a2a?ba1?b1?c1a2?b2c2a?ba1?b1a2?b2a?ba1?b1?b1a2?b2b2a1?b1?c1a2?b2c22

b?b1b2cc1c2acaa1a2babb1b2cc1.c2a1?c1a2c2b1?2a1b2a2xy0?000xy?00n?11?nn?1n1?nn17.1)???????xx?(?1)yy?x?(?1)y.

0y2)

0000??x0yxa1?b1a2?b1?a1a1?b2?b2?bn?a1?bn?b1??a1a2?ana1?b2??a1?bn???b1b1?b1a1?b2??a1?bn??

a2?b2?a2?bna2?b2?a2?bnan?b2?an?bna2?b2?a2?bnan?b2?an?bnan?b1an?b2?an?bna1?b1n?1?a2b2?bnb1a2?a2?????(a1?a2)(b1?b2)n?2.?????????0n?3?ab?bba?an2n1nna1?a1x1?m3)

x2?x2??xnxn??x1?mm?mx2?0??xn0??x1?x1nx2?m??m??x?mii?1nx2?0??xn0

?0?0?m??xn?m??m??m?(?xi?m)(?m)n?1.

i?1122?2?120230021???020??2(n?2)!

222?24)223?2??????222?n?????00?n?2115)02?130?n?1?00?n0?i?i?ii?1nnn?2n?1???00?0n00?1?n?

2?2?????0000?0???n?11?n00i?2?10?0i?30?2??03

1(?1)n?1(n?1)!

2a011???100?1a10181)10a2???1002)

a0??i?1n???an00?01ai1a101?1n01?a1a2?an(a0??).0i?1ai0?a2?????00?anx?10x00??00a0a1a2?an?2?0000000???000xn?an?1xn?1???a0xn?1?an?1xn?2???a1xn?2?an?1xn?3???a2?x2?an?1x?an?2x?an?1?

0?1x?0?????000000?x?100?10000??????0??1??1x?an?1?xn?an?1xn?1???a0.

???3).

10?0?????1?00?000?Dn

????????0??1???Dn?(???)Dn?1???Dn?2,即Dn??Dn?1??(Dn?1??Dn?2).

从而Dn??Dn?1??(Dn?1??Dn?2)??2(Dn?2??Dn?3)????n?2(D2??D1)??n.同理Dn??Dn?1??n.则(???)Dn??n?1??n?1.

?n?1??n?1故???时,有Dn?.

???cos?4)

12cos?1?00?000000?cosn?

10?01?2cos???0????12cos?证明:n?1时,

cos?11?2cos2??1?cos2?.

2cos?4

设结论对?n成立.

则Dn?2cos?Dn?1?Dn?2?2cos?cos(n?1)??cos(n?2)?

?2cos?cos(n?1)??cos[(n?1)???]

?2cos?cos(n?1)??cos(n?1)?cos??sin(n?1)?sin??cos?cos(n?1)??sin(n?1)?sin??cosn?.

1?a115)

11?a21?111??11111?1?an1???1i?1ai0000n100?0?01a100011?a111?110a20231111??111111111?1?an

1?11?a3?1???1?11?a21?11?a3??1??1100001???11?1a1??1010a2100???100100?100a3?00????????100n0?0?a3?0?000an0?0an????????a1a2?an(1??i?11).ai

补充题.

123?n123?n123?n234?1111?1?n111?