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互不相同,
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?babb1b2cc1 c2a?bcc?ac1?a1c2?a2a?ba1?b1 a2?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?b1 a2?b2a?ba1?b1?b1a2?b2b2a1?b1?c1a2?b2c2 2
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?120020021???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??0 3
1(?1)n?1(n?1)!
2a011???100?1a1018 1) 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?110a20011111??111111111?1?an
1?11?a3?1???1?11?a21?11?a3??1??1100001???11?1a1??1010a2100???100100?100a3?00????????100n0?0?a3?0?00 0an0?0an????????a1a2?an(1??i?11). ai
补充题.
123?n123?n123?n234?1111?1?n111?1?n2?111?1?111?1 4.1) 345????????????????n12?n?111?n1?111?n1?1111020?n?1??n0??0?100?????1?n0?i?i?1n00?0n100200?n?1???ni??(?n)n?1(?1)0i?1n?0n(n?1)(n?2)2
????n0? 5