利用对角线法则计算下列三阶行列式

第一章 行列式

1? 利用对角线法则计算下列三阶行列式?

201(1)1?4?1? ?183

解

2011?4?1 ?183 ?2?(?4)?3?0?(?1)?(?1)?1?1?8 ?0?1?3?2?(?1)?8?1?(?4)?(?1) ??24?8?16?4??4?

abc(2)bcacab?

解

abcbcacab ?acb?bac?cba?bbb?aaa?ccc ?3abc?a3?b3?c3?

111(3)abca2b2c2?

解

111abca2b2c2 ?bc2?ca2?ab2?ac2?ba2?cb2 ?(a?b)(b?c)(c?a)?

(4)

xyx?yyx?yxx?yxy?

解

xyx?yyx?yxx?yxy

?x(x?y)y?yx(x?y)?(x?y)yx?y3?(x?y)3?x3 ?3xy(x?y)?y3?3x2 y?x3?y3?x3 ??2(x3?y3)?

2? 按自然数从小到大为标准次序? 求下列各排列的逆序数?

(1)1 2 3 4? 解 逆序数为0 (2)4 1 3 2?

解 逆序数为4? 41? 43? 42? 32? (3)3 4 2 1?

解 逆序数为5? 3 2? 3 1? 4 2? 4 1, 2 1? (4)2 4 1 3?

解 逆序数为3? 2 1? 4 1? 4 3? (5)1 3 ? ? ? (2n?1) 2 4 ? ? ? (2n)? 解 逆序数为 3 2 (1个) 5 2? 5 4(2个) 7 2? 7 4? 7 6(3个) ? ? ? ? ? ?

n(n?1)2?

(2n?1)2? (2n?1)4? (2n?1)6? ? ? ?? (2n?1)(2n?2) (n?1个)

(6)1 3 ? ? ? (2n?1) (2n) (2n?2) ? ? ? 2? 解 逆序数为n(n?1) ? 3 2(1个) 5 2? 5 4 (2个) ? ? ? ? ? ?

(2n?1)2? (2n?1)4? (2n?1)6? ? ? ?? (2n?1)(2n?2) (n?1个) 4 2(1个) 6 2? 6 4(2个) ? ? ? ? ? ?

(2n)2? (2n)4? (2n)6? ? ? ?? (2n)(2n?2) (n?1个) 3? 写出四阶行列式中含有因子a11a23的项? 解 含因子a11a23的项的一般形式为

(?1)ta11a23a3ra4s?

其中rs是2和4构成的排列? 这种排列共有两个? 即24和42? 所以含因子a11a23的项分别是

(?1)ta11a23a32a44?(?1)1a11a23a32a44??a11a23a32a44? (?1)ta11a23a34a42?(?1)2a11a23a34a42?a11a23a34a42? 4? 计算下列各行列式?

41(1)100125120214207?

解

41100125120214c?c4?123212??????0c?7c103300742?104?1?1002?122?(?1)4?3 2?14103?1410

23(2)154?110c2?c39910?12?2??????00?2?0?

c17171410314c1?123

1?1202315423611224236?

1c?c421?????224230 解

1?12023151?12042360r?r422?????0223121?12142340200

r4?r1213?1?????120002?0? 00

?abacae(3)bd?cddebfcf?ef?

解

?bce?abacaebd?cdde?adfb?cebc?ebfcf?ef?111?adfbce1?11?4abcdef11?11b?1001c?1001d?

a1(4)?00?

解

a?1001b?1001c?10r?ar201?????1d01?aba?1b10?1c00?1001d

1?aba0c3?dc21?abaad2?1c1?cd?(?1)(?1)?1c1??????10?100?1d?(?1)(?1)3?21?abad?11?cd

5? 证明:

?abcd?ab?cd?ad?1?

a2abb2 (1)2aa?b2b111?(a?b)3;

证明

a22a1aba?b13?1b2c2?c1a2ab?a2b2?a22b?????2ab?a2b?2a001c3?c11

?(a?b)3 ?

?(?1)

ab?a2b?ab2?a2?(b?a)(b?a)ab?a122b?2aax?byay?bzaz?bxxyz(2)ay?bzaz?bxax?by?(a3?b3)yzxaz?bxax?byay?bzzxy;

证明

ax?byay?bzaz?bxay?bzaz?bxax?byaz?bxax?byay?bz

xay?bzaz?bxyay?bzaz?bx?ayaz?bxax?by?bzaz?bxax?byzax?byay?bzxax?byay?bz

?

ann?1?i?j?1?(i?j)?

bn? ? ??? ? (4)D2n?cna1b1c1d1?? ? ? ? ?;

dn 解

an? ? ?? ? ?an?1 D2n?cna1b1c1d1? ? ?? ? ?bn(按第1行展开)

dn?? ? ? ? ?bn?10? ? ?? ? ? ?ancn?10a1b1c1d1

dn?100dn?0an?1? ? ??? ? a1b1c1d1?? ? ? ? ?bn?1 ?(?1)2n?1bncn?1cn?

dn?10 再按最后一行展开得递推公式

D2n?andnD2n?2?bncnD2n?2? 即D2n?(andn?bncn)D2n?2?

于是 D2n??(aidi?bici)D2?

i?2n而 D2?a1b1?a1d1?b1c1? c1d1n所以 D2n??(aidi?bici)?

i?1 (5) D?det(aij)? 其中aij?|i?j|; 解 aij?|i?j|?

01231012101Dn?det(aij)?23210? ? ?? ? ?? ? ?? ? ?n?1n?2n?3n?4? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?n?1n?2n?3n?4? ? ?0

?111111r1?r2?1?11??????1?1?1r2?r3?1?1?1?1? ? ?? ? ?? ? ?? ? ? ? ? ? n?1n?2n?3n?4? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?111 1? ? ?0? ? ?0? ? ?0? ? ?0 ? ? ?0? ? ?? ? ?? ? ?n?1

?1000?200c2?c1?1?2?20??????1?2?2?2c3?c1?1? ? ?? ? ?? ? ?? ? ? ? ? ? n?12n?32n?42n?5 ?(?1)n?1(n?1)2n?2? 解

1?a11(6)Dn?11?a2? ? ?? ? ?11? ? ?1? ? ?1? ? ?? ? ?? ? ?1?an, 其中a1a2 ? ? ? an?0?

1?a1111?a2Dn?? ? ?? ? ?11? ? ?1? ? ?1? ? ?? ? ?? ? ?1?an

a10?a2a2c1?c20?a3?????c2?c3? ? ?? ? ?00 ? ? ? 0000a3? ? ?00? ? ?001? ? ?001? ? ?001? ? ?? ? ?? ? ?? ? ?? ? ??an?1an?11? ? ?0?an1?an

1?1?a1a2? ? ?an0? ? ?00100?a1a2? ? ?an? ? ?0001?1? ? ?00010? ? ?00n001? ? ?00001? ? ?0? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?? ? ?000? ? ??10000? ? ?00a1?1?10a2?10a3? ? ?? ? ??11an?1?1?11?an000? ? ?1a1?1?1a2?1a3? ? ??1an?1ni?1

0? ? ?001??ai?1 ?(a1a2?an)(1??1)?

i?1ai

8? 用克莱姆法则解下列方程组?

?x1?x2?x3?x4?5?x?2x2?x3?4x4??2(1)?1?

2x1?3x2?x3?5x4??2?3x?x?2x?11x?0?1234

解 因为

11D?2312?311?1?1214??142?511?

151D2?1?2?12?2?114??284?5

5D1??2?212?31?1?114??142?5? ?

0121130211111111

D?123?2?5?

D?2??5?234?1142?13?22?4??42601152?313?12102所以 xD1D2D3DDD?3?D41??1? x2??2? x3?? x4D??1?

?5x1?6x?1(2)??x1?5x2

?x2?6x3?02?5x3?6x

??5x4?0? x?34?6x5?0?x4?5x5?1

解 因为 56000

D?1560001560?66500156?

000151600051000

D056001?01560?1507?060000560??114500156?

D1200156?

1001501015

51D3?00051D5?000651006510010001065100065100651000?70365100?21201?

51D4?000651000651010001000??39565?

所以

x1??

1507665? x2??1145? x3?703? x4??395? x4?212?

665665665665 9? 问?? ?零解？

?x?x2?x3?0??1取何值时? 齐次线性方程组?x1??x2?x3?0有非

??x1?2?x2?x3?0 解 系数行列式为

11D?1?1?????12?1??

令D?0? 得 ??0或??1?

于是? 当??0或??1时该齐次线性方程组有非零解?

10? 问?取何值时? 有非零解？

解 系数行列式为

(1??)x1?2x2?4x3?0??齐次线性方程组?2x1?(3??)x2?x3?0??x1?x2?(1??)x3?0

1???241???3??4D?23??1?21??1111??101??

?(1??)3?(??3)?4(1??)?2(1??)(?3??) ?(1??)3?2(1??)2???3? 令D?0? 得

??0? ??2或??3?

于是? 当??0? ??2或??3时? 该齐次线性方程组有非零解?

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