线性代数第五版答案（全）

第一章 行列式

1? 利用对角线法则计算下列三阶行列式?

(1)2?11?084?131?

解 201?11?84?31

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

(2)abbcccaab?

解 abbcccaab

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

(3)1aa12bb12cc2?

解 111aa2bb2cc2

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

xyx?y (4)yx?yxx?yxy?

xyx?y 解 yx?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?

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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n(n?1)2?

3 2 (1个) 5 2? 5 4(2个) 7 2? 7 4? 7 6(3个) ? ? ? ? ? ?

(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??at11a23a32a44? (?1)a11a23a34a42?(?1)2a11a23a34a42?a11a23a34a42? 4? 计算下列各行列式?

4 (1)101122042? 05121074 解 101122042c2?c34?12?10??????1202?41?21?102?0512107c4?7c(?1)4?3 31003021?140103?14 2

?4?110c2?c3991010123?142??????c00?2?0?

1?12c31717142 (2)31?114211?5203 62221414 解 3c1?121??4???c?223?114202r??4?r??221?3?1205203622152036021221320 40 ?r?4???r12?31?114202?0? 0203000 (3)?bdabbf?accfcdae?deef?

解 ?bdabbf?accfcdae?deef?adf?bbceb?cc?ee

?adfbce?111?1111?11?4abcdef?

a100 (4)?01b10? 0?01?c11da100 解 ?r1?ar010?b011?c011?????2?0d011?0?baba00110 ?c11d)(?1)2?11??aba0c3?dc ?(?121?abaad01?c11d??????01?c11?0cd

?(?1)(?1)3?21?abad ?abcd?ab?cd?ad?1? ?11?cd 5 ? 证明

:

(1)2a21aaab?1bb2 21b?(a?b)3;

证明

a2abb2 21aa?1b21b?cc?2???c?12a21aabb??aa22b2b??a2 2a

3?c100 ?(?1)3?1abb??aa22b2b??a22a?(b?a)(b?a)a1b?2a?(a?b)3 ?3

ax?byay?bzaz?bxxyz (2)ay?bzaz?bxax?by?(a3?b3)yzxaz?bxax?byay?bzzxy;

证明

ax?byay?bzaz?bx ay?bzaz?bxax?byaz?bxax?byay?bz

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

xay?bzzyzaz?bx ?a2yaz?bxx?b2zxax?byzax?byyxyay?bz

xyzyzx ?a3yzx?b3zxyzxyxyz

xyzxyz ?a3yzx?b3yzxzxyzxy

xyz ?(a3?b3)yzxzxy?

a2(a?1)2(a?2)2(a?3)2 (3)b2(b?1)2(b?2)2(b?3)2c2(c?1)2(c?2)2(c?3)2?0; d2(d?1)2(d?2)2(d?3)2 证明

a2(a?1)2(a?2)2(a?3)2 b2(b?1)2(b?2)2(b?3)2c2(c?1)2(c?2)2(c?3)2(c4?c3? c3?c2? c2?c1得) d2(d?1)2(d?2)2(d?3)2a222 ?b22a?12adc2222bdc??1?12b?c?3?322ab??55(c?c? 1?3322dc??543c3?c2得) 2d5a22 ?b22ab??11222dc2222dc??1122?0? 222

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