工程数学线性代数(同济大学第六版)课后习题答案(全)

第一章 行列式

1? 利用对角线法则计算下列三阶行列式?

201 (1)1?4?1?

?183201 解 1?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)bca?

cababc 解 bca

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

111 (3)abc?

a2b2c2111 解 abc

a2b2c2 ?bc2?ca2?ab2?ac2?ba2?cb2 ?(a?b)(b?c)(c?a)?

xyx?y (4)yx?yx?

x?yxyxyx?y 解 yx?yx

x?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

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??a11a23a32a44? (?1)ta11a23a34a42?(?1)2a11a23a34a42?a11a23a34a42?

4? 计算下列各行列式?

41 (1)1001251202142? 0741 解 100125120214c2?c342??????10c?7c103074?12302021?104?1?102?122?(?1)4?3 ?14103?1404?110c2?c3991000?2?0? ?12?2??????110314c1?2c317171423 (2)151?120423611? 2242361c4?c221?????321251?120423023 解 151?1201?12042360r4?r222?????310221?121423402 00r4?r123 ?????1002?0? 00?abacae (3)bd?cdde?

bfcf?ef?abacae?bce 解 bd?cdde?adfb?ce

bc?ebfcf?ef?111 ?adfbce1?11?4abcdef?

11?1