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)?
n(n?1) 解 逆序数为?
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 101252024c2?c342??????1010?123202?104?1?102?122?(?1)4?3 ?140117c4?7c30010103?14 ?4?110c2?c3991010123?142??????c00?2?0?
1?12c31717142 (2)31?114152203162? 224211440 解 31?1212?c?4???c3?223?1202?r?4???r22??1506215210360321221213240 0r4?140 ?????r12?3121220300?0? 00 (3)?bdabacaebf?cfcd?deef?
解 ?bdabbf?accfcdae?deef?adf?bbb?ccce?ee
?adfbce?11111?11?11?4abcdef?
a1 (4)?001b?1001c?100? 1d0r1?ar201?ab0??????1b10?1d00a1c?100 1da1 解 ?001b?1001c?11?aba0c3?dc21?abaad ?(?1)(?1)2?1?1c1??????1c1?cd
0?100?1d
5? 证明:
abad?abcd?ab?cd?ad?1? ?(?1)(?1)3?21??11?cda2abb2 (1)2aa?b2b?(a?b)3;
111 证明
a2abb2c2?c1a2ab?a2b2?a2 2aa?b2b?????2ab?a2b?2a
00111c3?c11222ab?ab?aab?a?(a?b)3 ? ?(b?a)(b?a)1 ?(?1)2b?a2b?2a3?1ax?byay?bzaz?bxxyz (2)ay?bzaz?bxax?by?(a3?b3)yzx;
az?bxax?byay?bzzxy 证明
ax?byay?bzaz?bx ay?bzaz?bxax?by
az?bxax?byay?bz
(完整版)同济大学工程数学线性代数第六版答案(全)
