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)?
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 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)
n(n?1)? 2 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)10012512021125142? 0720214c?c4?123122??????0c?7c103300742?104?1?1002?122?(?1)4?3 2?14103?141041 解 100
23 (2)154?110c2?c39910?12?2??????00?2?0? 10314c1?1c17171423
1?120423611? 2242361c?c421?????22423023151?12042360r?r422?????0223121?1214234020023 解 151?120
r4?r121?????3?1120002?0? 00
?abacae(3)bd?cddebfcf?ef?
解
?abacae?bcebd?cdde?adfb?cebc?ebfcf?ef?111?adfbc1e?11?4abcde? f11?1
a1(4)?001b?1001c?1001d?
0r?ar120?????1d01?aba0?1b100?1c100?1d 解
a?1001b?1001c?1
1?aba0c3?dc21?abaad2?1?(?1)(?1)?1c1??????1c1?cd0?1d0?10
5? 证明:
?(?1)(?1)3?21?abad?abcd?ab?cd?ad?1? ?11?cda2abb2 (1)2aa?b2b?(a?b)3;
111 证明
a22a1aba?b13?1b2c2?c1a2ab?a2b2?a22b?????2ab?a2b?2a001c3?c11
?(?1)ab?a2b?ab2?a23
?(b?a)(b?a)ab?a?(a?b) ? 122b?2aax?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?bzxay?bzaz?bxyay?bzaz?bx ?ayaz?bxax?by?bzaz?bxax?by
zax?byay?bzxax?byay?bzxay?bzzyzaz?bx ?a2yaz?bxx?b2zxax?by
zax?byyxyay?bzxyzyzx ?a3yzx?b3zxy
zxyxyzxyzxyz ?a3yzx?b3yzx
zxyzxyxyz ?(a3?b3)yzx?
zxy
a2b2 (3)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)2(b?3)2?0; (c?3)2(d?3)2 证明
a2b2 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)2(b?3)2(c4?c3? c3?c2? c2?c1得) (c?3)2(d?3)2a22b ?c2d2a22b ?c2d22a?12b?12c?12d?12a?12b?12c?12d?12a?32b?32c?32d?322222a?52b?5(c?c? c?c得) 2c?54332
2d?522?0? 22