第一章 行列式
1 利用对角线法则计算下列三阶行列式 (1)201?11?84?31
解 2?11?084?131
2(4)30(1)(1)1 0
132(1)8
1(
4)
248164
4
(2)abbcccaab
解 abbcccaab
acbbaccbabbbaaaccc
3abca3b3c3
(3)11aa2bb12cc2
解 111aa2bb2cc2
bc2ca2ab2ac2ba2cb2
(ab)(bc)(ca)
18 (1)
xyx?y (4)yx?yxx?yxyxyx?y 解 yx?yx
x?yxy 序数
x(xy)yyx(xy)(xy)yxy3(xy)3x3
3xy(x2(x3
y)y33x2 yx3y3x3 y3)
求下列各排列的逆
2 按自然数从小到大为标准次序 (1)1 2 3 4 (2)4 1 3 2 (3)3 4 2 1 (4)2 4 1 3
解 逆序数为0
解 逆序数为4 41 43
3 1 4 1
4 2 4 1, 2 1 4 3
(2n)
(2n1) 2 4
解 逆序数为5 3 2 解 逆序数为3 2 1 (5)1 3 n(n?1) 解 逆序数为
2 42 32
3 2 (1个) 5 2 5 4(2个) 7 2 7 4
7 6(3个)
(2n1)2 (2n1)4(2n1)(2n
(6)1 3
(2n2) (n1个)
(2n1)6
1) (2n) (2n2) 2
解 逆序数为n(n1) 3 2(1个) 5 2 5 4 (2个) (2n1)(2n 2) (n
(2n1)2 (2n1)4
1个)
4 2(1个) 6 2 6 4(2个) (n1个)
(2n1)6
(2n)2 (2n)4 (2n)6 (2n)(2n
2)
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
(
42
tt1)a11a23a3ra4s
这种排列共有两个
即24和
t其中rs是2和4构成的排列所以含因子a11a23的项分别是 (1)a11a23a32a44 (1)a11a23a34a42
(1)a11a23a32a44(1)a11a23a34a42
2
1
a11a23a32a44 a11a23a34a42
4 计算下列各行列式
41 (1)100125120244207
?123202?104?1?102?122?(?1)4?3 ?1441 解 101252024c2?c342??????10100117c4?7c30010103?14 ?4?110c2?c3991010123?142??????c00?2?01?12c3171714
2 (2)31?1421
52103162 22 解3?14 1211c4??11402r4?r2205210362????c?22322?1422150360?????3212212130 40?140 ?r?4???r12?3121220300?0
00 (3)?bdabacaebf?cfcd?deef
解 ?bdabbf?accfcdae?deef?adf?bbb?cecc?ee
?adfbce?11111?11?11?4abcdef
?a10 (4)011000?b01?c11
da1000r1?ar201?aba10 解 ?0100?b110?c11??????d01b0?011 ?c1d ?(?1)(?1)2?11??aba0c3?dc 21?abaad01?c11d??????01?c11?0cd
?(?1)(?1)3?21??ab11?adcdabcdabcdad 5 证明:
2a2ab1aa?b2 (1)1b21b(ab)3
;
证明
a2 21aaab?1bb221bc?c?2?c???12a2aabb??aa22b2b??a2 3?c11002a
?(?1)3?1abb??aa22b2b??a22a?(b?a)(b?a)a1b?2a
(2)axay?byay?bzaz?bxxyzaz??bzbxazax??bxbyaxay??bybz?(a3?b3)yzzxx;
y 证明
axay?byay?bzaz?bxaz??bzbxazax??bxbyaxay??bybz
(ab)31
工程数学线性代数课后习题答案
