《线性代数》同济大学版 课后习题答案详解
第一章 行列式
1 利用对角线法则计算下列三阶行列式
111 解 abca2b2c2 (a
bc2ca2ab2ac2ba2cb2
b)(bc)(ca)
201 (1)?11?84?31
201 解 ?11?84?31
2(4)30(1)(1)1 013
2(
1)
81(
4)
24
8164
4
abbccc (2)aab
abc 解 bccaab
acbbaccbabbbaaaccc
3abca3b3c3
111 (3)aa2bb2cc2
8 1)
xyx?y (4)
x?yyx?xyxyxyx?y 解 x?yyx?xyx
y x(xy)yyx(xy)(xy)yxy3(xy)3x3
3xy(xy)y3
3x2
yx3y3x3
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
(2n1) 2 4
(2n)
解 逆序数为n(n?1)2
3 2 (1个)
1(
5 2 5 4(2个) 7 2 7 4 7 6(3个)
(2n1)2
(2n1)4
(2n1)6
(2n1)(2n2) (n1
个)
(6)1 3
(2n1) (2n) (2n2)
2
解 逆序数为n(n1)
3 2(1个) 5 2 5 4 (2个)
(2n1)2
(2n1)4
(2n1)6
(2n1)(2n2) (n1
个)
4 2(1个) 6 2 6 4(2个)
(2n)2 (2n)4 (2n)6
(2n)(2n2) (n1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)2
a11a23a34a42a11a23a34a42
4 计算下列各行列式
41 (1)10122040512210
7 解
4c??1010112204??????2?c34121205122107c4?7c310030021?2140?41?21?102?(?1)4?3103?14 4?110c103?142??????2?c39c09010 ?12?2?01?12c3171714
24 (2)31?12115210362
22 解 31?141521212?c?4???c?223?14240362152120r4?r0360????22?31?122212021340 0r4?r?140????12 ?312122
00300?00?bdabacae (3)bf?cfcd?deef
?bdabbf?accfcdae?de?bce 解 ef?adfbb?cc?ee
adfbce?111?1111 ??11?4abcdef
?a10 (4)011000?b01?c11
d1b0100r1?ar201?aba10 解 ?a010?01?c11??????1bd00?01c0?11 d ?(?1)(?1)2?11??ab01a?c0c11d?3?dc????21??abaad01?c11?0cd?(?1)(?1)3?21??ab11?adcdad1
cdabcdab 5 证明:
2a21aaab?b2 (1)1b21b(ab)3
;
证明
a21aaab?1bb221bc?c?2?c???12a2ab?a2b2?a2
21ab?a2b?2a3?c100
?(?1)3?1abb??aa22b2b??a22a?(b?a)(b?a)a1b?2a(ab)3
ax??bybzay?bzaz?bxxybxazax??bxbyaxay??bybz?(a?b)yzzz (2)ay33xx;
az?y 证明
ax ay??by?bzaybxaz?
ax?bz?bxaz?bxbyaxay??by azbzxay?bzaz?bxyay?bzaz?bx ?ayzazax??bxbyaxay??bybz?bzxazax??bxbyaxay??by
bzxay?bzzyzaz?bx ?a2yazax??bxbyxy?b2zxxyaxay??by
zbzxyzyzx ?a3y3zzxxy?bzxxyy
z
线性代数同济大学 课后习题答案详解
