大学-线性代数习题答案01
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大学数学-线性代数习题答案
第一章行列式
1利用对角线法则计算下列三阶行列式
(1)
381141102
解
381141102
2(4)30(1)(1)118
0132(1)81(4)(1)
2481644
(2)
bacacbcba
解
bacacbcba
acbbaccbabbbaaaccc
3abca3b3c3
(3)
222111
cbacba
解
222111
cbacba
bc2ca2ab2ac2ba2cb2
(ab)(bc)(ca)
(4)
yxyxxyxyyxyx
解
yxyxxyxyyxyx
x(xy)yyx(xy)(xy)yxy3(xy)3x3
3xy(xy)y33x2yx3y3x3
2(x3y3)
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)1234
解逆序数为0
(2)4132
解逆序数为441434232
(3)3421
解逆序数为532314241,21
(4)2413
解逆序数为3214143
(5)13(2n1)24(2n)
解逆序数为
2)1(nn
32(1个)
5254(2个)
727476(3个)
(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1个)
(6)13(2n1)(2n)(2n2)2
解逆序数为n(n1)
32(1个)
5254(2个)
(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1个)
42(1个)
6264(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a
11a
23的项
解含因子a
11a
23的项的一般形式为
(1)ta
11a
23a
3ra
4s
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a
11a
23的项分别是
(1)ta
11a
23a
32a
44(1)1a
11a
23a
32a
44a
11a
23a
32a
44
(1)ta
11a
23a
34a
42(1)2a
11a
23a
34a
42a
11a
23a
34a
42
4计算下列各行列式
(1)
71100251020214214
解
71100251020214214
0100142310202110214
732
34
cc
c
c34)1(
143102211014
143102211014
0
1417172001099
32
321
1
cc
cc
(2)
2605232112131412
解
2605232112131412
2605032122130412
24
cc
0412032122130412
24
rr
0
0000032122130412
14
rr
(3)
efcfbfdecdbdaeacab
解
efcfbfdecdbdaeacab
ecbecbecb
adf
abcdefadfbce4
111111111
(4)
dcba
100110011001
解
dcba
100110011001
dcbaab
arr
100110011010
21
dcaab
101101
)1)(1(12
010111
23
cdcadaabdcc
cdadab
111
)1)(1(23abcdabcdad1
5证明:
(1)
1112222
bbaababa
(ab)3;证明
1112222
bbaababa
0012222222
12
13ababaabaabacc
cc
abababaab
22)1(222
13
21))((aba
abab
(ab)3
(2)
yxzxzyzyx
ba
bzaybyaxbxazbyaxbxazbzaybxazbzaybyax
)(33
;证明
bzaybyaxbxazbyaxbxazbzaybxazbzaybyax
bzaybyaxxbyaxbxazzbxazbzayy
b
bzaybyaxzbyaxbxazybxazbzayx
a
bzayyxbyaxxzbxazzy
b
ybyaxzxbxazyzbzayx
a
22
zyxyxzxzy
b
yxzxzyzyx
a33
yxzxzyzyx
b
yxzxzyzyx
a33
yxzxzyzyx
ba)(33
(3)0
)3()2()1()3()2()1()3()2()1()3()2()1(
2222222222222222
ddddccccbbbbaaaa
;证明
2222222222222222
)3()2()1()3()2()1()3()2()1()3()2()1(
ddddccccbbbbaaaa
(c
4c
3c
3c
2c
2c
1得)
523212523212523212523212
2222
ddddccccbbbbaaaa
(c
4c
3c
3c
2得)
0
2212221222122212
2222
ddccbbaa
(4)
444422221111
dcbadcbadcba
(ab)(ac)(ad)(bc)(bd)(cd)(abcd);证明
444422221111
dcbadcbadcba
)()()(0)()()(001111
222222222addaccabbaddaccabbadacab
)()()(111
))()((
222addaccabbdcbadacab
))(())((00111
))()((
abdbddabcbccbdbcadacab
)()(11
))()()()((
abddabccbdbcadacab
=(ab)(ac)(ad)(bc)(bd)(cd)(abcd)
(5)
1221 1 000 00 1000 01
axaaaaxxx
nnn
xna
1xn1a
n1xa
n
证明用数学归纳法证明
当n2时
212
1221
axax
axax
D
命题成立