线性代数(同济四版)答案

线性代数(同济四版)习题参考答案

黄正华

Email:huangzh@http://doc.guandang.net

武汉大学数学与统计学院,湖北武汉430072

WuhanUniversity

目录

第一章行列式第二章矩阵及其运算

第三章矩阵的初等变换与线性方程组第四章向量组的线性相关性第五章

相似矩阵及二次型

i

117334870

ii目录

第一章

1.利用对角线法则计算下列三阶行列式: (1)

(3)

201

1 4 1 ;

183

111

abc ;

a2b2c2

01 2

1 4 1 183

行列式

(2)

(4)

abc

bca ;

cab xyx+y

yx+yx

x+y

x .

y

解:(1)

=2×( 4)×3+0×( 1)×( 1)+1×1×8 0×1×3 2×( 1)×8 1×( 4)×( 1)= 24+8+16 4= 4.(2)

abc

bca =acb+bac+cba bbb aaa ccc=3abc a3 b3 c3.

cab

111

222222

abc =bc+ca+ab ac ba cb=(a b)(b c)(c a).

a2b2c2

yx+yx

x+y

333x =x(x+y)y+yx(x+y)+(x+y)yx y (x+y) x

y

=3xy(x+y) y3 3x2y 3y2x 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)1234;(2)4132;(3)3421;

(5)13···(2n 1)24···(2n);(6)13···(2n 1)(2n)(2n 2)···2.解

(1)逆序数为0.

(2)逆序数为4:41,43,42,32.(3)逆序数为5:32,31,42,41,21.(4)逆序数为3:21,41,43.(5)逆序数为

n(n 1)

:(3)

(4)

x y

x+y

(4)2413;

1

32...........................................................................1个52,54.....................................................................2个72,74,76..................................................................3个..................................................................................(2n 1)2,(2n 1)4,(2n 1)6,...,(2n 1)(2n 2)..............(n 1)个

(6)逆序数为n(n 1):

32...........................................................................1个52,54.....................................................................2个72,74,76..................................................................3个..................................................................................(2n 1)2,(2n 1)4,(2n 1)6,...,(2n 1)(2n 2)..............(n 1)个42...........................................................................1个62,64......................................................................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:由定义知,四阶行列式的一般项为

( 1)ta1p1a2p2a3p3a4p4,

其中t为p1p2p3p4的逆序数．

由于p1=1,p2=3已固定,p1p2p3p4只能形如13 ,即1324或1342.对应的逆序数t分别为

0+0+1+0=1,或0+0+0+2=2.

所以, a11a23a32a44和a11a23a34a42为所求.

4.计算下列各行列式: 4124

1202

(1) 10520 ;

0117

ae abac

;(3) bd cdde

bfcf ef

(2)

(4)

21511000

02212100

215a120143611220 ;

3 12

解:(1)

r2 r4 ======

411001000

1 124

4202 r r12 ====== 10 520

0117

202

117 r+7r42 ======= 152 20 r3+15r2

72 4

0

1b10 .0 1c1

00 1d

1 22

0 74 r2 4r1

======= 0 15 r 10r310

017

12 02

17 =17×9 01 001785

00945

02210111

2

4 20

7 2 7 =0.5 5

(2)注意到r1+r2=r4,所以原行列式值为0.

(3) abacae e bd cdde bc 11

1 =adf b ce

=adfbce 1 11 bf

cf ef bc 1

e 1 1

111 =r2+r1 r===r=adfbce 002 = adfbce 02 =4abcdef.3+1

020 20 (4)

a

1

0 01+aba0

1b10

0 1c1 =r=1=+=ar=2= 1b10 1c1

00 1d 0

00 1d

1+aba

ad ==按第===1=列

==( 1)( 1)2+1 0 1+aba

c3+dc2 展开 1c1 ====== 1c1+cd 0 1d

0 10

==按第=展开==3=行 ==( 1)( 1)3+2

1+abad 11+cd

=abcd+ab+cd+ad+1.5. 证明: 2abb2 (1)

a

2aa+b2b =(a b)3; 111

证明

2 aabb2 2222 aab ab a

c2 c 2aa+b2b =c===1=

2ab a2b 2a 3 c1

1

11

1 00 1)3+1 ab a2b2 a2

=(

b a2b 2a =(b a)(b a)

ab+a 12 =(a b)3.

ax+byay+bzaz+bx (2) xyz

ay+bzaz+bxax+by =(a3+b3) yzx ; az+bxax+byay+bz

zxy 证明:

ax+byay+bzaz+bx

x yay+bz

ay+bzaz+bxax+by ay+bzaz+bx az ==按第=1列 分裂开=====a yaz+bxax+by +b zaz+bx

+bxax+by

ay+bz

zax+byay+bz xax+by xay+bzz

…… 此处隐藏：8836字，全部文档内容请下载后查看。喜欢就下载吧 ……