同济大学线性代数第六版答案(全)

同济大学线性代数第六版答案(全)

1 利用对角线法则计算下列三阶行列式201

(1)1 4

*****

解1 4

183

2 ( 4)

3 0 ( 1) ( 1) 1 1 8 0 1 3 2 ( 1) 8 1 ( 4) ( 1) 2

4 8 16 4 4 abc

(2)bca

cababc

解bca

cab

acb bac cba bbb aaa ccc 3abc a3 b3 c3

111

(3)abc

a2b2c2111

解abc

a2b2c2

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)

n(n 1)

解逆序数为

2 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) 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)***-*****14

2 07

41 解***-*****c2 c***** 1

***** 104 1 10

2 122 ( 1)4

3 *****c

4 7c*****

3 1

4 4 110c2 c*****

123 142c00 2 0

1 2c*****

2 (2)31 1***** 22

4 解31 ***** c 4 c3 223 1202r 4 r ***-*****06

***-*****

6

2

***-***** 40

r4 140

r12

***-*****30

0 0 0

(3) bdabbf acae

cfcd deef

解bdabbf accfcdae

deef

ad bbb ccce ee

adfbce 1111

1 11 1 4abcdef

a1 (4) 001b 1001c 100 1d

a1 解001b 1001c 10r1 ar201 ab0 1b10 1d00a 1c 100 1d

aba0c3 dc2 abaad

( 1)( 1)2 1 1c1 1c1 cd

0 100 1d

5 证明:

abad abcd ab cd ad 1 ( 1)( 1)3 2 11 cd

a2abb2

(1)2aa b2b (a b)3;

111 证明

a2abb2c2 c1a2ab a2b2 a2

2aa b2b 2ab a2b 2a

00111c3 c11

ab ab aab a (a b)3 (b a)(b a)1 ( 1)2b a2b 2a 3 1

ax 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 bz

xay 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)2

cd2

(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