工數(shù)第1章行列式習(xí)題

第1章

習(xí)題1.

計(jì)算下列排列的反序數(shù)，從而判斷奇偶性。（3）

t(n(n

-

1)321)=

(n

-

1)

+

(n

-

2)

+

+

2

+

1=

n(n

-

1)2（4）

t(135(2n

-

1)246(2n))=

0

+

1

+

2

+

+

(n

-

1)=

n(n

-

1)22.已知排列i1i2

in

的反序數(shù)，求inin-1

i1的反序數(shù)。，設(shè)排列中有l(wèi)(i

j

)個(gè)解：對(duì)于排列i1i2

in

中的數(shù)字i

j21

2

nt(i

i

i

)

=

n(n

-

1)

-t(i

i

i

)n n-1

1小于它的數(shù)字，設(shè)這些小于它的數(shù)字中，位于其右邊的有

r(i

j

)

個(gè)，則位于其左的有

l(i

j

)

-

r(i

j

)

個(gè)。n

n

n則：t(in

in-1

i1

)=[l(i

j

)-r(i

j

)]=

l(i

j

)-

r(i

j

)j

=1

j

=1

j

=1對(duì)于任意n個(gè)不相等的自然數(shù)，其中最大的數(shù)字有n-1個(gè)小于它的，次大的數(shù)字有n-2

個(gè)小于它的，……因此，2n(n

-

1)nj

=1

l(i

j

)

=

(n

-

1)

+

(n

-

2)

+

+

1

+

0

=5.寫出四階行列式中含因子a23

且?guī)ж?fù)號(hào)的項(xiàng)。解：四階行列式中的項(xiàng)為a1

j

a2

j

a3

j

a4

j1

2

3

4j1

j2

j3

j4

是數(shù)字1、2、3、4的組合。含因子

a23

時(shí)，令

j2

=

3則j1

j2

j3

j4

可能的組合有：1324，1342，2314，2341，4312，4321其中奇排列為：1324，2341，4312則含因子a23

且?guī)ж?fù)號(hào)的項(xiàng)為：a11a23a32a44

,

a12a23a34a41

,

a14a23a31a42如何組合，在中都至少有一個(gè)數(shù)字≥3，使得中出現(xiàn)，使得因此該行列式的值為0.6.利用行列式的定義計(jì)算424132311

j1

2

j2

3

j3

4

j4

5

j5a

a

a

aa51

a52a21

a22（2）a11

a12j1

j5a13

a14

a15a23

a24

a25a

a

0

0

0

=

(-1)t(

j1

j5

)

aa

a

0

0

00

0

0分析

a1

j

a2

j

a3

j

a4

j

a5

j1

2

3

4

51

2

3

4

5，無論

j

j

j

j

j3

4

5j

j

j543211

j

2

j

3

j

4

j

5

ja

a

a

a

aa

(i

?

3,

j

?

3)ij5431

21

j

2

j

3

j

4

j

5

ja

a

a

a

a

=

06.利用行列式的定義計(jì)算（4）=1

5(-1)a1

j

a2

j

a3

j

a4

j

a5

j1

2

3

4

5xy0000xy0000xy0000xyy000xj1

j5t(

j

j

)其中非0項(xiàng)為：(-1)t(12345)

a

a

a

a

a

+

(-1)t(

23451)

a

a

a

a

a11

22

33

44

55

12

23

34

45

51=

x5

+

y5（3）8.利用行列式的性質(zhì)計(jì)算1214=

01

=

21a

b

c

1b

c

a

1c

a

b

10

0

0

0abc1bca1cab1b

+

cc

+

aa

+

b2r4

-r2

-r3=b

c

1c

a

1

r

?a bb

+

c

c

+

a

a

+

b2

2

2（1）9.不展開行列式，證明下列等式成立。a

b

cb'

c'a''

b''

c''=

2

a'a

+

b

a'+b'

a''+b''c

+

a

c'+a'

c''+a''b

+

c

b'+c'

b''+c''證明：a''+b''+c''2

a'+b'+c'a

+

ba'+b'a''+b''c

+

ac'+a'c''+a''a

+

b

+

cc3

-c1c2

-c1=

2a

+

b

+

ca'+b'+c'bb'cc'=

2a

+

b

+

ca'+b'+c'bb'cc'=右邊a''+b''+c''-

b''-

c''a''+b''+c''b''c''(c1

+c2

+c3

)?2左邊

=cos

2b

=

0sin2

bsin2

gcos

2a（2）

sin2

a證明：=0

=右邊sin2

g=

sin2

bcos2

acos2

bcos2

gcos2

acos2

bcos2

gsin2

asin2

g左邊=sin2

bcos2

a

-

sin2

acos2

b

-

sin2

bcos2

g

-

sin2

gcos

2gcos2

acos2

bcos2

gcos2

acos2

bcos2

gsin2

ac3

+c1（3）0x

2

,

(

xyz

?

0)y2x

y

z

0

1

1

10

z

y

1

0

z

2z

0

x

=

1

z

2

0y

x

0

1

y2

x

2

0證明：0112222xyz0

1

1

1xyz

0

z

yz

0

xy2

x2xyz

xyz0

1

12

1

0

z

xy1

z

xy

0

x

yz1

y

xz

x

yz

0y xz

r4

·

xyzr3

·

xyzr2

·

xyz=r2

?

x

r3

?

y

r4

?

z左邊=(xyz)c2

?

x

c3

?

yc4

?

z=右邊=0

1

1

10

z2

y21

z2

0

x21

y2

x2

0c1

?

xyz

110.計(jì)算行列式。（1）

ada

+

b

+

c

+

d4a

+

3b

+

2c

+

d10a

+

6b

+

3c

+

db

ca a

+

b a

+

b

+

ca

2a

+

b

3a

+

2b

+

ca

3a

+

b

6a

+

3b

+

c解：=a

b

c

d0

a a

+

b a

+

b

+

ca

2a

+

b

3a

+

2b

+

c0

a

3a

+

b

6a

+

3b

+

cri

-ri-1i

=4,3,2

0a a

+

b a

+

b

+

c=

a

a

2a

+

b

3a

+

2b

+

ca

3a

+

b

6a

+

3b

+

c按第一列展開a a

+

b a

+

b

+

c=

a

0

a

2a

+

b0

a

3a

+

bri

-ri-1i

=3,2a

3a

+

b按第一=

a列展開原式42

a

2a

+

b

=

a（2）123

n-103

n-1-20

n

-1-2-3

0解：1

2

3

n0

2

6

2n0

0

3

2n

=

n!

0

0

0

nci

+c1原式

=i

=2,3n（3）a(

n-1)nxna1n-1

a1na2n

-1

a2na3n

-1

a3n

x1

a12a13

x1

x2

a23

x1x2x3x1x2x3x1x2x3解：a1n

-1a(

n

-1)n

-

a(

n

-2)nxn

-

a(

n

-1)nxn

-1

-

a(

n

-2)(

n

-1)0a1na2n

-

a1na3n

-

a2na2n

-1

-

a1n

-1a3n

-1

-

a2n

-1xn

-1xn

-1a13a23

-

a13x1

a120

x2

-

a1200x3

-

a23000000ri

-ri

-1=i

=n,n

-1,2=

x1

(

x2

-

a12

)(

x3

-

a23

)(

xn

-

a(

n-1)n

)（4）

a1

-

b1a1

-

b2

a1

-

bna2

-

b1a2

-

b2

a2

-

bn

an

-

b1an

-

b2

an

-

bn解：000013121110

n

>

2(a2

-

a1

)(b2

-

b1

)

n

=

2a3

-

a1an

-

a10

0

=

a1-

b1b1-

b2b1-

b3

b1-

bna2-

a1000a-

ba-

ba-

b

a-

bna2a3a1a1a2a3a1a1a2a3a1a1

a2

a3a1a1an-

a1an-

a1an-

a1an

-

a1ci

-c1=i

=2,3nri

-r1原式

=i

=2,3,n（5）

1

+

x1

y11

+

x1

y2

1

+

x1

yn1

+

x2

y11

+

x2

y2

1

+

x2

yn

1

+

xn

y11

+

xn

y2

1

+

xn

yn解：原式==

A

+

B11

+

x1

y2

1

+

x1

ynx1

y11

+

x1

y2

1

+

x1

yn11

+

x2

y2

1

+

x2

yn+x2

y11

+

x2

y2

1

+

x2

yn11

+

xn

y2

1

+

xn

ynxn

y11

+

xn

y2

1

+

xn

yn=

01

22

22

2n

>

2n

=

2x

y

-

x

yA

=xnyn1

xny

22

nri

-r11

x1y

2

x1yn1

x

y

x

y

i

=2,3n

∴原式=

0n

=

2n

>

3

y1

(

x1

-

x2

)B

=

y1x2

1

1

xn

1

1x1

1

1x11

+

x1

y2

1

+

x1

ynx21

+

x2

y2

1

+

x2

yn

xn1

+

xn

y2

1

+

xn

ynci

-

yi

c1=

y1i

=2,3nn

=

2n

>

3(

x1

-

x2

)(

y1

-

y2

)=

A

+

B

=

0（6）xnxnxn

-

m=

x

-

mx

-

mx

-

mn

inixn

-

mn

i

i

=1n

-1x2

i

=1=

(-m)1x1

-

mx1x2x2

-

mx1x21ci

-c1

·

xi

1

0

0

1

-

m

01

0

-

m

x2

xn

1

x2

-

m

xni

=2,3n

i

=1c1

+c2

++cn

解：原式

=

11.

利用行列式的性質(zhì)求方程：=

0

,

n

>

1111

1111

-

x1

11112

-

x

11

111

n

-

2

-

x1111

1n

-

1

-

x解：左邊=

(-

x)(1

-

x)(n

-

3

-

x)(n

-

2

-

x)

=

0111

110-

x0

00001

-

x

0

0000

n

-

3

-

x0000

0n

-

2

-

xri

-r1=i

=2,3n則方程的根為x

=

0,1,2n

-

212.計(jì)算下列n

階行列式。（1）

x000y解：00

xy00

y000

0x00

xyy000xy000x0000xy000xxy00y0000x00xy00x

+

(-1)n

+1

y

按第一列=展開原式=

xn

+

(-1)n

+1

yn123

n

-

1n1-

10

0002-

2

00

000

2

-

n0000

n

-

11

-

n解：（2）n(n

+

1)202-

130

n

-

1

0n002-

2

00

000

2

-

n0000

n

-

11

-

nc1

+c2

++cn原式

=222=(n

+

1)!(-1)n

-1=

n(n

+

1)

(-1)n

-1

(n

-

1)!-

10

002-

2

00

00

2

-

n000

n

-

11

-

n按第1列n(n

+1)=展開an

-1a2a1a

-

1(a

-

1)2

(a

-

1)n-11a

-

2(a

-

2)2

(a

-

2)n

-1

1a

-

n

+

1(a

-

n

+

1)2

(a

-

n

+

1)n

-1（3）

1(a

-

n

+

1)n

-1(a

-

2)n

-1(a

-

1)n-1an

-11aa21

1

1a

-

1

a

-

2

a

-

n

+

1(a

-

1)2

(a

-

2)2

(a

-

n

+

1)2轉(zhuǎn)置原式==

[(a

-

1)

-

a]

[(a

-

2)

-

a][(a

-

n

+

1)

-

a]范德蒙行列式2=

(-1)

(n

-

1)!(n

-

2)!1![(a

-

2)

-

(a

-

1)][(a

-

n

+

1)

-

(a

-

1)][(a

-

n

+

1)

-

(a

-

n

+

2)]=

(-1)n

-1

(n

-

1)! (-1)n

-2

(n

-

2)!

(-1)11!n(

n

-1)212

21

1221

1221

121in+1n+1

n

+1n

+1

n

+1n

+1

n

+1n

+1,

(a

?

0

)bnbn

-1

bnaa

bn-1a

bn

-1

bnan

-2b2an

-2b2an

-2b2an-1ban

-1ban

an

-1ban

（4）

anibbbann

b

b

b

n

b

nr

?an

n+1

an

+1

2

n+1

an

+1

n+1

an

+1

2

a2

b

2

2

a2

1

a1

b

2

1

a1

1

a1

2

a2n

+1i

=1111

i

i原式

=i

=1,2n

+1nbn

-

a

a

b

b

b

b

b

b

n

+1

n

bn

+1b2

bn

+1

b3

b2

a2

a1

a3

a1

an

+1

a1

3 2

n

+1 2

-

-

a

a

a

a

2

-

1

3

-

1

n+1

-

1

范德蒙n

+1=

ai行列式i

=1=

(b2a1

-

b1a2

)(b3a1

-

b1a3

)(bn+1a1

-

b1an+1

)(b3a2

-

b2a3

)(bn+1a2

-

b2an+1

)(bn+1an

-

bnan+1

)13.證明下列等式22

22cos

g

+

a2

2cos

g

-

a

sin

g

+

a2

2cos

b

+

g

=

1

[sin(b

-

g)

+

sin(a

-

g)

+

sin(g

-

b

)]2

2cos

b

-

g

sin

b

+

gcos

a

-

b

sin

a

+

b

cos

a

+

b（1）左邊=sin

b

+g

cos

g

+a

cos

a

-b

+sin

a

+b

cos

b

+g

cos

g

-

a2

2

2

2

2

2+

sin

g

+

a

cos

a

+

b

cos

b

-

g

-

cos

a

+

b

sin

b

+

g

cos

g

-

a2

2

2

2

2

2-

cos

g

+

a

sin

a

+

b

cos

b

-

g

-

cos

b

+

g

sin

g

+

a

cos

a

-

b2

2

2

2

2

242

2

2

22

2

2=

1

[sin(b

+

g)

+

sin(g

+

a

)

+

sin(b

-

a

)]2

=

1

sin

b

+

2g

+

a

cos

a

-

b

+

sin

b

-

a

cos

a

-

b

2

=

1

sin

b

+

2g

+

a

+

sin

b

-

a

cos

a

-

bsin

b

+

g

cos

g

+

a

cos

a

-

b2

2

2其中其中42

2

2

22

2

2=

1

[sin(a

+

b

)

+

sin(b

+

g)

-

sin(a

-

g)]2

=

1

sin

a

+

2b

+

g

cos

g

-

a

-

sin

a

-

g

cos

g

-

a

2

=

1

sin

a

+

2b

+

g

-

sin

a

-

g

cos

g

-

acos

a

+

b

sin

b

+

g

cos

g

-

a2

2

224

4=

1

[sin(b

-

a

)

+

sin(a

-

g)

+

sin(g

-

b

)]+

1

[sin(g

+

a

)

+

sin(a

+

b

)

+

sin(g

-

b

)]-

1

[sin(a

+

b

)

+

sin(b

+

g)

-

sin(a

-

g)]4

4-

1

[sin(g

+

a

)

+

sin(a

+

b

)

-

sin(g

-

b

)]-

1

[sin(b

+

g)

+

sin(g

+

a

)

-

sin(b

-

a

)]4

4=

1

[sin(b

+

g)

+

sin(g

+

a

)

+

sin(b

-

a

)]+

1

[sin(a

+

b

)

+

sin(b

+

g)

+

sin(a

-

g)]左邊b2c2=

4(a-

b)(a

-

c)(b

-

c)(a

+

2)2(b

+

2)2(c

+

2)2(a

+

1)2(b

+

1)2(c

+

1)2（2）

a2223

2c2c2a2c2a2c2a2a2c2a2=

4(a

-

b)(a

-

c)(b

-

c)a2

-

c2

a

-

cb2

-

c2

b

-

ca2

-

c2

a

-

c

0b

-

c

0

=

4c

1=

4

b2

-

c2c22a

1

a

12b

1

=

4

b2

b

12c

1

c

1=

2

b2=

2

b22a

+

1

22b

+

1

22c

+

1

22a

+

12b

+

12c

+

12a

+

1

12b

+

1

12c

+

1

1r1

-r3r2

-r3c2

-c32a

+

3c3

-c22b

+

3

=

b2b

+

3c

-c=

bc2

-c1左邊aaaaaa1

=

ax1

x2

xna

+

xn

aa

aa

aa a

+

x2

aa

+

x（3）1x

0

0

a0

x2

0

a

=

ax1

x2

xn0

0

xn

a0

0

0

ai

=1,2,,nci

-

cn

+1

左邊證明：=右邊n

-1an

-2an

-1a0a1a2=

a

xn

-1

+

a

xn

-2

+

+

a0

1-

10

00x-

1

000x

00

00

x-

100

0x（4）證明：n

-1n

+1+

an

-1

(-1)=

a

xn

-1

+

a

xn

-2

+

+

a0

1-

100

0x-

10

00x-

1

0

000

-

1x-

1

0

0-

1

0

0

00x-

1

00x-

1

0a000x

0-

a100x

0+

000

x000

x按第1列左邊

=展開14.利用拉普拉斯定理計(jì)算行列式。a100-

1b100-

1c100-

1d（1）a1c1+

(-1)1+

2+1+

3a0-

11-

1b-

1d-

110d解：按前兩行展開，非0項(xiàng)有：D

=

(-1)1+

2+1+

2=

(ab

+

1)(cd

+

1)

+

ad0ab0bc00cd0da00

a

0

ab

0

c

0

db2

0

c2

0

d

2（3）

a解：按前兩行展開，非0項(xiàng)有：D

=

(-1)1+

2+1+3

abac0abcd0bcdad

200+

(-1)1+

2+1+5

abad0abcdc2000bcda+

(-1)1+

2+

3+5

acadb2000abcd0bcdab2

1

1 a

cc

d

c a=

abd

(a2

-

c2

)[d

2

(b

-

c)

-

c2

(b

-

d

)

+

b2

(c

-

d

)]1

a

cd

c

a

-

abdd

2

1

1

a c

+

abd

c2

1b

c

c

a

b=

-abdabbba

bb

aaaD

=2n（4）解：n行b

a

n行baba

bb

abaaa