哈工大版線性代數(shù)與空間解析幾何課后題答案高等教育

1/1哈工大版線性代數(shù)與空間解析幾何課后題答案-高等教育

習(xí)題一

1．按自然數(shù)從小到大的自然次序，求解各題.(1)求1至6的全排列241356的逆序數(shù).解：t(241356)0021003.

(2)求1至2n的全排列135(2n1)246(2n)的逆序數(shù).

解：t(13(2n1)242n)000(n1)(n2)210(3)選擇i與j，使由1至9的排列，91274i56j成偶排列.

解：由91274i56j是從1至9的排列，所以i,j只能取3或8.

當(dāng)i8,j3時，t(912748563)01112133618，是偶排列.當(dāng)i3,j8時t(912743568)01112322113，是奇排列，不合題意舍去.

(4)選擇i與j，使由1至9的排列71i25j489成奇排列.

解：由71i25j489是從1至9的排列，所以i,j只能取3或6.

當(dāng)i3,j6時，t(713256489)0112113009，是奇排列.當(dāng)i6,j3時，t(716253489)01122330012，是偶排列，不合題意舍去.

2．計算下列行列式(1)

n(n1)

.2

9a18b3215332053

；(2)；

26b13a7528475184

10

812

2

abbf

accdcf

aede.ef

3；(4)bd

(3)15

203212

解：(1)

9a18ba2b

913117(a24b2).

26b13a2ba

32153320533205310032053320533205332053

(2)

75284751847518410075184751847518475184

0751840032053004313100.

10

(3)15

812

222230.

35433

203212

4812

1

ab

(4)bd

accdcf

aeef

11

111

11

111

02200

deabcdef11abcdef0

bf

111

abcdef0

204abcdef.02

x

3．已知3

yz

021，利用行列式性質(zhì)求下列行列式.11y3yy2x

y3yy201

z

3z2；(2)z2

zz2

xxyz

3z230

yz

x1y1z134x

y01z

2.3

1x

(1)3x3

x2

解：(1)3x3223022.

111111413111111

x2

(2)

222

x1y1z134

3

2302302

413x

yz

302302101.11

1

4．用行列式定義計算：

0100

20230

(1)；(2).3

4000n15n000

12

(1)t(p1p2p3p4p5)a1p1a2p2a3p3a4p4a5p5解：(1)3

45

(1)t(54321)a15a24a33a42a51(1)1012345120.

2

1

0102

(1)t(p1p2pn)a1p1a2p2anpn(2)0

n1

n0

(1)

t(23n1)

a12a23a(n1)nan1

(1)n1123n(1)n1n!5．用行列式的定義證明：

a11a21

(1)0

a12a22000a12a22a32a42

a13a2300000a33a44

a14a24a34a44a5400a34a45

a15a25

a350；

a45a55a11a21

a12a33

a22a43

a34a44

00a11

(2)

a21a31a41

.

a11a21

證：(1)D0

a12a22000

a13a23000

a14a24a34a44a54

a15a25

a35(1)t(p1p2p3p4p5)a1p1a2p2a3p3a4p4a5p5a45a55

00

假設(shè)有a1P1a2P2a3P3a4P4a5P50，由已知p3,p4,p5必等于4或5，從而p3,p4,p5中至少有兩個相等，這與p1,p2,p3,p4,p5是1,2,3,4,5的一個全排列沖突，故全部項

a1P1a2P2a3P3a4P4a5P50，因此D0.

a11

(2)

a12a22a32a42

00a33a43

00a34a44

(1)t(p1p2p3p4)a1p1a2p2a3p3a4p4，由已知，只有當(dāng)p1,p2

a21a31a41

取1或2時，a1p1a2p2a3p3a4p40，而p1,p2,p3,p4是1,2,3,4的一個全排列，故p3,p4取

3或4，于是

3

D(1)t(1234)a11a22a33a44(1)t(1243)a11a22a34a43(1)t(2134)a12a21a33a44

(1)

t(2143)

a12a21a34a43

a11a22a33a44a11a22a34a43a12a21a33a44a12a21a34a43從而

a11a21a12a33

a22a43

a34a44

(a11a22a12a21)(a33a44a34a43)

a11a22a33a44a11a22a34a43a12a21a33a44a12a21a34a43D6．計算

a0

(1)

10

121102111

；

214421110

123n

xaa

1100

axa

；(4)Dn1010；(3)Dn

aax

1001

a0011110a00111

(5)Dn00a0；(6)Dn111.

100a111a305

a30

按第4按第30b02a344

解：(1)(1)d0b0(1)33dcabcd.

12c3行綻開0b列綻開

12c

000d121112111211021110211102111

(2)

21440366036621110033120036

3b2000c052

；(2)3d

4

10

3

00

2111211121112202320232

339.

21110037003703600360001

xaar1r2(n1)ax(n1)ax(n1)ax

(3)Daxar1r3axnaaxr1rn

aa111ax

a[(n1)ax]aa

aa

a

11110xa

00[(n1)ax]00

xa0

000

xa

[(n1)ax](xa)n1.

1

23n1

100cc23n

12(4)Dn1

010c1c

30

1

00

1c1c

n0

123nn(n1)

2

.

a0010a00

(5)Dn00a0

100a

23100100a

x

n00

1

5

a00

按第1行綻開0a011

(1)a(1)1n

00a

an(1)1n(1)n11an2anan2.

0001a00a

0000a0

1111

(6)Dn11

1111111110200020001002

(2)n1(1)n12n1.7．證明

a2

(1)2a

abb2

1a2

證：2a

ab2b(ab)211ab

b2

ab2b2aabab2b2b

c(1)c3

112023

33

c1(1)c2

a2ababb2b2

1

a(ab)b(ab)

(1)

ababab

(ab)2(ab)3

11

a2

(2)

(a1)2(b1)2(c1)2(d1)2

a2b2c

2

(a2)2(b2)2(c2)2(d2)2

(a3)2(b3)2(c3)2(d3)2

a24a4b24b4c4c4

2

b2c2d2

0

a22a1b22b1c2c1

2

a26a9b26b9c6c9

2

證：等式左端

d2

6

d22d1d24d4d26d9

2

2a1ac2(1)c1

b22b1

c3(1)c12

c2c1

c4(1)c12

d2d1

4a46a94c4

a2

2

2a1262b1262c1262d126x1

x12

2x22x32x4

4b46b9c3(2)c2b26c9c4(3)c3c

4d46d9d2

x13c1x12c2x1c3

32

x2c1x2c2x2c3

0

x1a1

(3)

x12b1x1b2

2

x2b1x2b22x3b1x3b22x4b1x4b2

x13

3

x2

3x33x4

x2a1x3a1x4a1

x232

x3c1x3c2x3c3x3

x4

x12b1x1

2

x2b1x22x3b1x32x4b1x4

32x4c1x4c2x4c3

c2(a1)c1x1

x2

證：等式左端c(b)c321

x3

c4(c3)c1

x4

c3(b1)c2x

2c4(c2)c2x3

x4

x1

x12

2x2

x13c1x12c2x1

32x2c1x2c2x232x3c1x3c2x332x4c1x4c2x4

x13c1x12xcx

3

334

213214

x1x3x4

x12

2x2

x13

3x2

32c(c)cc1x2x2x2413x

x

2

324

x

xcxx

2324

x

x

3334

等式右

端.

8．解關(guān)于未知數(shù)x的方程

x

(1)3

1x201x20

26x12

6(x1)x1

x3

1x2

0

0x

解：3

(x1)[x(x2)3](x1)[x22x3](x1)(x3)(x1)0所以x11,x23,x31.

a

(2)m

ax

xb

mm0(m0)

b

7

a

解：m

ax

xb

aab

x00b

x

xa1b

mmm111m11

xb

b

11

m(xa)m(xa)(xb)0

bx

因m0，所以x1a,x2b.

a11

9．設(shè)

a12a1na22a2nan2ann

a22a2na12a1na1p1

a1p2a2p2anp2

a1pna2pn

anpn

，其中“”是對1,2,,n的全部全排列p1p2pn

；(2)

a21an1

a，求下列行列式：

an2ann

a1n

a12

a11a21an1

；

an1

(1)

a21a11

a2na22annan2

(3)

p1p2pn

a2p1anp1

取和，n2.

解：（1）經(jīng)行的交換得

a11an1

原式(1)n1

a12a1nan2anna32

a3n

a31a21

a22a2n

a11

a12a1na22a2n

an2ann

(1)

(n1)(n2)21

a21an1

(1)

n(n1)2

a.

(2)與(1)類似，經(jīng)列的交換得

8

n(n1)原式(1)

2

a.

(3)經(jīng)列的交換，得

a1p1

a1p2a1pna11a12a1n

a2p1a2p2a2pn

pn)

a21a22a2n(p1p2pn)(1)(p1p2

(1)a

anp1

anp2

anpn

an1

an2ann

111故原式(1)(p1p2pn)

aa

111

p1p2pn

0.

1

11

10．計算行列式

aaa000100b111aa00

(1)

0a2b200b；(2)011aa0；

3a30b0011aa4

a4

00011a6111110001

61110100

(3)1

1611；(4)0010.

11161000111

11

6

k000

a100b1a100b1a1b100解：(1)0a2b200b3a30

b400a4b4a4000b

3a30

00a3b3b4

a4

a2b20

b2

a2

a1b1a3b3b(a1a4b1b4)(a2a3b2b3).

4

a4b2

a2

(2)將前4行依次加到第5行，再按第5行綻開得

aa000

11aa0

aa00

原式011aa

0a5

11aa0

0011aa

011aa

a000

1

0011a

9

aa00

a5

11aa

aa

04011aa

a5a11aaa00

1

011a

a

a

a5

a4

1

1aaa5a4a3aa

a

01

11a

1aa2a3a4a5

61111101010

10101

611116111(3)1

16111161111161111611111

611

116111111111116

111050001011

6

111000500

111610005011116

00005

10546250.(4)按最終一行綻開得

10000100

1

0001000010k

1

00010

00010

1

0001

k000

1000

k5

11．計算行列式

1111x1

111x11

x1mx1x1(1)11x111

；(2)x2x2mx21x1111

x3x3x3mx11111

x4x4x4解：(1)依次將第2,3,4,5列加到第1列得

10

x1

x2

x3

x4m

x1111x1x111x11

原式x11x111

x1x1111x11111

111x111x11

(x1)1x111

x11111111000x00x0

(x1)0x00

x0000000

4(41)(1)

2

(x1)x4x4(x1)

(2)依次將第2,3,4行加到第1行得

44

4

4

xi

m1

xi

mi1

xi

mi1

xi

m

ii1

原式

x2x2mx2x2x3x3x3mx3x4x4

x4

x4m

11114

(

xm)x2

x2mx2x2i

i1

x

3

x3x3mx3x4

x4

x4

x4m

1111

4(x0m00

im)

i1

00m0000m

4

(m

x3

i)m

i1

12．計算行列式

11

a1b1

(1)

a1b2a2b2a3b2a4b21a1b2

a1b3a2b3a3b3a4b31a1b3

a1b4a2b4a3b4a4b41a1b4

；

a2b1a3b1a4b11a1b1

(2)

a2b11a2b21a2b31a2b4a3b11a3b21a3b31a3b4a4b11a4b21a4b31a4b41100

(3)

；(4)

11a301a401

00

a1a2

a1b1

a1b2a2a1a3a2a4a3

a1b3a2a1a3a2a4a3

a1b4a2a1a3a2a4a31a1b3

11123x

2

49x827x3

解：（1）依次將第3,2,1行乘1加到第4,3,2行得

原式

a2a1a3a2a4a31a1b1

0

(2)依次將第3,2,1行乘1加到第4,3,2行得

1a1b21a1b4

原式

b1(a2a1)b2(a2a1)b3(a2a1)b4(a2a1)b1(a3a2)b2(a3a2)b3(a3a2)b4(a3a2)b1(a4a3)b2(a4a3)b3(a4a3)b4(a4a3)

a1b11a1b21a1b31a1b4

(a2a1)(a3a2)(a4a3)

b1b1b1

b2b2b210

00

b3b3b301

b4b4b410

110

0

(3)按最終一列綻開得

1

原式a41

01

010

00011

0a3110a2011a10

01101

a1a2a3a4

(4)由Vandermonde行列式的計算公式得

12

原式(x3)(x2)(x1)(32)(31)(21)2(x1)(x2)(x3)13．證明

a11000a2

x10

0(1)Da30x0

n

a11xna22xnan1xan

an100x1an

0

x

a110000a2

x1000rx)ra0x0

00證：等式左端

n(n1

3

an200x10an1000x1anan1x

0x20

a110000a2x1

000rn(x2)rn2a3

0x000

r3

n(x)rn3

r1n(xn)rnan200

x10an1000x1f(x)00000

1x1

(1)n1

f(x)

f(x)

1x

1(n1)階

其中f(x)a11xnan1xan.

13

210

(2)D

00

1210001200000000

n12112

證：1n1時，D1211

21

2假設(shè)當(dāng)nk時結(jié)論成立，當(dāng)nk1時，若k12，D2

12

41321結(jié)論成立.若k13，將Dk1按第一行綻開得

21Dk1

121

2DkDk12(k1)(k11)(k1)1

1

12

11

1

11

ai(1

i1

i1

n

n

由數(shù)學(xué)歸納法，對一切自然數(shù)n結(jié)論成立.

a1

(3)D

11

1a21

1

ai0,i1,2,,n.ai

11an

證：(用加邊法)

1

等式左端0

111

111a2

1

111

111

111

100

100

100

01a10

1a1

a2

11anan

111a1a2an

00

111

a100

0a2000an

14

nn

1111

a1a2anai(1等式右端.(1a1a2ani1aii1xyxy0001xyxy0001xy00xn1yn1

(4)Dn，其中xy.

xy000xyxy0001xy

x2y2

，等式成立.證：當(dāng)n1時，D1xy

xy

假設(shè)nk時等式成立，當(dāng)nk1時，若k12，則

x3y3

Dk1D2xxyy，等式成立.若k13，將Dk1按一列綻開，得

x