線性代數(shù)-同濟四版-習題參考答案

線性代數(shù).同濟四版—習題參考答案

線性代數(shù)(同濟四版)習題參考答案

黃正華

Emai1:huangzh@whu.edu.cn

武漢大學數(shù)學與統(tǒng)計學院,湖北武漢430072

WuhanUniversity

目錄第一章行列式第二章矩陣及其運算

第三章矩陣的初等變換與線性方程組

第四章向量組的線性相關(guān)性

第五章相似矩陣及二次型117334869

第一章

您發(fā)現(xiàn)有好的解法,請不吝告知.

行列式

課后的習題值得我們仔細研讀.本章建議重點看以下習題：5.(2),(5);7;8.(2).(這幾個題

號建立有超級鏈接.)若

1.利用對角線法則計算下列三階行列式:????????(1)????

999999999999(3)9999

????

????

201??

??

1?4?1????;

??

?183??

????

111??

??

abc????;

??

a2b2c2??

99Q99999⑵999Q

999999999999⑷Q999

????

????

abc??

??

bca????;

??

cab??xyx+y

yx+yx

????x+y??

??x????.

??y??

解：⑴

????

????

01????2

????

??1?4?1??

????

????

???183??

=2X(?4)X3+0X(?1)X(?1)+1X1X8?OX1X3?2X(?1)X8?1X(?4)X(?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.按自然數(shù)從小到大為標準次序,求下列各排列的逆序數(shù)：(1)1234;(2)4132;(3)3421;

(5)13???(2n?l)24???(2n);(6)13???(2n?l)(2n)(2n?2)???2.解

(1)逆序數(shù)為0.

(2)逆序數(shù)為4:41,43,42,32.

1

(4)2413;

(3)

99999999999??999

(4)

??????x????y????

??x+y

2

(3)逆序數(shù)為5:32,31,42,41,21.(4)逆序數(shù)為3:21,41,43.(5)逆序數(shù)為

n(n?l)

:第一章行列式

32................................................................................1

個52,54.........................................................................2

個72,74,76......................................................................3

個................................................................................

.....(2n?l)2,(2n?l)4,(2n?l)6,...,(2n?l)(2n?2)...............(n?l)個

(6)逆序數(shù)為n(n?D:

52

54.........................................................................2

72

74,76......................................................................3

(2n?l)2,(2n?l)4,(2n?l)6.....(2n?l)(2n?2)(n?l)個

.....(2n)2,(2n)4,(2n)6,(2n)(2n?2)................................(n?l)個

3.寫出四階行列式中含有因子alla23的項.

解：由定義知，四階行列式的一般項為

(?1)talpla2p2a3p3a4p4,

其中t為plp2p3p4的逆序數(shù).

由于pl=l,p2=3已固定,plp2p3p4只能形如13????,即1324或1342.對應(yīng)的逆序數(shù)t分別

為

0+0+1+0=1,或0+0+0+2=2.

所以，？alla23a32a44和alla23a34a42為所求.

4.計算下列各行列式:？??？

??4124??

????

????

??1202??

??(1)????10520??;????

????

??0117??

????

????

ae?????abac

????

??;(3)??bd?cdde????

????

??bfcf?ef??

??

9QQ9999Q(2)99QQ

99Q99999999999Q9⑷9Q9Q

999999

2021

4207

????

????

????

????

????rl?r2

99Q9=====?99Q9

????

????

????

140

21102212407

??

1999999

3?121????;1232????

??

5062??2

1

4

a

1

??0??????

?lblO????.O?lcl????

??

OO?ld??

1000

??

202????

??

?72?4?????152?20????

??

117??

解：⑴

??

999999999999999999

410

121

105105

????

????

????

????

????r2?4rl

99QQ=======999Q9p3^1Or1

????

????

????

線性代數(shù)(同濟四版)習題參考答案

????????r2?r4??======????

99Q999

(2)

9999999999999999

0117??r4+7r2????=="=??????r+15r320?152?20????

????????0?72?41

2

2

215210

120120

43643011220200

999999999999999999C^9C2Q9QQ--=99QQ=9999QQ99Q9999Q99999Q9999

??=0.????????

1000215

2100120

01179436

????

??2??????

????

9979999=17X9999985999999

????????450202

999999999999999999r49r29QQ9----9QQQ-Q9999Q999Q9Q9Q

21212010000121

0111434

??

299999979999=0599999Q59QO2OO99999999999999999999

3

9999999Q99999Q99999999999999r4Qr1QQ====9Q99

999999

3?123?123?12

3?12

(3)

????

ae???abac

??

??bd?cdde??????bfcf?ef

????

???Hl??r2+rl

=====adfbce????002r3+rl

??

??020

99Q99999999999Q9Q999a9100

lb?10

9999999999999999

e??l?????bc???ll

99999999

??=adf??b?ce??=adfbce??1?11????????????????????b??1c?e??1?1????????????02?????

???=?adfbce????=4abcdef.????????20????001

999999999999999999999

9Q9999Q999999Q9QQ9r1+ar2===^=

0?10

l+abb?l

ale

001??

999999999999999999

9QQQ999999Q9999?

(4)

01c

?ld????

??l+abaO??按第1列

=======(?1)(?l)2+l??cl???l展開?？

??0?ld????ad按第3行3+2??l+ab====(?l)(?1)??

???1展開1+cd

00?ld

????

????

ad??l+aba??

??c3+dc2????===??c1+cd???1??=

????

????0?10??????

??=abcd+ab+cd+ad+l.??

99Q9999999999999

5.??證明：？？?？22??abb????a

????

??=(a?b)3;(l)??2aa+b2b????

????

??111??

??????證明？？2??22??abb????aab?a2b2?a2??a

??c2?cl????

??2aa+b2b??=====????2ab?a2b?2a??c??

??3?cl????

??l??10011??

99Q999

??a??ab?a2b2?a2??

????3+l??=(?l)??=(b?a)(b?a)????

??l??b?a2b?2a??

99999999

9QQ9999Q

??xyz????ax+byay+bzaz+bx??

99999999

??=(a3+b3)??yzx??;(2)??ay+bzaz+bxax+by????????

99999999

??zxy????az+bxax+byay+bz??

??

99Q99999999999

??b+a????

??=(a?b)3.2??

4第一章行列式

-99999999

??ax+byay+bzaz+bx????xay+bz

??按第1列???？

??ay+bzaz+bxax+by??=====a????=？？yaz+bx??

??分裂開???？

??az+bxax+byay+bz????zax+by

9QQ9999Q

99Q99999

??yzaz+bx????xay+bzz??

??????再次2????+0+0+b2??zxax+by??==a??yaz+bxx????????裂開？？？?？？?？

??xyay+bz????zax+byy??

999?9999?999999999xy79999yZX99

??????再次3??3????==a??+byzx??zxy????????裂開？？？？？？？？

99zxy9999xyz9999^w99999w999^9^9999xyz9999xyz9999xy999ww999

??+b3(?l)2??yzx??=(a3+b3)??yz=a3??yzx??????????

999999999999/

????

????

az+bx????y

??????ax+by??+b????z????

??xay+bz??

??

??

az+bx??

??

ax+by????

??

ay+bz??

ay+bzaz+bxax+by

????z????x????.??y??

此題有一個“經(jīng)典”的解法：

999Q99

999Q99

??ax+byay+bzaz+bx????ax

999999

??ay+bzaz+bxax+by??二？？ay??????

99Q999

??az+bxax+byay+bz????az

9999999999

999999999999xyz9999yzx99999999999999

??+b3??zxy??=a3??=a3??yzx??????????

????????????zxy????xyz????

????

????

??xyz??

????

??.=(a3+b3)??yzx????

??????zxy??

ayazaxxyz

yzx

azaxayzxy

99Q9999?

????bybz????

??+??bzbx????????

99?Qbxby999999999999999999+b31)

??

??

xyz??

??

yzx????

??

zxy??

????bx??

??by??????bz??

這個解法“看上去很美”，實則是一個錯解!我們強調(diào),行列式不能作這種形:.式.上的加法.

??????????????all...aln????bll...bln????all+bll...aln+bln????????????????.????

9Q9999999QQ9999QQ9+=9QQ9999999999Q99999Q99Q9Q9anl?

??ann????bnl???bnn????anl+bnl???ann+bnn??

99Q99999(3)99QQ

999999

a2b2c2d2

(a+l)2(b+l)2(c+l)2

(a+2)2(b+2)2(c+2)2

(a+3)2????

??

(b+3)2????=0;

2??(c+3)????

(d+3)2??

(a+3)

2

證明：

(d+l)2(d+2)2??

??a2(a+l)2(a+2)2????2??b(b+1)2(b+2)2??

??c2(c+l)2(c+2)2????2

??d(d+l)2(d+2)2

??

??a22a+12????2b2b+12c3?2c2??

二二二二二二？？c4?3c2????c22c+12

??2

??d2d+12

(b+3)2(c+3)2

(d+3)2??6??????6??兩列成比例??======0.??6????6??

????

????

????

????

99cj?c1Q9Q9===：=99Q999j=2349999

????

????

a

2

2a+12b+12c+12d+l

4a+44b+44c+44d+4

6a+96b+96c+96d+9

b2c2d2

??

99Q99999999999Q9Q9

線性代數(shù)(同濟四版)習題參考答

案？？？？？？？？？？(4)??????????laa2a41bb2b41CC2c4??l??????d????=(a?b)(a?c)(a?d)(b?c)

(b?d)(c?d)(a+b+c+d);2??d????4??d5

證明：

????????????????????laa2a41bb2b41cc2c41dd2d4??????????????????cj?cl????======?

???j????=2,3,4??????????laa2a40b?ab2?a2b4?a40c?ac2?a2????0????d?a????22??d?a??

??d4?a4??c4?a4

????????b?ac?ad?a??????展開

rl??222222??===??c?ad?a??b?a????22??2222222??b(b?a)c(c?a)d(d?a)??????????!!

1??????????=(b?a)(c?a)(d?a)??b+ac+ad+a??????2????b(b+a)c2(c+a)d2(d+a)?????????

?100????????c2?cl??====(b?a)(c?a)(d?a)??b+ac?bd?b????c3?cl????22222??b(b+a)c(

c+a)?b(b+a)d(d+a)?b(b+a)??

????11展開

rl??=====(b?a)(c?a)(d?a)(c?b)(d?b)??2??(c+bc+b2)+a(c+b)(d2+bd+b2)+a(d+b)

=(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a+b+c+d).????????????????????????(5)??????????

??xO...Oan?lx...0an?10?l...0an?2........................00...xa2??????????0???

?..??=xn+alxn?l+??,+an?lx+an..?????l????x+al??0

證明：方法一.設(shè)法把主對角線上的x變?yōu)?,再按第一列展開.

9999999999999999pn_999999999999?999

??????????????cn?i+xcn??=====????????????????xO...OOanxO...00an?lx...00an?l

?lx...00an?10?l...00an?20?l...00an?2..............................................................................

................................00...x0a300...x0a3????00??????00????.???????10??x?l??

????a2x+al??00...?10x2+alx+a2????0??????0????..??.??????0???l??????x+al??

6第一章行列式

99999999999

9999999999x910???000"W0xl???000QQ

=cn=？=2=+=xc=n=?=1=？？？？…

9999999999QQQ00???091"000???0091^9^99

??anan?lan?2,??x3+alx3+a2x+a3x2+alx+a2x+a??

1??

9Q9999QQJ?.?9999QQ...99

??....????00?,???????00??,??xn+alxn?l+??,+an?lx+anxn?l+alxn?2+???+

an?2x+an?l???

999Q9999Q1???0Q999999

??0???0??

=(xn+alxn?l+???+an?lx+an)(?l)n+l??0??????...????

99

9999Q9

????

??0????10??

??0???0?l????

=(xn+alxn?l+???+an?lx+an)(?l)n+l(?l)n?l

=xn+alxn?l+???+an?lx+an.

方法二.設(shè)法把?1全部變?yōu)?,得到一個下三角矩陣.若x=0,則Dn=an.等式成立.

若x=0,則

9QQ999

????

??XOO???00??

??0x?l???00??????

Dc2+1??

n===c=l=?99999

9999

9Q99

????

??000???X?l??

??a??

nan?l+anan?2???a2x+al??????????x00???00????????0x0???00????==c3=+l??

==c=2=99?9

99

??.........??..??????????000???x?l??????anan?l+anan?2+an?l+an???a2x+a??

1??

99999999x00???

????0x0???00????

=9999

99

9999999999

????000???xO????

??anan?1+anan?2+an?1

+an???P2P??

1??

這里，

P2=a2+a3

x+a4an

x+???++x,00...?10x2+alx+a2??0????0????.??..??????0???????l????x+a??l??

線性代數(shù)（同濟四版）習題參考答案

Pl=x+al+

a2a3an

++???++.xxx

7

得到下三角陣，所以

Dn=xn?l?Pl=xn+alxn?l+?,?+an?lx+an.

方法三.用遞歸法證明.記

??

9999999Q9999I）n=99

9999999999

xO...Oan

?lx

…Oan?l

0?1

...0an?2

00

...xa2

??

???????[??

??

x+al??0

99Q99999999999Q9Q999Q9

則

一？？_

展開cl??.Dn===x??

??0????

??an?l

?1...0an?2

…xa2

????

9999910???0999999999999x91???99n+1"為夕+an…？？.？1??..????????

??0x+al??0???

00

00

X?1

=xDn?l+an(?l)n+l(?l)n?l=xDn?l+an.

所以，Dn=xDn?l+an.由此遞歸式得

Dn=xn+alxn?l+?,?+an?lx+an.

方法四.按最后一行展開.先看an?i

9QQ9999QQQ999Q99Q999999Qaaa

n

n?l

n?2

的代數(shù)余子式.因為

99

?????????????????????????????????????????]????

x?l??

a2x+

al....

?lx

?lx

?lx

an?(i?l)

an?ian?(i+l)

劃掉an?i所在的行和所在的列，左上角是iXi的方塊，右下角是(n?i?l)X(n?i?l)的方塊,

余下全為0.

則an?i的代數(shù)余子式為(注意到an?i處在第n行、i+1歹！J)::::::::::

????

??????????…？？？？？？.X…？？？？丁+「+1"""=xi(91

??..?1??????????..??????.?1??????

??????x?l??

(n?i?l)X(n?i?l)??x??

iXi所以,Dn按最后一行展開,得到

Dn=an+an?lx+an?2x2+???+an?ixi+?,,+a2xn?2+(x+al)xn?l

8第一章行列式

=xn+alxn?l+,??+an?lx+an.

方法五.針對cl作變換.

9999999999999999pn_9999

99999999999999999999999999

cl+xc2??===????

9999999999999999999999Q99?

cl+x2c3??======????

9QQ9999QQQ99

x00...Oan

0x20

???0

00x3

...0

an+an?lx+an?2x2

?lx0...0an?l

0?lx

…0an?2

?lx0...0

0?lx

…0an?2

?lx0...0an?1000...xa2

??

??0??????0????0??

999999

???!????

??

x+al??................

000...xa20?lx...0an?2

an+an?lxan?l

??

??0??????0????0??

9999QQ

???!????

??

x+al??.......................................................

000...xa2

??

??0??????0????0??

999999

???1????

??

x+al??

=?..????0??????0??????0二？？？？.？？.一

?lx0...0an?l

0?lx

…0an?2

000

…xa2

??

??0??????0????0??

9?9......9...9.....9...9..

???1????

??

x+al??

這里，P=an+an?lx+an?2x2+???+alxn?l+xn.

再按第一列展開,得

Dn=xn+alxn?l+?,?+an?lx+an.

6.設(shè)n階行列式D二det(aij),把D上下翻轉(zhuǎn)、或逆時針旋轉(zhuǎn)90?、或依副對角線翻轉(zhuǎn)，依次

得

999999999Q99

??ann???aln????aln???ann????anl???ann??

999999999999

9999999999999999D3=9999D2="Dl=99999999999999

9999999Q9999

??anl???all????all???anl????all???aln??

證明D1=D2=(?1)證明：？？

??anl???ann????....Dl=??..????

??all???aln

n(n?l)

????a????ll????????anl??n?l次行的相鄰互換??n?l??=========(?1)??.??使r換

到第一行??.n????.??????a21

D,D3=D.

alnann...a2n

9999999999999999999Q99

線性代數(shù)(同濟四版)習題參考答案

99Q999all???aln999999999999a21???

n?2次行的相鄰互換？？n?ln?2??a=========(?l)(?l)??nl???ann??=???

??.??使rn換到第二行.??.??..????.??????a31???a3n??

??????all???aln??????????n(n?l)..??=(?1)1+2+???+(n?2)+(n?l)D=(?l)D...=(?

l)n?l(?l)n?2???(?1)??.????.??????anl???ann??

9

同理可證

D2=(?l)

r)(n?l)

????all????.??.??.????aln

anl...ann

n(n?l)????????n(n?l)n(n?l)

??=(?1)DT=(?1)D.??????(?1)

n(n?l)

D3=(?l)

n(n?l)

D2=(?l)

D=(?l)n(n?l)D=D.

7.計算下列各行列式(Dk為k階行列式)：

????

????al????

????..??,其中對角線上元素都是a,未寫出的元素都是0;(l)Dn=??.??????????la??

解:方法一.將cn作n?l次列的相鄰對換,移到第二列：

9QQQ9999

??a0???01????a0???01??

99Q9999?

9QQ9999Q

??10???Oa????Oa???00??

999Q9999

??????.Dn二外20分???009^"=C?l)999999999999Q99999?

?..??00???a0??

99999999

??00???a0????10???0a??

再將rn作n?l次行的相鄰對換,移到第二行：

??

??alO???0????

??laO???0??

n?ln?l??00a???ODn=(?l)(?1)??

9Q9Q9999

??000???a

方法二.

9Q9Q99Q999999QDn=99

9999999Q99

a0...01

???

??

l??????00??

9Q9Q999Q99a099

??0a??0

9QQ9999QQ9

99999999a???

99QQa199Q9999Q?99QQQ

9QQQ=Q999991a99999Q9Q990???99

????

=(a2?l)an?2.

(n?2)X(n?2)

a???

...00

????a????展開cl

======a????

????

99

9999999Q99999Q999999n+1994-lX⑴1)????????????/??1)X(n^l)

0???00

10

a,

...0

aO

??

999Q99Q9QQ999Q9Q99Q9

(n?l)X(n?l)

10第一章行列式

????a????展開rlnn+1(n?l)+l??=：：a+(?l)X1X(?l)??????=an?an?2.

99Q999999999

(2)Dn=??

99999999

xa...

a

???

aa

Q9999999999999?99999999

(n?2)X(n?2)

Y??????

aa,?,x

解：方法一.將第一行乘(?1)分別加到其余各行,得

????xaa????

??a?xx?aO????Ox?aDn=??a?x??...??......????

??a?xOO

??

??????0??

??0????,

??.??..????x?a??a

再將各列都加到第一列上，得

??

??x+(n?l)aaa??????Ox?aO????OOx?aDn=????...??......??????000

方法二.將各列都加到第一列得

??

??x+(n?l)aa???????

??x+(n?l)ax?????Dn=??.

??

??x+(n?l)aa???再將第一行乘以(?1)分別加到其余各行，得

9999999Q99

????????Dn=x+(n?l)a??

999Q999999

10...0

a0

...0

aO

??

??????O??

??????O????=x+(n?l)a(x?a)n?1.

??.??..????x?a??a

aa...x

????

????

????

????

9999999Q9999

??=x+(n?l)a??

????

????

????

????

la,??aa

...x

lx???

..........1

a

??

9QQ9999QQQ999Q99Q999

Ox?a

x?a???...0

???

??

a????

??0??

??????0????=x+(n?l)a(x?a)n?1.

??.??..????x?a??

方法三.升階法.

??

??la????

??0x????

Dn-??Oa

9999Q999

??0a

aa

aaa...x

x???

..a

??

99Q999999999999999999999

(n+1)X(n+1)

999Q99Q9QQ99

ri?rl??======??i=2,3,?????

99999999

1?1...?1

aO

...0

aO

?lx?a

x?a???

...0

???

??

a????

??0??

??0????

??.??..????x?a??

(n+1)X(n+1)

線性代數(shù)（同濟四版）習題參考答案

若x=a,則Dn=O.若x=a,則將

1

cj

11

加到cl,j=2,3,???,n+1:

????

????l+anaa???a????????

??Ox?aO???0??

????

??00x?a???0??Dn=????

9999Q9999999

????

??000???x?a??

(n+1)X(n+1)

????

????nan

=l+(x?a)=x+(n?l)a(x?a)n?l.

x?a

??

??an??

二？？.？？.

(a?l)n(a?l)n?l

...a?ll

(a?n)n(a?n)n?l

...a?nl

??

999Q9999QQ99

??；(提示:利用范德蒙德行列式的結(jié)果.)？????????？

(3)Dn+l

解:從第n+1行開始,第n+1行經(jīng)過n次相鄰對換,換到第1行;第n行經(jīng)(n?l)次對換換到

第2

+1)

行.經(jīng)n+(n?l)+???+l=n(n次行交換，得(或者直接由題6的結(jié)論)

??

??an

la?l...(a?l)n?l(a?l)n

la?n

...(a?n)n?l(a?n)n

??

??????????????,??????????

Dn+l=(?l)

n(n+l)

此行列式為范德蒙德行列式.

對照范德蒙德行列式的寫法知,這里的a=xl,a?l=x2,...,a?(n?l)=xn,a?n=xn+l.則

xi=a?(i?l),xj=a?(j?l).所以

Dn+l=(?l)

=(?1)=(?1)=(?1)=

二？？

n(n+l)

??

n+l??i>j??l

??

xi?xj

??

n(n+l)

????

????(a?i+l)?(a?j+l)??

??

?(i?j)

X

??

n+l??i>j??l

n+l??i>j??l

n(n+l)

n+l??i>j??l

n(n+l)

X(?1)(i?j).??bn????

????????????()??:

n+(n?l)+???+1

(i?j)

??

n+l??i>j??l

alcl

Obldl

(4)D2n

解:方法一.將c2n作2n?l次列的相鄰對換,移到第二列;再將r2n作2n?l次行的相鄰對換,

移到

12第一章行列式

第二行：

9Q9Qa9Qn9Q9?cn9QQ99QQ9999Q

=(?l)2n?l(?l)2n?l????

9QQQQ9Q999999Q9999

bndn

an?l

???alcl

cn?l

bldl

...dn?l

??,bn?l

9999999999999999999999

??=(andn?bncn)D2(n?l),????????????????????

D2n

????a??l

又n=l時D2=??

??cl??bl????

??=aldl?blcl,所以dl??

D2n=(andn?bncn)???(aldl?blcl)=

n??i=l

(aidi?bici).

這個方法與教材P.15的例11相同.本題的第⑴小題也用到了此方法.方法二.

??

99anQ19Q99999Q9999990

展開rl??======an??

999Q999999

??cn?l????O

展開c2n?l

alcl

0???bldl

bn?10...

D2n

dn?10

Odn

????

????

????

????

????

????

????

????

????

????2n+l

bn????+(?l)

????

????

????

????

????

????

????

????

Oan?l

alcl

..bldl

bn?l

00

Ocn

cn?10

dn?10

??

999999999999999999999999999999999999

=======:==andnD2n?2?bncnD2n?2.

由此得遞推公式：

D2n=(andn?bncn)D2n?2,

即D2n=(aidi?bici)D2.

????i=2

n??ab????l????l

(aidi?bici).而D2=????=aldl?blcl,得D2n=

??cldl??

i=l

(5)Dn=det(aij),其中aij二|i?j|;解：由aij=|i?j|得

????0123??????1012????

101??2

??Dn=det(aij)=？？.......??....??

??n?2n?3n?4n?5????

??n?ln?2n?3n?4

??n?l????

??n?2????

??n?3??

??.??.??.

??0??

n??

線性代數(shù)(同濟四版)習題參考答案

??

???1111???????

???1?111???????

?1?11???ri?ri+l???l

===="1-l2???9999

??

???1?1?1?1???????

??n?ln?2n?3n?4?????

???1000????

???l?200????

?2?20???lcj+cl

====??....j=2,3,?????....??,

??

???1?2?2?2????

??n?12n?32n?42n?5

??

??l+a??l????l??

(6)Dn=??..??.??

????1

解：升階法.

=(?l)n?l(n?l)2n?2.

999Q???1999Q99??.p9

0.??,其中ala2???an=..??.??

??

???l+an??100.,.01?l...?l

11+al

1

...110...0

10a2

...0

lll+a2

...1...........................................

.............................................100

...an

??????????????????????????????01111+an

??

999999999999999999999999

(n+1)X(n+1)

13

111...10

??????????????????????????????????()????

??0????

??0??

??一？？一？？一

??n?l??

Il+a2

…1??

99Q999999999Dn=99

999999999999999QQ9999Q

ri?rl??======??i=2,3,?????

99Q9999?

?lal

??

??1+111?????l????OalO???1

cl+c2????l

0a2???=======???1

99999999

???100?????

??1+1+1+???+l??12n????01

cl+cj+l??j??0========??j=2,3,?????.??..??

????0

(n+1)X(n+1)

(n+1)X(n+1)

10a2

...0

??

(n+1)X(n+1)

n????l??

=(ala2???an)1+.

aii=l

14第一章行列式

8.用克萊姆法則解下列方程

組：？？xl+x2+x3+x4=5,????x+2x?x+4x=?2,1234(1)?2x1?3x2?x3?5x4=?2,????3xl+x2+2x3+l

1x4=0;

解：

999999999999999999HQ999H9999H999?HP^lllllll999999999999999999999999999999

????12?14????01?23????01?23????01?2??3??=????=????=????=????2?3?l?5????0?5?3?7

99990091389999009l954"=914299999999999QQ999999Q999999QQ999Q9Q3H42992119Q9Q0

92?]1111191910W^W?99999Q9999Q9Q9

?????5?1?10????5?1?10?????22?14????0????????5?5?18????=????=??0?5?18??=???40??

二？142?=9910999293?1959Q9??2935999999999928999999999999999999999999999252899992

30?22????01211????0100??

9999Q9999Q99999Q999Q999999999915H99H999????]??????????????????

9Q9Q991929149Q9Q097923WW09132"990913299:Z9284-99999999999999=99:Z99:::99=999999

9999992929P59Q990912^3^7^^00231P^9OO919199999^9999WW"99^9^9^9W999999^

30000?284??0039310?15?18211??????????????1115??51??????ll????????????12?l?2???

?12?24??????D4=??=????2?3?1?2??=142.??2?3?2?5??=?426;??????????????????31??312

O??O11??DD1D2D3

所以，

xl二

??5xl+6x2???????xl+5x2+6x3

(2)????????D1=1,Dx2==l,=0,=0,D2=2,Dx3=D3=3,Dx4=D4=?l.Dx2+5x3+6x4x3+5x4+6x5=0,

x4+5x5=l.

555966夕569990=56夕夕9515??????1??????5??二5r)4X????0??????????60

999Q9999999915展開

c??l=====5D4X4?????=??01????????????00????????0????6????=5D4X4?6D3X3.??5???

???????????????????解：系數(shù)行列式06510065D5X565651

由遞歸式D5義5=5D4X4?6D3X3知，

D3X3=5D2X2?6D1X1=5(25?6)?6X5=65,

D4X4=5D3X3?6D2X2=5X65?6X19=211,

線性代數(shù)(同濟四版)習題參考答案D5X5=5D4X4?6D3X3=5X211?6X65=665.

又

99Q999

????16000????

??6000??????????05600????

????

????

D5600????1=????01560????

????

??1560????????00156??=展開=D??

4X4+??

????

????

????

????100150156??

????

=D??4X4+64=211+

????64=1507.

????51000??????

????600????????10600????????l????000??????D560????5

????

2=??????00560????

??二展開==c=2=????0??+(???1600????

????????????00156????156560??

??????0??1)5+2????

????0??

??010??????????15??00150156????=展開=c2

??===?D3X3?5X63=?65?1080=?1145

999Q

????56100????

????????0????????15000????????150????0D??0160????56

????

3=??????01060????=展開==c=3=(?1)1+3??????150

????06??????+(?1)5+3????6??00056??????????05

??00115??????0015????01

??????005

二二展開二c

??==2=D2X2?6X6XD2X2=703.

????

????56010??????

9Q9999i5609999999915600999999999999560D01509999994=99999999^Ff:c4"

??01500??======(?1)1+4????6??????001+(?1)5+4????15

6??

9999999999015W00106999999

??00015??????0005????????001二二展開二c

??==4=?5?6D3X3=?5?6X

????65二?395.

????56001????

9Q156(P-9m15609WW99^^9995600999^"D569999

????0????

5=??0????

夕—？展開

??01560??==c=5=(?1)1+5????01????156??????=??15??6????00150??????00??+??????01

5??????00011????

????0001????????0015????

=1+D4X4=1+211=212.

所以，

x2121=1507

665,x2=?1145

665,x3=703

665,x4=?395

665,x5=665.

????Xxl+x2+x3=0,

9.問人，u取何值時,齊次線性方程組??xl+ux2+x3=0,有非零解？

9xl+2Ux2+x3=0

16第一章行列式

解：系數(shù)矩陣

999999Q9999999入H99Q9入9999r39r19999=0=99==99991g199=9919999999912.19999Q9

?????11??????3+2??入1????U1??二U(?1)??11??U0??????????=U(1?入).？？

要使齊次線性方程組有非零解，則D=o,即

u(1?入)=0,

得u=0或X=1.???2x2+4x3=0,??(1?x)X1

10.問x取何值時,齊次線性方程組2x1+(3?X)x2+x3=o,有非零解????x+x+(l?X)x=0.123

解：系數(shù)矩陣

?????24??1?X??D=??3?X1??2????111?X????3+1???3+入=1X(?1)????!?X???????????

?4??????1今A93+X??c29C19??9?9=?9===??219X1??99=9??999??9?991019X???999?99?4

??3+3??1?x?3+X??+(1?X)(?1)????21?X1????????????=(X?3)?4(1?X)+(1?X)3?2(1

?X)(?3+X)=(l?X)3+2(l?X)2+X?3

=A(X?2)(3?X).

齊次線性方程組有非零解,則D=0,即

X(X?2)(3?X)=0.

得X=0,入=2或X=3.

第二章1.已知線性變換:矩陣及其運算????xl=2yl+2y2+y3,

x2=3yl+y2+5y3,???x=3y+2y+3y,3123

求從變量xl,x2,x3到變量yl,y2,y3的線性變換.解:方法一.用消元法解方程,得出

yl,y2,y3.略.方法二.解矩陣方程.由已知：

??????x221y?1????I??x2?=?315??y2?,??????

323y2x3

故??？

2

1

21??l??????y2?l???y2?=?3???y23?5??3yx?7?49??l??l?????x2?=?63?7????y2?.??y332?4

x3

即？？？？yl=?7xl?4x2+9x3,

y2=6xl+3x2?7x3,???y=3x+2x?4x.3123

方法三.用克拉默法則解方程.系數(shù)矩陣

??????221??D=????315????323

所

以11?llPxZy1

01????0??????c1?2c3?????7?95??=l.??=====????c2?2c3???????????3?43????212?????

???21???????x2??????23????????21??????+x3??????15????????=?7xl?4x2+9x3;??同理

得y2=6xl+3x2?7x3,y3=3xl+2x2?4x3.2.已知兩個線性變換

????xl=2yl+y3,

x2=?2yl+3y2+2y3,???x=4y+y+5y,3123????yl=?3zl+z2,y2=2zl+z3,???y=?z+3z,323

求從zl,z2,z3到xl,x2,x3的線性變換.解:方法一.直接代入.比如：

xl=2yl+y3

=2(?3zl+z2)+(?z2+3z3)=?6zl+z2+3z3.

17

18第二章矩陣及其運算

方法簡單，但我們應(yīng)盡可能使用本章學習的矩陣知識.

方法二.由已知

???x2?l???x2?=??2???x34

???y?3?l???y2?=?2???y20所以,

?

?

?

??

?

?

??0

1

?

y

??l???32???y2?,

15y2

???10z

??l???01???z2?.

?13z3

??

??

?

x201y201?310z

?l????l??????l??x2?=??232??y2?=??232??2??01???z2??????????x3415y24150?13z3

????

z?613

??1??

???=??12?49??z2?.

?10?116z3

即

?

???xl=?6zl+z2+3z3,x2=12zl?4z2+9z3,???x=?10z?z+16z.

3123

?

?

?

?

123111

????

???3.設(shè)A=?ll?l?,B=??l?24??,

0511?ll

求3AB?2A及ATB.

解：

?

??

3?

?

????

??20??.?222

12111

???

??3AB?2A=3??11?1???1?2

051?ll

???

11058

???

??=3??0?56??2?ll

l?1290

9

??

Ill

??

?4???2?11?11?111

??

?2131

??

?

4

29?

058123111

??????T?????AB=?11?1???1?24?=?0?56??.2900511?ll

4.?計算下列乘積:？?4

?(1)??1

5?2?(3)??1

3

3

1

?

?

?

7

??????23???2?;170

??

?(?1,2);?

3??

?(2)(1,2,3)??2?;1

線性代數(shù)(同濟四版)習題參考答案

?

??

1

3

l????;???xl

19

???

2140??0?12(4)

1?134??1?31

40?2

???aal2al3?ll????(5)(xl,x2,x3)?al2a22a23???

?

al3a23

??

12101????0101??0

??(6)??0021??0

???00030???431?????ft?:(l)?l?23???57015X7+7X2+0X1??3??

?(2)(1,2,3)??2?=(1X3+2X2+3Xl)=(10)=10.

?

?

1

?

?

?

4?

?222X(?1)2X2

99999

????(3)??l?(?l,2)=?ix(?l)ix2?=??1

?33X(?1)3X23

??

131?????????6?72140?0?12??=(4)?20?51?134??l?31?40?2????

(5)

aal2al3x?ll??l?

???(xl,x2,x3)??al2a22a23??x2?

a!3a23a33x3

?

x2??;

a33x3

?

031

?

12?1??.0?23??00?3??74X7+3X2+1X1??

?2??=?1X7+(?2)X2+3X1

???

35

????二？6?.???

49

?2??.68?6

??.

??

x?l?

?二(allxH-al2x2+al3x3,al2xl+a22x2+a23x3,al3xl+a23x2+a33x3)??x2?x3

22

=allx21+a22x2+a33x3+2al2xlx2+2al3xlx3+2a23x2x3.

99Q999

125210311210

999999

?0101??012?1??012?4?

?????(6)??0021??00?23?=?00?43?.??????

000?9000?30003

99Q999991210

5.設(shè)A=,B=,問:

1312

(DAB=BA嗎？

(2)(A+B)2=A2+2AB+B2嗎？(3)(A+B)(A?B)=A2?B2嗎?

解：(1)因為

AB=

??

34

46

??,

BA=

??

1238

??,

20第二章矩陣及其運算

所以，

AB二BA.

(2)因為

(A+B)2=

??228??+25??6????

22825??+??=????

8

14??,??二

??,

14291034

??

但

A2+2AB+B2二

??

31015

1627

411812

故

(A+B)2=A2+2AB+B2.

(3)因為

(A+B)(A?B)二

??22811???25??13????

020104??=??=????06092817

??,??,

而

A2?B2=

??

34

故

(A+B)(A?B)=A2?B2.

當然,一個簡單的說法是,在得到AB=BA之后，直接有

(A+B)2=A2+AB+BA+B2=A2+2AB+B2,(A+B)(A?B)=A2?AB+BA?B2=A2?B2.

6.舉反例說明下列命題是借誤的：（1）若A2=0,則A=0;

(2)若A2=A,則A=(^A=E;

(3)若AX=AY=O,則X=Y.??,且A??

01

,A2=0,但A=0.解：(1)取人=

00

????11

(2)取A=,A2=A，但A=0且人=￡.

00W9999999999101111

(3)取A=,X=,Y=.

0071101

AX=AY且A=O但X=Y.

題5和題6看上去很簡單,實則是再次提醒我們注意矩陣運算不滿足交換律,不滿足零律，

不滿足消去律.這是線性代數(shù)初學者最容易犯的幾個錯誤之」為數(shù)不少的人會一直犯這個

錯誤.

我們要注意,雖然矩陣也有所謂的“加法”、“乘法”，但是這和我們熟知的實數(shù)加法、乘

法是完全不同的.運算的對象不同,運算的內(nèi)容不同，當然,運算的規(guī)律也不同.這是兩個不

同的討論范圍里的不同運算,相同的只不過是沿用了以前的稱謂或記號而已我們不要被這

一點“相同”而忘記二者本質(zhì)的不同.

這種不同的討論范圍里的“加法”、“乘法”，還有很多很多,在現(xiàn)代數(shù)學里非常廣泛和一

般.

??

7.設(shè)A=

1X

01??

.求A2,A3,,?,,Ak.

線性代數(shù)(同濟四版)習題參考答案

解：由計算

A2二

??

????

99=01??=??

??,01??.

21

IX??

011

1X????

011

12X??

011

A3=A2A=

??

1

0??

01

2人入3人

猜測:An=

.下用數(shù)學歸納法證明.

n入1

當n=l時，顯然成立.假設(shè)n=k時成立,則n=k+l時

??

Ak+l=AkA=

??

lkX?lAO

?kl??,求AX

?X?A2=??0

1X0

o??x???i???oox

1X0?

0?

?

?

2人入20??

3人？?.X3

2

IkX

01

????

IX

01

??二

??

l(k+l)X

01

??.

由數(shù)學歸納法知:Ak二

?

入？

8.設(shè)A=??0

01

??.

解:方法一.首先計算

X

??

?1??=?0

ox

3

2

2

?

2X??,X2

1

?

?

猜測:An=？入n?0

00An

下用數(shù)學歸納法證明：當n=2時,顯然成立.

假設(shè)n二k時成立，則n=k+1時

入nn入n?l

入3人？

A3=A2?A=?X3?0

00

?

n(n?l)n?2

X?n?l?,(n??2).nX?

3X

?

Ak+1

XXkX

???

??0=Ak?A=?XkkX???0

k

000X

?

kk?l

入k+l(k+l)Xk?l(k+l)X?=?Xk+l(k+l)Xk?l?0

?

k(k?l)k?2

Xk?l

kk?l

k(k?l)k?2

Xk?l

??

1X0???.?

??l??x

Xk+1

99Q9

X?

由數(shù)學歸納法得證:Ak=??0

k

kA

k?lk

XkX

入k

上面的猜測其實是不容易得到的.這里另有一個解法.

22第二章矩陣及其運算

方法二.記

?X?A=??0

0X0

0?

?

0?

?1????XE+B.0

01

??

?0??+?00

xoo

則1

An=(入E+B)n

12

=(XE)n+Cn(XE)n?lB+Cn(XE)n?2B2+???+Bn.

注意到

?0?B=??0

100

??1??,0

?00?B2=??00

OO?O?Bk=??O

000

0??0??,0

1

??0??,0

?00?B3=??00

00

??0??.0

則

(k??3).

所以

An=(XE+B)n

12

=(XE)n+Cn(入E)n?lB+Cn(XE)n?2B2

???

n

OnXn?10X00

???

n?=?O?OnXn?l?+?O?OX

??

?

OXnO

OXnO

入n