线性代数习题解答(同济大学（第四版）)

第一章 行列式

1.利用对角线法则计算下列三阶行列式：

2（1）1?11（3）aa201abc?4?1； （2）bca 83cab11xyx?ybc； （4）yx?yx. b2c2x?yxy201解 （1）1?4?1?2?(?4)?3?0?(?1)?(?1)?1?1?8

?183?0?1?3?2?(?1)?8?1?(?4)?(?1) =?24?8?16?4 =?4

abc（2）bca?acb?bac?cba?bbb?aaa?ccc

cab?3abc?a3?b3?c3

111bc?bc2?ca2?ab2?ac2?ba2?cb2 （3）aa2b2c2?(a?b)(b?c)(c?a)

xyx?yyx?yx （4）

x?yxy?x(x?y)y?yx(x?y)?(x?y)yx?y3?(x?y)3?x3 ?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）1 2 3 4； （2）4 1 3 2； （3）3 4 2 1； （4）2 4 1 3； （5）1 3 … (2n?1) 2 4 … (2n)；

（6）1 3 … (2n?1) (2n) (2n?2) … 2. 解（1）逆序数为0

（2）逆序数为4：4 1，4 3，4 2，3 2

1

（3）逆序数为5：3 2，3 1，4 2，4 1,2 1 （4）逆序数为3：2 1，4 1，4 3 （5）逆序数为

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）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 由定义知，四阶行列式的一般项为

n(n?1)： 2(?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t分别为

0?0?1?0?1或0?0?0?2?2

??a11a23a32a44和a11a23a34a42为所求.

4.计算下列各行列式：

?4?1（1）??10??04??214?3?122??； （2）??1230???7??506?a1ae???abac??1b??； （4）??cdde（3）bd???0?1?cf?ef???bf??00解

125120211?1??； 2??2?00?10?? c1???1d?41(1)

1001251202142c2?c30c4?7c374?112103002?1002

2?14102

4?1?10=122?(?1)4?3 103?144?1109910c2?c3=12?200?2=0

1c1?2c310314171714

213?1(2)

12502r4?r2312

412121c4?c23?1321262501402?122r4?r13

230114004022

306214?12230002=0 00?abacae?bce(3)bd?cdde=adfb?ce

bfcf?efbc?e?111?11=4abcdef =adfbce111?1

a10001?aba0?1b10r1?ar2?1b10(4)

0?1c10?1c100?1d00?1d1?aba01?abaadc3?dc22?1?1c1 ?1c1?cd =(?1)(?1)0?1d0?10ad3?21?ab=(?1)(?1)=abcd?ab?cd?ad?1

?11?cd

5.证明:

3

a2abb23(1)2aa?b2b=(a?b);

111ax?byay?bzaz?bxxyz33(2)ay?bzaz?bxax?by=(a?b)yzx;

az?bxax?byay?bzzxya2(a?1)2(a?2)2(a?3)2b2(b?1)2(b?2)2(b?3)2(3)?0;

2222c(c?1)(c?2)(c?3)d2(d?1)2(d?2)2(d?3)21111abcd(4)2 222abcda4b4c4d4?(a?b)(a?c)(a?d)(b?c)(b?d)?(c?d)(a?b?c?d); x?10?000x?1?00(5)???????xn?a1xn?1???an?1x?an.

000?x?1anan?1an?2?a2x?a1证明

a2ab?a2b2?a2c2?c12ab?a2b?2a (1)左边?c3?c1100ab?a2b2?a2?(?1)

b?a2b?2aab?a?(a?b)3?右边 ?(b?a)(b?a)12xay?bzaz?bxyay?bzaz?bx按第一列ayaz?bxax?by ?bzaz?bxax?by (2)左边分开zax?byay?bzxax?byay?bzxay?bzzyzaz?bx分别再分2ayaz?bxx?0?0?bzxax?by

zax?byyxyay?bz3?14

xy分别再分3ayzzxxyz?a3yzx?b3zxyzxyxyzyzx?b3zxy

xyzyzzx(?1)2?右边 xy(a?2)2(b?2)2(c?2)2(d?2)2(a?3)2(b?3)2 2(c?3)(d?3)2a2a2?(2a?1)b2b2?(2b?1)(3) 左边?c2c2?(2c?1)d2d2?(2d?1)a22a?14a?4c2?c1b22b?14b?4c3?c1c22c?14c?4c4?c1d22d?14d?46a?96b?9

6c?96d?9a2a4a?46a?9a214a?46a?9按第二列b2b4b?46b?9b214b?46b?9 22?2分成二项cc4c?46c?9c14c?46c?9d2d4d?46d?9d214d?46d?9c3?4c2a2a49a214a6a第一项c4?6c2b2b49b214b6b?2?0

2c3?4c2cc49c14c6c第二项c4?9c2d2d49d214d6d1000ab?ac?ad?a(4) 左边?2 222222ab?ac?ad?aa4b4?a4c4?a4d4?a4b?ac?ad?a22c2?a2d2?a2 =b?ab2(b2?a2)c2(c2?a2)d2(d2?a2)111c?ad?a =(b?a)(c?a)(d?a)b?ab2(b?a)c2(c?a)d2(d?a)=(b?a)(c?a)(d?a)?

5

100 b?ac?bd?bb2(b?a)c2(c?a)?b2(b?a)d2(d?a)?b2(b?a)=(b?a)(c?a)(d?a)(c?b)(d?b)?

11 2222(c?bc?b)?a(c?b)(d?bd?b)?a(d?b)=(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a?b?c?d)

(5) 用数学归纳法证明

x?1当n?2时,D2??x2?a1x?a2,命题成立.

a2x?a1假设对于(n?1)阶行列式命题成立，即

n?1 Dn?1?x?a1xn?2???an?2x?an?1,

则Dn按第1列展开:

?10?00?1?00n?1xDn?xDn?1?an(?1)?xDn?1?an?右边 ?????11?x?1所以，对于n阶行列式命题成立.

6.设n阶行列式D?det(aij),把D上下翻转、或逆时针旋转90?、或依

副对角线翻转，依次得

an1?anna1n?annann?a1nD1???， D2??? ，D3???a11?a1na11?an1an1?a11证明D1证明

，

?D2?(?1)?D?det(aij)

n(n?1)2D,D3?D.

a1nann ?a2na11?an1?annn?1an1??D1????(?1)?a11?a1na21?a11?a1na21?a2nann?? ?(?1)n?1(?1)n?2an1??a31?a3n6

a11?a1n?(?1)n?1(?1)n?2?(?1)??

an1?ann?(?1)1?2???(n?2)?(n?1)D?(?1)D

a?an1n(n?1)11n(n?1)n(n?1)同理可证D2?(?1)2???(?1)2DT?(?1)2D

a1n?ann D3

7.计算下列各行列式（Dk为k阶行列式）：

n(n?1)2?(?1)n(n?1)2D2?(?1)n(n?1)2(?1)n(n?1)2D?(?1)n(n?1)D?D

a(1)Dn1?,其中对角线上元素都是a，未写出的元素都是0；

?1aax?a????aa; ?x?(a?n)n?(a?n)n?1???a?n?1bn?0?;

xa(2)Dn??aan(a?1)nan?1(a?1)n?1(3) Dn?1???aa?111an;

提示：利用范德蒙德行列式的结果．

a1b10(4) D2n?0c1d1?0?cndn(5)Dn?det(aij),其中aij?i?j;

7

1?a1111?a2(6)Dn???11解

?1?1,其中a1a2?an?0.

???1?ana00(1) Dn??010a(?1)n?10?00a0?0000a?00??????000?a0100按最后一行展开

?0a00a?0000?0?????000?a1a0 ?(?1)2n?a?0a(n?1)(n?1)?0(n?1)?(n?1)(再按第一行展开)

a?(?1)n?1?(?1)n?a(n?2)(n?2)(2)将第一行乘(?1)分别加到其余各行，得

?an?an?an?2?an?2(a2?1)

xaaa?xx?a0Dn?a?x0x?a???a?x00再将各列都加到第一列上，得

?a?0?0 ??0x?a?a?0?0 ??0x?a8

x?(n?1)aaa0x?a0Dn?00x?a???000?[x?(n?1)a](x?a)n?1

(3)从第n?1行开始，第n?1行经过n次相邻对换，换到第1行，第n

n(n?1)行经(n?1)次对换换到第2行…，经n?(n?1)???1?次行

2交换，得

Dn?111n(n?1)aa?1?(?1)2??an?1(a?1)n?1an(a?1)nn(n?1)2?1?a?n?? ?(a?n)n?1?(a?n)n此行列式为范德蒙德行列式

Dn?1?(?1)?(?1)?

n(n?1)2n?1?i?j?1?[(a?i?1)?(a?j?1)]

n(n?1)2n?1?i?j?1?[?(i?j)]?(?1)?(?1)n?(n?1)???12?n?1?i?j?1?[(i?j)]

n?1?i?j?1?(i?j)

an?0a1c100?a1b1c1d10??0??dn?100dn

bnb1d1?

(4) D2n?0?cnan?1?dnbn?10按第一行0an展开cn?10?9

0an?1??(?1)2n?10?a1b1c1d1bn?10?dn?10bn0?cn?1cn

0都按最后一行展开andnD2n?2?bncnD2n?2

由此得递推公式： D2n即 D2n?(andn?bncn)D2n?2 ??(aidi?bici)D2

i?2na1b1而 D2??a1d1?b1c1

c1d1得 (5)aijD2n??(aidi?bici)

i?1n?i?j

012310122101Dn?det(aij)?3210????n?1n?2n?3n?4?1111?1?111r1?r2?1?1?11r2?r3,??1?1?1?1????n?1n?2n?3n?4????????????n?1n?2n?3

n?4?0111c2?c1,c3?c1

1c4?c1,??010

?100?1?20?1?2?2?1?2?2???n?12n?32n?41?a1111?a2(6)Dn???11a100??a2a20?0?a3a3?00?a4?????000?000?0?00?00?0n?1n?2=(?1)(n?1)2

?2?0???2n?5?n?1?1?1c1?c2,c2?c3

??c3?c4,??1?an001001001按最后一列001

展开（由下往上）????an?1an?110?an1?ana100?000?a2a20?0000?a3a3?0000?a4?000 (1?an)(a1a2?an?1)?0???????000??an?2an?20000?00?ana100?00?a2a20?000?a3a3?00????

??????000??an?1an?1000?0?an11

?a2a20?000?a3a3?0000?a4?00

??????000??an?1an?1000?0?an?(1?an)(a1a2?an?1)?a1a2?an?3an?2an???a2a3?an

n1?(a1a2?an)(1??)

i?1ai

8.用克莱姆法则解下列方程组：

?x1?x2?x3?x4?5,?x?2x?x?4x??2,?234(1)?1

?2x1?3x2?x3?5x4??2,??3x1?x2?2x3?11x4?0;?1,?5x1?6x2?x?5x?6x?0,123??(2)?x2?5x3?6x4?0,

?x3?5x4?6x5?0,??x4?5x5?1.?1111111112?1401?23(1)D? ?2?3?1?50?5?3?7312110?2?1810?00111?20?130?51130?801401111?23??142

0?1?540014251110509

?2?3?1?501211解

5111?22?14D1???2?3?1?50121112

1?5?1?90509??0?13?3?23012111?5?1?901211

05090?13?3?231?5?1?901211??00?10?4600231201?5?1?901211??142

00?138000142151115111?2?140?7?23 D2??2?2?1?50?12?3?7302110?15?18151115110?1320?132????284 00231100?1?19003931000?284115112?24D3???426

2?3?2?531011111512?1?2?142 D4?2?3?1?23120?D3D1D2D4x1??1,x2??2,x3??3,x4???1DDDD6510006510006510501按最后一行?5D?00展开60565100651

51(2)D?00000?5D??6D?? 06?5(5D???6D???)?6D???19D???30D???

13

?65D????114D?????65?19?114?5?665

?的余子式(D?为行列式D中a11的余子式,D??为D?中a11,D???,D????类推) 10D1?00151D2?0006510010001065100651000651006510605按第一列?D?01展开6050160按第二列05?001展开60050651006500 06?D??64?19D????30?????64?1507

0651050016?605501006500 06560?156?5?63??65?1080??1145 01551D3?0006510010001006510150按第三列010展开0060050651056015?601500006500 06160560?056?6150?19?6?114?703 01501651D4?0006510006510100010150按第四列01?000展开600565100501?60506510065100 06560??5?6156??395

01514

51D5?000?651000651000651110按最后一列000展开0015100651006?D??1?211?212 51x1?15071145703?395212;x2??;x3?;x4?;x4?． 665665665665665??x1?x2?x3?0?9.问?,?取何值时,齐次线性方程组?x1??x2?x3?0有非零解？

?x?2?x?x?023?11解 D3?1?1?????，

12?1齐次线性方程组有非零解，则D3即 ?得 ?不难验证，当?

?1?0

????0

?0或??1

?0或??1时,该齐次线性方程组确有非零解.

?(1??)x1?2x2?4x3?0?,齐次线性方程组10.问?取何值时?2x1?(3??)x2?x3?0

?x?x?(1??)x?023?1有非零解？ 解

1???241???3??4D?23??1?21??1

111??101???(1??)3?(??3)?4(1??)?2(1??)(?3??) ?(1??)3?2(1??)2???3

齐次线性方程组有非零解，则D?得 ? 不难验证，当?0

?0,??2或??3

?0,??2或??3时，该齐次线性方程组确有非零解.

15

第二章 矩阵及其运算

1．已知线性变换：

?x1?2y1?2y2?y3,??x2?3y1?y2?5y3, ?x?3y?2y?3y,123?3求从变量x1,x2,x3到变量y1,y2,y3的线性变换．

解

?x1??221??y1???????由已知：?x2???315??y2?

?x??323??y???2??3???1?y1??221??x1???7?49??y1???????????故 ?y2???315??x2???63?7??y2?

?y??323??x??3?y?2?4??2????3????3??y1??7x1?4x2?9x3??y2?6x1?3x2?7x3 ?y?3x?2x?4x123?3

2．已知两个线性变换

?x1?2y1?y3,?y1??3z1?z2,??x??2y?3y?2y, ?2?y2?2z1?z3, 123?x?4y?y?5y,?y??z?3z,12323?3?3求从z1,z2,z3到x1,x2,x3的线性变换．

解 由已知

?x1??201??y1??20????????x2????232??y2????23?x??415??y??41??2???3????613??z1???????12?49??z2? ??10?116??z????3??x1??6z1?z2?3z3?所以有 ?x2?12z1?4z2?9z3

?x??10z?z?16z123?3?111??1???1?1?, B???13．设A??1?1?11??0???1???310??z1??????2??201??z2? ????5???0?13??z3?23???24?, 51??16

求3AB?2A及A解

TB.

23??11?111??1?????3AB?2A?3?11?1???1?24??2?11?1?11??0?51??????1?1?058??111???213??????3?0?56??2?11?1????2?17?290??1?11??429?????23??058??111??1??????TAB??11?1???1?24???0?56?

?1?11??0??51??????290?

4．计算下列乘积：

1???1? 1??22??20? ?2???431??7??3??2?????????(1)?1?23??2?; (2)?1,2,3??2?; (3)?1???1,2?; ?570??1??1??3?????????1??13???2140??0?12?(4)??1?134???1?31?; ????40?2?????a11a12a13??x1?????(5)(x1,x2,x3)?a12a22a23??x2?;

?aa??x?a?132333??3?1??1210??103?????0101??012?1?(6)

?0021??00?23?. ??0003????000?3??????解

?431??7??4?7?3?2?1?1??35?????????(1)?1?23??2???1?7?(?2)?2?3?1???6? ?570??1??5?7?7?2?0?1??49??????????3???(2)?123??2??(1?3?2?2?3?1)?(10)

?1???17

?2?(?1)2?2???24??2???????(3)?1???12???1?(?1)1?2????12? ?3??3?(?1)3?2???36???????1??13???2140??0?12??6?78?(4)??1?134???1?31????20?5?6?? ???????40?2????a11a12a13??x1?????(5)?x1x2x3??a12a22a23??x2?

?a??x?aa?132333??3???a11x1?a12x2?a13x3a12x1?a22x2?a23x3a13x1?a23x2?a33x3? ?x1???222?a22x2?a33x3?2a12x1x2?2a13x1x3?2a23x2x3 ??x2??a11x1?x??3?1??1252??1210??103???????0101??012?1??012?4?(6)

?0021??00?23???00?43? ??0003????000?3????000?9????????

?12??10?5．设A???13??, B???12??，问：

????(1)AB?BA吗?

222(2)(A?B)?A?2AB?B吗?

22(3)(A?B)(A?B)?A?B吗?

解

?12??10???, B??????13??12??34??12?则AB???46?? BA???38?? ?AB?BA

?????22??22??814?2(2) (A?B)???25????25?????1429??

???????38??68??10??1016?22但A?2AB?B???411?????812?????34?????1527??

????????222故(A?B)?A?2AB?B

?22??02??06?(3) (A?B)(A?B)???25????01?????09??

??????(1)A???18

?38??10??28?A2?B2???411?????34?????17??

??????22故 (A?B)(A?B)?A?B

而

6．举反列说明下列命题是错误的:

?0,则A?0; 2（２）若A?A,则A?0或A?E; （３）若AX?AY,且A?0,则X?Y.

?01?2?解 (1) 取A?? A?0,但A?0 ?00????11?2A?A,但A?0且A?E ?(2) 取A?? ?00????10??11??11??????(3) 取A?? X?? Y?? ?????00???11??01?AX?AY且A?0 但X?Y

（１）若A

2?10?23k?,求. A,A,?,A???1??10??10??10?2解 A????1?????1?????2?1??

???????10??10??10?32 A?AA???2?1?????1?????3?1??

???????10?k利用数学归纳法证明: A???k?1??

??当k?1时,显然成立,假设k时成立,则k?1时

0??10??10??1kkA?AA???k?1?????1?????(k?1)?1??

???????10?k由数学归纳法原理知:A???k?1??

??7．设A???

??10???k8．设A??0?1?,求A.

?00????解 首先观察

2?2?1???10???10????????22A??0?1??0?1???0?2??

?00???00???00?2???????

19

??33?23????3232?3?? A?A?A??0?03?0???k(k?1)k?2??kk?1????k?2??kkk?1由此推测 A??0?k?? (k?2)

k?0?0?????用数学归纳法证明:

当k?2时,显然成立.

假设k时成立，则k?1时，

k(k?1)k?2??kk?1?k?????10??2????k?1kkk?1A?A?A??0?k???0?1?

k?0??00??0???????(k?1)kk?1??k?1(k?1)?k?1????2??k?1k?1??0?(k?1)??

k?1?0?0?????k(k?1)k?2??kk?1?k????2??kkk?1由数学归纳法原理知: A??0?k??

k?0?0?????

9．设A,B为n阶矩阵，且A为对称矩阵，证明B证明 已知：ATTAB也是对称矩阵.

?A

TTTTTT则 (BTAB)?B(BTA)?BAB?BAB

T从而 BAB也是对称矩阵.

AB?BA.

10．设A,B都是n阶对称矩阵，证明AB是对称矩阵的充分必要条件是

?A BT?B

TTT充分性：AB?BA?AB?BA?AB?(AB) 即AB是对称矩阵.

TTT必要性：(AB)?AB?BA?AB?BA?AB.

证明 由已知：A

11．求下列矩阵的逆矩阵：

T20

?12??cos?(1)??25??； (2)??sin?????1??1(4)?2??1??a1??(6)??0?解

02120031a2?5??2(5)?0??0??0??(??0) ?a1a2an??an?0??0?; ?0?4???12?1????sin???4?2?; ； (3)?3?cos???5?41???200??100?; ?083?052???12?? A?1 ??25?A11?5,A21?2?(?1),A12?2?(?1),A22?1

?A11A21??5?2?1??1????A?A A??? ???A?A?12A22???21??5?2??1故 A????21??

???1(2)A?1?0 故A存在

A11?cos?A21?sin?A12??sin?A22?cos?

?cos?sin???1从而 A????sin?cos???

???1(3) A?2, 故A存在

A21?2A31?0 A11??4A22?6A32??1 而 A12??13A23?14A33??2 A13??32??210?1??131??1A???3?? 故 A?2?A?2??167?1??1000????1200?(4)A??2130?

??1214???? A?24 A21?A31?A41?A32?A42?A43?0 A11?24A22?12A33?8A44?6

(1)A???21

100120A12?(?1)3230??12 A13?(?1)4210??12

114124120100A14?(?1)5213?3 A23?(?1)5210??4

121124100100A24?(?1)6213??5 A34?(?1)7120??2

1211211A?1?A?

A000??1?1?100???2?2??111故A??1??0?

?2?63?1511?????24124??8?1(5)A?1?0 故A存在 而 AA21??2A31?0A41?0 11?1 AA22?5A32?0A42?0 12??2 AA23?0A33?2A43??3 13?0 AA24?0A34??5A44?8 14?00??1?20??00???25?1从而A?

?0?02?3???00?58????a1??0?a2??

(6)A?????0?an???1???0??a11??a2?1由对角矩阵的性质知 A???

????01???an??

22

12．解下列矩阵方程：

(1)

?25??4??13??X???2????14??2???12??X??????1?010??1????100?X?0?001??0????1(3)

(4)

?21?1????1?13??6???; (2) X?2; 10??????1??1?11??432???0??31???; ?????1??0?1?00??1?43????01???20?1?.

?1?20?10????解 (1)

(2)

(3)

(4)

?25??4?6??3?5??4?6??2?23?X???13????21??????12????21?????08??

???????????1?101??21?1????1?13??1?1?13???X??0? ????23?2? ?432???21??3?432?????1?11??????330???221????85?2? ??3??3?1?1?14??31??20?1?2?4??31??10?X????12????0?1?????11???12??11????0?1????12??

????????????11?1?66??10???1? ?????????0?12??30??12???4??1?1?010??1?43??100???????X??100??20?1??001?

?001??1?20??010????????010??1?43??100??2?10???????????100??20?1??001???13?4? ?001??1?20??010??10?2?????????

13．利用逆矩阵解下列线性方程组:

?x1?2x2?3x3?1,?x1?x2?x3?2,??(1) ?2x1?2x2?5x3?2, (2) ?2x1?x2?3x3?1,

?3x?5x?x?3;?3x?2x?5x?0.2323?1?1?123??x1??1???????解 (1) 方程组可表示为 ?225??x2???2?

?351??x??3????3???23

?x1??123?????故 ?x2???225??x??351???3???x1?1?从而有 ?x2?0

?x?0?3?1?1?(2) 方程组可表示为 ?2?1?32??1?1??1??????2???0? ?3??0??????1??x1??2???????3??x2???1?

?x??0??5???3????1?x1??1?1?1?????故 ?x2???2?1?3??x??32?5???3???x1?5?故有 ?x2?0

?x?3?3

14．设Ak?2??5??????1???0? ?0??3??????O(k为正整数),证明

(E?A)?1?E?A?A2???Ak?1.

?1证明 一方面， E?(E?A)(E?A)

k另一方面，由A?O有

E?(E?A)?(A?A2)?A2???Ak?1?(Ak?1?Ak) ?(E?A?A2???Ak?1)(E?A)

?12k?1故 (E?A)(E?A)?(E?A?A???A)(E?A)

?1两端同时右乘(E?A)

?12k?1就有(E?A)?E?A?A???A

15．设方阵A满足A2?A?2E?O,证明A及A?2E都可逆,并求A?1及

(A?2E)?1.

22证明 由A?A?2E?O得A?A?2E

2两端同时取行列式: A?A?2

AA?E?2,故 A?0

2所以A可逆,而A?2E?A

22A?2E?A?A?0 故A?2E也可逆.

即 由A?A?2E?O?A(A?E)?2E

1?A?1A(A?E)?2A?1E?A?1?(A?E)

224

2?A?2E?O?(A?2E)A?3(A?2E)??4E

?(A?2E)(A?3E)??4E

?(A?2E)?1(A?2E)(A?3E)??4(A?2E)?1

1?(A?2E)?1?(3E?A)

4又由A

2?033???16．设A??110?,AB?A?2B,求B.

??123???解 由AB?A?2B可得(A?2E)B?A

??233??033??033????????1?10??110????123? 故B?(A?2E)A??1??121???123??110???????

?1??1?4???10?11A???17．设PAP??,其中P??,,求. ???11??02??????1?11111?1解 PAP??故A?P?P所以A?P?P

?14?1?14???1 P?3 P????11?? P?3???1?1??

?????1??10???10?11而 ????02?????0211??

????4??1???27312732??1?4?10????1133????????故A?? 11??1????111??02????683?684???????3??3

11f(x)?a0?a1x?a2x2???amxm,记

f(A)?a0E?a1A?a2A2???amAm

f(A)称为方阵A的m次多项式.

k??0???10??f(?1)1k??(1)设????0???,证明: ???0?k?,f(?)???0??2??2??1kk?1?1(2)设A?P?P,证明: A?P?P,f(A)?Pf(?)P.

18．设m次多项式证明

(1) i)利用数学归纳法.当k0??; ?f(?2)??2时

2??0?0?0?????1112 ????0?????0??????0?2??

?2??2??2?命题成立,假设k时成立,则k?1时

25

?k?1k??1??????0?kk?10???10???10???? ?????k??k?1??2??0?2??0?2?故命题成立. ii)左边?f(?)?a0E?a1??a2?2???am?m

m??10???10??10???a0??01???a1??0??????am?0?m??

???2??2?2m?a0?a1?1?a2?1????am?10?? ??2m?0a0?a1?2?a2?2???am?2??0??f(?1)?=右边 ???0?f(?2)??(2) i) 利用数学归纳法.当k?2时

A2?P?P?1P?P?1?P?2P?1成立

假设k时成立,则k?1时

Ak?1?Ak?A?P?kP?1P?P?1?P?k?1P?1成立,故命题成立,

kk?1即 A?P?P

ii) 证明 右边?Pf(?)P?1

?P(a0E?a1??a2?2???am?m)P?1

?a0PEP?1?a1P?P?1?a2P?2P?1???amP?mP?1 ?a0E?a1A?a2A2???amAm?f(A)=左边

?

19．设n阶矩阵A的伴随矩阵为A，证明： (1) 若(2) 证明

(1) 用反证法证明．假设由此得A?这与有

A?0,则A??0;

n?1A??A.

A??0则有A?(A?)?1?E

AA?(A?)?1?AE(A?)?1?O?A??O

A??0矛盾，故当A?0时

A??0

(2)

1?由于A?A, 则AA??AE

A?1取行列式得到: 若若

AA??An

A?0 则A??An?1

A?0由(1)知A??0此时命题也成立

故有

A??An?1

26

?120．取A?B??C?D???0?101010AB检验: ?CD?1010?10AB11??0 而

CD11故

AB0??,验证??CD1?02001020?0?10110?10ACBD

002010??4 002011ABA?CDCBD

?34??O??4?3?,求A8及A4

21．设A??20??O?22???34??O??34??20?4?3??解 A?，令A1???4?3?? A2???22?? ?20??????O?22???A1O?则A???OA??

?2?8?AOAO???118?故A???OA????OA8??

?2??2?8888A8?A1A2?A1A2?1016

8

4?A1A4???O??540??O?4?O?05? ??44??20?A2??O?64??22???1

?OA?22．设n阶矩阵A及s阶矩阵B都可逆,求??BO?????1?C1C2??OA?解 将??BO??分块为??CC??

??4??3其中 C1为s?n矩阵, C2为s?s矩阵

C3为n?n矩阵, C4为n?s矩阵

．

27

An?n??C1C2??EnO????? ?E???????O??C3C4??OEs??AC3?En?C3?A?1??1AC?O?C?O(A存在)?44由此得到?

?1?BC1?O?C1?O(B存在)?BC?E?C?B?1s2?2?1?OB?1??OA?故 ??BO?????A?1O??．

?????O则??B?s?s

第三章 矩阵的初等变换与线性方程组

1．把下列矩阵化为行最简形矩阵：

(1)

(3)

?1??2?3??1??3?2??3?02?1??0??031?; (2) ?0?004?3????13?43??2???35?41??1; (4) ??3?23?20????2?34?2?1??2?31??3?43?; 4?7?1??31?320?2?283?374?7???4?. ?0?3??

解

?102?1?r2?(?2)r1?102?1?????1?(1) ?203~?00?13?

?304?3?r3?(?3)r1?00?20?????r2?(?1)?102?1?r3?r2?102?1?????~?001?3?~?001?3? r3?(?2)?0010??0003?????r3?3?102?1?r2?3r3?102?1?????~?001?3?~?0010?

?0001??0001?????r1?(?2)r2?1000???~?0010? r1?r3?0001???

(2)

?02?31?r2?2?(?3)r1?02?31?????~?0013? ?03?43?

?04?7?1?r3?(?2)r1?00?1?3?????28

0r3?r2??~?0r1?3r2?0??1?1??3?3(3)

?2?2??3?3?2003534010?r1?2?010??13?~?001?00000????43??1?r2?3r1??41??0 ~??20r3?2r1?0????2?1?r4?3r1??03111?4?2?2?23??r1?3r22?~?2r3?r2?2??r4?r25??3? 0???10003?43???48?8? ??36?6??510?10??02?3??1?22? ?000?000???1?1r2?(?4)??00~?00r3?(?3)??00r4?(?5)?

?1?1??00?00??00?(4)

?23??12?3?2??2?3??0r2?2r1??1~?0r3?8r1??0r4?7r1??1r2?r3??0~?0??0?1?3?7??0?1?r1?2r2?0?2?4??12~ ?830r3?3r2?0?8???743?r4?2r2??0?7?1111??10?r1?r2?020?2??01~?0014r2?(?1)?00???0014?r4?r3??00020?2??1?103? ?0014?0000??111??0?2?4? ?8912?7811??20?2???1?1?1? ?014?000??

2．在秩是r的矩阵中,有没有等于0的r?1阶子式?有没有等于0的r阶 子式?

解 在秩是r的矩阵中,可能存在等于0的r?1阶子式,也可能存在等 于0的r阶子式.

?1000????0100?例如，???0010?

???0000??0000???R(?)?3同时存在等于0的3阶子式和2阶子式.

3．从矩阵A中划去一行得到矩阵B,问A,B的秩的关系怎样?

29

R(A)?R(B)

设R(B)?r，且B的某个r阶子式Dr?0.矩阵B是由矩阵A划去一行得 到的，所以在A中能找到与Dr相同的r阶子式Dr，由于Dr?Dr?0， 故而R(A)?R(B).

解

4．求作一个秩是4的方阵,它的两个行向量是(1,0,1,0,0),(1,?1,0,0,0) 解 设?1,?2,?3,?4,?5为五维向量,且?1?(1,0,1,0,0), ??1?????2??2?(1,?1,0,0,0),则所求方阵可为A???3?,秩为4,不妨设

????4?????5???3?(0,0,0,x4,0)???4?(0,0,0,0,x5)取x4?x5?1 ???(0,0,0,0,0)?5?10100????1?1000?故满足条件的一个方阵为?00010?

???00001??00000???

5．求下列矩阵的秩,并求一个最高阶非零子式：

(1)

(3)

02??31???1?12?1?; (2) ?13?44????21837????2?307?5??3?2580?. ??10320????1?13?14402?42?6?6?32?1?3?1???1?3?； ?2?13?705?1?8???

?3?解 (1) ?1?1??1?3r2r1?~?0r3?r1?0?2???1?r1r2??1?~?3?14????1??1??r3r2?5?~?0?05????12103?4?124?600?1??2? 4???1??5?秩为2 0??30

二阶子式

31??4．

1?1r1?r2?2??13?4?41??r2?2r1???3?~?0?7119?5?

??7??8??r3r1?0?213327?15??41??9?5?秩为2. 00???32?1?3?(2) ?2?131?705?1??13?4?r3?3r2?0?711~?0?0032二阶子式??7．

2?12?17??21837??01r?2r??1?4??2?307?5??0?3?63?5?(3)

?3?2580?~?0?2?42?0r2?2r4?????10320??10320???r3?3r4??r1?r2?012?17??10320?r2?3r1??r4?r1???000016??012?17?~?000014?~?00001?秩为3

r3?14?r3?2r1???10320??00000????r4?16??r4?r307?5580??5?70?0． 三阶子式5832320

6．求解下列齐次线性方程组:

?x1?x2?2x3?x4?0,?x1?2x2?x3?x4?0,??(1) ?2x1?x2?x3?x4?0, (2) ?3x1?6x2?x3?3x4?0,

?2x?2x?x?2x?0;?5x?10x?x?5x?0;234234?1?1?2x1?3x2?x3?5x4?0,?3x1?4x2?5x3?7x4?0,?3x?x?2x?7x?0,?2x?3x?3x?2x?0,?1?1234234(3) ? (4)?

?4x1?x2?3x3?6x4?0,?4x1?11x2?13x3?16x4?0,???x1?2x2?4x3?7x4?0;?7x1?2x2?x3?3x4?0.解 (1) 对系数矩阵实施行变换:

31

4?x?x41?3??112?1????10?10?????x2??3x4? ?211?1?~?013?1?即得??2212??001?4??x?4x???3343???x?x?44?4?x?1?????3???x2???3? ?k故方程组的解为

?x??4?3??3??x???1??4???

(2) 对系数矩阵实施行变换:

?121?1??120?1??????36?1?3?~?0010??5101?5??0000????? ?x1???2??1????????x2??1??0?故方程组的解为

?x??k1?0??k2?0?

3????0???1???x???????4?

(3) 对系数矩阵实施行变换:

?x1??2x2?x4?x?x?22即得?

?x3?0??x4?x4?23?15??1???2?7??0?31?41?36?~?0??1?24?7???????0?x1?0?x?0?2故方程组的解为?

?x3?0??x4?0

(4) 对系数矩阵实施行变换:

01000010?x1?00???x?00??2即得? ?0?x3?0??1???x4?032

313??10???34?57??1717????19202?33?2???01??~?411?1316??1717????0000??7?21?3??? ?0??000313?x?x??117317x4??x?19x?20x即得?234

1717??x3?x3?x?x?44?3??13???????x1????17??17?19?20??x2???故方程组的解为

?x??k1?17??k2??17?

3??1??0??x???????4??0??1?

7．求解下列非齐次线性方程组:

?2x?3y?z?4,?4x1?2x2?x3?2,?x?2y?4z??5,??(1) ?3x1?1x2?2x3?10, (2) ?

?11x?3x?8;?3x?8y?2z?13,?12??4x?y?9z??6;?2x?y?z?w?1,?2x?y?z?w?1,??(3) ?4x?2y?2z?w?2, (4) ?3x?2y?z?3w?4,

?2x?y?z?w?1;?x?4y?3z?5w??2;??

解 (1) 对系数的增广矩阵施行行变换,有

?42?12??13?3?8????3?1210?~?0?101134?

??11308??000?6????R(A)?2而R(B)?3,故方程组无解．

(2) 对系数的增广矩阵施行行变换:

14??1?23????1?24?5??0?38?213?~?0??4?19?6????0???02?1??1?12? ?000?000??33

?x??2z?1?x???2???1????????即得?y?z?2亦即?y??k?1???2?

?z??1??0??z?z???????

(3) 对系数的增广矩阵施行行变换:

?21?111??21?111??????42?212?~?00010? ?21?1?11??00000?????111??1??1??1?xx??y?z?????????????22222??2?????y???1??k?0???0? ??k即得?y?y 即12?z??0??1??0??z?z???w????????????????000?w?0???????

(4) 对系数的增广矩阵施行行变换:

1??14?35?2??21?11???595????3?21?34?~?01?777??14?35?2?????00000?116??10????777??595~?01??? ?777??00000?????116??1??1??6?x?z?w?????????x?777777?????????5?9??5??y?5z?9w?5?y?????k?k???? 即得? 即1??2????777777z????1??0??0???z?z????????w??w?w010???????

8．?取何值时,非齐次线性方程组

??x1?x2?x3?1,??x1??x2?x3??, ?2x?x??x??23?1(1)有唯一解；(2)无解；(3)有无穷多个解?

34

解 (1)

?111?1?0,即??1,?2时方程组有唯一解. 11?

R(A)?R(B)

1??2??111??1???1???(1??)? B??1?1??~?0??12??11??2??00(1??)(2??)(1??)(??1)????2由(1??)(2??)?0,(1??)(1??)?0 得???2时,方程组无解.

(2)

(3)

R(A)?R(B)?3,由(1??)(2??)?(1??)(1??)2?0,

得??1时,方程组有无穷多个解.

9．非齐次线性方程组

当?取何值时有解？并求出它的解．

??2x1?x2?x3??2,??x1?2x2?x3??, ?2x?x?2x??23?1?1?21?1?2????21????2?21??~?01?1?(??1)? 解 B??13??2??1?1?2?0(??1)(??2)????00?方程组有解，须(1??)(??2)?0得??1,???2

?x1??1??1???????当??1时,方程组解为?x2??k?1???0?

?x??1??0??3??????x1??1??2???????当???2时,方程组解为?x2??k?1???2?

?x??1??0??3?????

?(2??)x1?2x2?2x3?1,?10．设?2x1?(5??)x2?4x3?2,

??2x?4x?(5??)x????1,123?问?为何值时,此方程组有唯一解、无解或有无穷多解？并在有无穷多解

时求解．

35

2?21??2????解 ?25???42?

??2?45?????1????5???1?21??初等行变换2??01??1??1??~??

(1??)(10??)(1??)(4??)??00??22??(1??)2(10??)?0 ???1且??10时，有唯一解. 当A?0,即

2(1??)(10??)(1??)(4??)当?0且?0，即??10时，无解.

22(1??)(10??)(1??)(4??)当?0且?0，即??1时，有无穷多解.

22?12?21???00? 此时，增广矩阵为?00?0000????x1???2??2??1?????????原方程组的解为?x2??k1?1??k2?0???0? (k1,k2?R)

?0??1??0??x????????3?

11．试利用矩阵的初等变换，求下列方阵的逆矩阵：

?3?20?1????321???21??02(1) ?315?; (2)

?1?2?3?2?.

?323??????0121????321100??32110???解 （1）?315010?~?0?14?11?323001??002?10???31??7?0???300?32022??2?~?0?1011?2?~?0?101?002?101??001?1???2???723?????100632??~?010?1?12?

1??001?10??22??0??0? 1??9?2??2?1?2?

1?02??36

23??7???632??故逆矩阵为??1?12?

1???10?2?2???3?20?11000???210100??02(2)

?1?2?3?20010? ???01?210001???1?2?3?20010???210001??01~?049510?30? ???02210100???0??1?2?3?2001??210001??01~?001110?3?4? ??00?2?1010?2????0??1?2?3?2001??210001??01~?001110?3?4? ???00?0121?6?10???1?200?1?1?2?2????010001`0?1?~?0010?1?136? ??000121?6?10?????100011?2?4???0?1??010001~?0010?1?136? ??000121?6?10?????11?2?4???0?1??01故逆矩阵为

??1?13?6??21?6?10????

37

?4?12．(1) 设A??2?3??0?(2) 设A??2??3?解

1?2??1?3????21?,B??22?,求X使AX?B;

?3?1?1?1????21???123??13?,B???2?31??,求X使XA?B.

??3?4???41?21?3?初等行变换?10???(1) ?AB???22122?~?01?31?13?1??00????102????1?X?AB???15?3?

?124???21??0?10????2?13?初等列变换?01?00?A???33?4???(2) ????~123?B????2?1?2?31???47??????????2?1?1??1?X?BA????474??．

??

0102??0?15?3? 1124??0??0?1?? ?1?4????

第四章 向量组的线性相关性

?(1,1,0)T,v2?(0,1,1)T,v3?(3,4,0)T,

求v1?v2及3v1?2v2?v3.

TT解 v1?v2?(1,1,0)?(0,1,1)

?(1?0,1?1,0?1)T?(1,0,?1)T

3v1?2v2?v3?3(1,1,0)T?2(0,1,1)T?(3,4,0)T

?(3?1?2?0?3,3?1?2?1?4,3?0?2?1?0)T ?(0,1,2)T

1．设v1

?a)?2(a2?a)?5(a3?a)其中a1?(2,5,1,3)T,

a2?(10,1,5,10)T,a3?(4,1,?1,1)T,求a

解 由3(a1?a)?2(a2?a)?5(a3?a)整理得

11a?(3a1?2a2?5a3)?[3(2,5,1,3)T?2(10,1,5,10)T?5(4,1,?1,1)T]

662．设3(a138

?(1,2,3,4)T

3．举例说明下列各命题是错误的:

(1)若向量组a1,a2,?,am是线性相关的,则a1可由a2,?am,线性表示. (2)若有不全为0的数?1,?2,?,?m使

?1a1????mam??1b1????mbm?0

成立,则a1,?,am线性相关,

b1,?,bm亦线性相关.

(3)若只有当?1,?2,?,?m全为0时,等式 才能成立,则a1,?,am线性无关,

?1a1????mam??1b1????mbm?0

b1,?,bm亦线性无关.

(4)若a1,?,am线性相关, b1,?,bm亦线性相关,则有不全为0的数, ?1,?2,?,?m使?1a1????mam?0,?1b1????mbm?0

同时成立.

?e1?(1,0,0,?,0)

a2?a3???am?0

满足a1,a2,?,am线性相关,但a1不能由a2,?,am,线性表示.

(2) 有不全为零的数?1,?2,?,?m使

?1a1????mam??1b1????mbm?0

解 (1) 设a1原式可化为

?1(a1?b1)????m(am?bm)?0

取a1?e1??b1,a2?e2??b2,?,am?em??bm 其中e1,?,em为单位向量,则上式成立,而

a1,?,am,b1,?,bm均线性相关

(3) 由?1a1????mam??1b1????mbm?0 (仅当?1????m?0) ?a1?b1,a2?b2,?,am?bm线性无关 取a1?a2???am?0 取b1,?,bm为线性无关组

满足以上条件,但不能说是a1,a2,?,am线性无关的.

TTTT(4) a1?(1,0) a2?(2,0) b1?(0,3) b2?(0,4) ?1a1??2a2?0??1??2?2??3? ??1??2?0与题设矛盾.

?1b1??2b2?0??1???2?4? 4．设b1?a1?a2,b2?a2?a3,b3?a3?a4,b4?a4?a1,证明向量组

b1,b2,b3,b4线性相关.

证明 设有x1,x2,x3,x4使得

x1b1?x2b2?x3b3?x4b4?0则

x1(a1?a2)?x2(a2?a3)?x3(a3?a4)?x4(a4?a1)?0

39

(x1?x4)a1?(x1?x2)a2?(x2?x3)a3?(x3?x4)a4?0 (1) 若a1,a2,a3,a4线性相关,则存在不全为零的数k1,k2,k3,k4, k1?x1?x4;k2?x1?x2;k3?x2?x3;k4?x3?x4;

由k1,k2,k3,k4不全为零,知x1,x2,x3,x4不全为零,即b1,b2,b3,b4线性相

关.

?x1?x4?0?1?x?x?0?1?1?2??(2) 若a1,a2,a3,a4线性无关,则??x2?x3?0?0?0?x?x?0?4?310011100?0知此齐次方程存在非零解 由

01100011则b1,b2,b3,b4线性相关.

综合得证. 5．设b1011000111??x1????0??x2??0

???0x3?????1??x4???a1,b2?a1?a2,?,br?a1?a2???ar,且向量组

a1,a2,?,ar线性无关,证明向量组b1,b2,?,br线性无关. 证明 设k1b1?k2b2???krbr?0则

(k1???kr)a1?(k2???kr)a2???(kp???kr)ap???krar?0

因向量组a1,a2,?,ar线性无关,故

?k1?k2???kr?0?1??1??k1??0????????k2???kr?0?01?1??k2??0????? ????????????????????????????????k0?010k?0???r????r1??101?1?1?0故方程组只有零解 因为

????0?01则k1?k2???kr?0所以b1,b2,?,br线性无关

6．利用初等行变换求下列矩阵的列向量组的一个最大无关组:

?25??75(1)

?75??25?

319494321753542043??1??132??0; (2)

?2?134???1?48??1201221??15?1?. ?3?13?04?1??40

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