线性代数习题解答(同济大学（第四版）)(5)

?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??

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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

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