同济第五版线性代数 课后题解析第二章

??k?k?1 A?Ak?A??0?0????k?1(k?1)?k?1??0 ??k?1?0?0?由数学归纳法原理知?

k?k?1k(k?1)k?2?????10?2??0?1? ?kk?k?1??00??0?k????(k?1)kk?1???2(k?1)?k?1??

??k?1??

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

?00??k???? 9? 设A? B为n阶矩阵，且A为对称矩阵，证明BTAB也是对称矩阵? 证明 因为AT?A? 所以

(BTAB)T?BT(BTA)T?BTATB?BTAB?

从而BTAB是对称矩阵?

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AB?BA? 证明 充分性? 因为AT?A? BT?B? 且AB?BA? 所以 (AB)T?(BA)T?ATBT?AB?

即AB是对称矩阵?

必要性? 因为AT?A? BT?B? 且(AB)T?AB? 所以 AB?(AB)T?BTAT?BA? 11? 求下列矩阵的逆矩阵? (1)

1??2?2?? 5??1

解

12? |A|?1? 故A?存在? 因为 A???25????A11A21??5?2?A*???AA????21??

??1222??5?2?

A?1?1A*????21??|A|???cos??sin?? (2)??? sin?cos?????sin??? |A|?1?0? 故A?存在? 因为 ?cos 解 A???cos???sin??A11A21???cos?sin???

A*?????sin?cos??AA??1222??1A*??cos?sin??? ?1所以 A???sin?cos??|A|??故

1

?12?1? (3)?34?2??

?5?41????12?1? 解 A??34?2?? |A|?2?0? 故A?存在? 因为

?5?41????A11A21A31???420????? A*?A12A22A32??136?1? ????3214?2?AAA??132333????210??13?11?1所以 A?3??? A*???22?|A|??167?1???a1a?0??2 (4)??(aa? ? ?a ?0) ?

???0an??1

12

n

解

?a1?0?a?2A???? 由对角矩阵的性质知

??0?a?n?

?1??a1?0?1a?2?? A?1?????1?0???a?n? 12? 解下列矩阵方程? (1)

2??1?5?X??4?6??

?21?3?????15??4?6??3?5??4?6??2?23??

?21????12??21???08?3???????????21?1?1?13 (2)X?210????432???

?1?11????? 解

2X???1??21?1?1?13???210? 解 X????432??1?11????101?1?131????23?2? ???3?432???330?????221?2? ??8??5??? 3??320???31?? ?14?? (3)?X???????12???11??0?1? 解

?11X????1?4??31??2?0?1???12??????10?1???1

2?4??31??10?

?1???????12?11??0?1??12?10????1?2???4?010??100??1 (4)?100?X?001???2?001??010??1?????1?6 ??12?36??1?0???11??0?? ??43?0?1?? ?20???010??1?43??1 解 X??100??20?1??0?001??1?20??0??????010??1?43??10 ??100??20?1??00?001??1?20??01????? 13? 利用逆矩阵解下列线性方程组?

?10?1? 0??0??2?10?1???13?4??

??0???10?2?001?1x?2x2?3x3?1??1 (1)?2x1?2x2?5x3?2?

??3x1?5x2?x3?3 解 方程组可表示为

?123??x1??1??? ?225?x2??2??

?351??x??3????3????1?x1??123??1??1???故 x2??225??2???0?? ?x??351??3??0???????3??x?1??1从而有 ?x2?0?

??x3?0x?x?x?2??123 (2)?2x1?x2?3x3?1?

??3x1?2x2?5x3?0 解 方程组可表示为

?1?1?1??x1??2??? ?2?1?3?x2??1??

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

??x3?3 14? 设Ak?O (k为正整数)? 证明(E?A)?1?E?A?A2?? ? ??Ak?1? 证明 因为Ak?O ? 所以E?Ak?E? 又因为 E?Ak?(E?A)(E?A?A2?? ? ??Ak?1)?

所以 (E?A)(E?A?A2?? ? ??Ak?1)?E? 由定理2推论知(E?A)可逆? 且

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

证明 一方面? 有E?(E?A)?1(E?A)? 另一方面? 由Ak?O? 有

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

故 (E?A)?1(E?A)?(E?A?A2?? ? ??Ak?1)(E?A)? 两端同时右乘(E?A)?1? 就有

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

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? 证明 由A2?A?2E?O得 A2?A?2E? 即A(A?E)?2E?

或

A?1(A?E)?E?

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

2由定理2推论知A可逆? 且 由A2?A?2E?O得

A2?A?6E??4E? 即(A?2E)(A?3E)??4E?

第二章 矩阵及其运算

1? 已知线性变换?

??x1?2y1?2y2?y3?x2?3y1?y2?5y3? ??x3?3y1?2y2?3y3求从变量x1? x2? x3到变量y1? y2? y3的线性变换?

解 由已知?

?x1??221??y1??? ?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??32?4?????3????y3??2????y1??7x1?4x2?9x3 ?y2?6x1?3x2?7x3?

??y3?3x1?2x2?4x3 2? 已知两个线性变换

???y1??3z1?z2?x1?2y1?y3 ?x2??2y1?3y2?2y3? ?y2?2z1?z3?

???y3??z2?3z3?x3?4y1?y2?5y3求从z1? z2? z3到x1? x2? x3的线性变换?

解 由已知

?x1??201??y1??201???31?? ?x2????232?y2???232??20?x??415??y??415??0?1??2?????3????613??z1??? ??12?49?z2?

??10?116??z????3?0??z1?1??z2? ?z?3???3???x1??6z1?z2?3z3所以有?x2?12z1?4z2?9z3?

??x3??10z1?z2?16z3?111??123? 3? 设A??11?1?? B???1?24?? 求3AB?2A及AB?

?1?11??051??????111??123??111? 解 3AB?2A?3?11?1???1?24??2?11?1?

?1?11??051??1?11????????058??111???21322? ?3?0?56??2?11?1????2?1720??

?290??1?11??429?2????????111??123??058?T AB??11?1???1?24???0?56??

?1?11??051??290???????T

4? 计算下列乘积?

?431??7? (1)?1?23??2??

?570??1??????431??7??4?7?3?2?1?1??35? 解 ?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? 解 (123)?2??(1?3?2?2?3?1)?(10)?

?1????2? (3)?1?(?12)?

?3????2?(?1)2?2???24??2?? 解 ?1?(?12)?1?(?1)1?2????12??

?3??3?(?1)3?2???36????????131??0?12?2140?? (4)???1?31? ?

1?134????40?2???131??0?12??6?78?2140?? 解 ???1?31???20?5?6??

1?134???????40?2??a11a12a13??x1????x??

(5)(x1x2x3)a12a22a23???2??a13a23a33??x3? 解

?a11a12a13??x1????x?

(x1x2x3)a12a22a23???2??a13a23a33??x3? ?(a11x1?a12x2?a13x3 a12x1?a22x2?a23x3 a13x1?a23x2?a33

?x1???x)x2 ?x??3?3

22?a11x12?a22x2?a33x3?2a12x1x2?2a13x1x3?2a23x2x3?

5? 设1A???1?2?? B??1?13???0?? 问?

2?? (1)AB?BA吗? 解 AB?BA? 因为3AB???4?4?? BA??1?36???2?? 所以AB?BA?

8?? (2)(A?B)2?A2?2AB?B2吗? 解 (A?B)2?A2?2AB?B2?

因为2A?B???2?2?? 5??2??2?25???

2(A?B)2???2?但

2???814??

?1429?5????38??68???1A2?2AB?B2???411???812??3?????0???1016??

?1527?4????所以(A?B)2?A2?2AB?B2?

(3)(A?B)(A?B)?A2?B2吗? 解 (A?B)(A?B)?A2?B2? 因为2A?B???2?2?? A?B??0?05???2??

1??22?02???06??

(A?B)(A?B)???25???01??09???????22?38??10??28?而 A?B?????34???17??

411??????

故(A?B)(A?B)?A2?B2?

6? 举反列说明下列命题是错误的? (1)若A2?0? 则A?0? 解 取0A???0?1A???0?1?? 则A?0? 但A?0? 0??2

(2)若A2?A? 则A?0或A?E? 解 取1?? 则A?A? 但A?0且A?E? 0??2

(3)若AX?AY? 且A?0? 则X?Y ? 解 取

1A???0?0?? X??11?? Y??1??11??00?????2

3

k

1??

1??则AX?AY? 且A?0? 但X?Y ?

7? 设10? 求A? A? ? ? ?? A?

A????1???? 解

10?10???10??

A2????1????1??2?1???????10?10???10??

A3?A2A???2?1????1??3?1??????? ? ? ? ? ? ??

10??

Ak???k?1?????10? 8? 设A??0?1?? 求A?

?00????

k

解 首先观察

??10???1A2??0?1??0??00???00?????33?2A3?A2?A??0?3?00???44?3A4?A3?A??0?4?00???55?4A5?A4?A??0?5?00?0???22?1?1???0?22???

?00?2??????3??3?2?? ?3??6?2?4?3?? ?4??10?3?5?4?? ?5????? ? ?? ? ? ? ? ? ??

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

或

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

4?1由定理2推论知(A?2E)可逆? 且(A?2E)

?1(3E?A)? 4 证明 由A2?A?2E?O得A2?A?2E? 两端同时取行列式得 |A2?A|?2?

即 |A||A?E|?2? 故 |A|?0?

所以A可逆? 而A?2E?A2? |A?2E|?|A2|?|A|2?0? 故A?2E也可逆? 由 A2?A?2E?O ?A(A?E)?2E

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

2又由 A2?A?2E?O?(A?2E)A?3(A?2E)??4E

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

所以 (A?2E)?1(A?2E)(A?3E)??4(A?2 E)?1?

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

41 16? 设A为3阶矩阵? |A|?? 求|(2A)?1?5A*|? 21A*? 所以 ?1 解 因为A?|A|

|(2A)?1?5A*|?|1A?1?5|A|A?1|?|1A?1?5A?1|

222 ?|?2A?1|?(?2)3|A?1|??8|A|?1??8?2??16? 17? 设矩阵A可逆? 证明其伴随阵A*也可逆? 且(A*)?1?(A?1)*? 证明 由A?1?1A*? 得A*?|A|A?? 所以当A可逆时? 有

|A|1

|A*|?|A|n|A?1|?|A|n?1?0?

从而A*也可逆?

因为A*?|A|A?1? 所以 (A*)?1?|A|?1A?

又A?1(A?1)*?|A|(A?1)*? 所以

?1|A| (A*)?1?|A|?1A?|A|?1|A|(A?1)*?(A?1)*? 18? 设n阶矩阵A的伴随矩阵为A*? 证明? (1)若|A|?0? 则|A*|?0? (2)|A*|?|A|n?1? 证明

(1)用反证法证明? 假设|A*|?0? 则有A*(A*)?1?E? 由此得 A?A A*(A*)?1?|A|E(A*)?1?O ?

所以A*?O? 这与|A*|?0矛盾，故当|A|?0时? 有|A*|?0?

(2)由于A?1?1A*? 则AA*?|A|E? 取行列式得到

|A| |A||A*|?|A|n? 若|A|?0? 则|A*|?|A|n?1?

若|A|?0? 由(1)知|A*|?0? 此时命题也成立? 因此|A*|?|A|n?1?

?033? 19? 设A??110?? AB?A?2B? 求B?

??123??? 解 由AB?A?2E可得(A?2E)B?A? 故

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

??121???123??110????????101? 20? 设A??020?? 且AB?E?A?B? 求B?

?101????12

?1 解 由AB?E?A2?B得 (A?E)B?A2?E?

即 (A?E)B?(A?E)(A?E)?

001 因为|A?E|?010??1?0? 所以(A?E)可逆? 从而

100?201? B?A?E??030??

?102??? 21? 设A?diag(1? ?2? 1)? A*BA?2BA?8E? 求B? 解 由A*BA?2BA?8E得 (A*?2E)BA??8E? B??8(A*?2E)?1A?1 ??8[A(A*?2E)]?1 ??8(AA*?2A)?1 ??8(|A|E?2A)?1 ??8(?2E?2A)?1 ?4(E?A)?1

?4[diag(2? ?1? 2)]?1

?4diag(1, ?1, 1)

22010?300100?0?? 0?8?? ?2diag(1? ?2? 1)?

?1?0 22? 已知矩阵A的伴随阵A*??1?0?且ABA?1?BA?1?3E? 求B?

解 由|A*|?|A|3?8? 得|A|?2? 由ABA?1?BA?1?3E得 AB?B?3A?

B?3(A?E)?1A?3[A(E?A?1)]?1A

?3(E?1A*)?1?6(2E?A*)?1

2?1?0 ?6??1?0?1

0??6000?0???0600?? 0??6060????6???030?1???1?4???10? 23? 设P?AP??? 其中P??? ?????? 求A

1102???? 解 由P?1AP??? 得A?P?P?1? 所以A11? A=P?11P?1. |P|?3?

01030010?111

?

1P*????1?4?? P?1?1?14??

??1?3??1?1????111而 ????00????10 ??

?11?2???02?11?14????27312732??1?4?10????1133故 A????0211??11????683?684??

11???????????33???1??111??1??

24? 设AP?P?? 其中P??10?2?? ?????1?11?5????求?(A)?A8(5E?6A?A2)?

解 ?(?)??8(5E?6???2)

?diag(1?1?58)[diag(5?5?5)?diag(?6?6?30)?diag(1?1?25)] ?diag(1?1?58)diag(12?0?0)?12diag(1?0?0)? ?(A)?P?(?)P?1

?1P?(?)P*

|P|?111??100???2?2?2???2?10?2??000???303??1?11??000???12?1????????111? ?4?111??

?111???

25? 设矩阵A、B及A?B都可逆? 证明A?1?B?1也可逆? 并求其逆阵? 证明 因为

A?1(A?B)B?1?B?1?A?1?A?1?B?1?

而A?1(A?B)B?1是三个可逆矩阵的乘积? 所以A?1(A?B)B?1可逆? 即A?1?B?1可逆?

(A?1?B?1)?1?[A?1(A?B)B?1]?1?B(A?B)?1A?

?1?0 26? 计算?0?0?031?12?1?? 0?23?00?3??12?? A??21?? B??31?? B???23?? ? 解 设A???????1?01???2?03?1?2?1?2?0?3??AE??EB1???A1A1B1?B2??

则 ?1??OB??OAB?OA??2??2?22?210010200??11??01??0?03???12??31????23???52?? ?AB?B????2?1??0?3??2?4?11201????????21???23????43?? ? AB????0?3??0?9?2203???????1252?A1E??EB1??A1A1B1?B2??012?4??所以 ???OB???OAB???00?43?? OA??2??2?22??000?9????1210??1031??1252??0101??012?1??012?4??即 ??

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

27? 取 解

10? 验证AB? |A||B|? A?B??C?D???01??CD|C||D|??10102000AB?0101?0200?2010?4? CD?1010?101002010?1010?101而

|A||B|11? ??0|C||D|11AB? |A||B|CD|C||D|?

8

4

故

?34O??4?3? 28? 设A???? 求|A|及A?

20?O22???34?? A??20?? ? 解 令A???1?4?3???2?22?A1O??A?则 ?OA?? ?2?

O??A18O?8?A1故 A?????OA8?? OA?2??2?8888816 |A|?|A|A1||A2|?1||A2|?10?

8?540O?4?054?O?4?A1A???? 4???4OA20?2??O64?22???1 29? 设n阶矩阵A及s阶矩阵B都可逆? 求 (1)

OA??BO?????

?1OA???C1C2?? 则 ??BO??CC????34??C1C2???AC3AC4???EnO?? ?OA? ???CC??BCBC??OE?BO???34??1s?2???AC3?En?C3?A?1?AC4?O?C4?O由此得 ???

BC1?O?C1?O?BC?E?C?B?1?2s?2 解 设?1OAOB??? ?????1所以 ???BOAO?????1?AO? (2)???CB??1?

?1?D1D2?? 则 ?AO?? 解 设?????CB??D3D4?AD2??EnO??D1D2???AD1?AO? ???DD??CD?BDCD?BD???OE??

CB???34??1324??s??D1?A?1?AD1?En?D2?O?AD2?O由此得 ???

CD1?BD3?O?D3??B?1CA?1?CD?BD?E?D?B?1?24s?4?1AOA???????1?1O所以 ???1?? ?CB???BCAB??1 30? 求下列矩阵的逆阵?

?5?2 (1)?0?0?0?0?? 3?2???52??8 解 设A??? B????21??5

210000853?? 则 2??3???2?3??

??58?2?????15A?1???2?2???1?2?? B?1??8??25??51??????1?1?5200??1?200??1?1?2100??A??A??2500??于是 ?? ??????1???00083??02?3B??B??0052???00?58??????1000??1200? (2)??

2130??1214????10??30??21? 解 设A??? B??? C?????? 则

141212??????

?1?1 ?2?1?021200310??10???AO???A?1O?

??B?1CA?1B?1?0??CB??????4??100?1?11?0?2211 ??1???263?15??1?0??0??? 0?1???824124?

百度搜索“70edu”或“70教育网”即可找到本站免费阅读全部范文。收藏本站方便下次阅读，70教育网，提供经典综合文库同济第五版线性代数 课后题解析第二章在线全文阅读。