工程数学线性代数课后习题答案

?1而 ?11???0?0????10 ??0211?2????11

?14????27312732??1?4?10????1133故 A????0211??11????683?684?11???????????33???1?111? 24 设APP 其中P??10?2? ???1??1?11????求

(A)A(5E6A()

8

8

8

??5??

A2)

(5E6

5

2

解 diag(1(1 diag(1

)

630)

diag

15)[diag(55)diag(6

125)]

15)diag(12

8

00)12diag(100)

(A)P()P1

?1P?(?)P* |P|?111??10

??2?10?2??00?1?11??00????111? ?4?111?

?111??? 25 设矩阵A、B及AB 并求其逆阵 证明 因为

0???2?2?2?0???303? ??0????12?1?都可逆 证明A1

B1也可逆

A(A1

1

1

B)B1

B1

A1

A1

B1

1

1

而A(AB)B是三个可逆矩阵的乘积逆

即A11

所以A(AB)B可

1

B1可逆 B1)

21001

(A[A(A1

B)B1]B(AB)1A

?1?0 26 计算?0?0?10200??11??01??0?03???031?12?1?0?23?00?3??

12? 解 设A1???01????23B2???0?3????

21? A2???03???31? B1???2?1???

A1E??EB1??A1A1B1?B2?则 ??OA??OB???OAB???2??2?22?

1而 A1B1?B2???0?2 A2B2???0?2??31????23???52???????1???2?1??0?3??2?4?

1???23????43??0?3??0?9?3???????1A1E??EB1??A1A1B1?B2??0?所以 ???OB???OAB???0OA??2??2?22??0??1?0即 ?0?0? 27

252?12?4?0?43?00?9??

210010200??11??01??0?03???031??112?1???00?23??0?000?3???252?12?4?0?43?00?9??1 取A?B??C?D???0?0?1??|A||B| 验证AB? CD|C||D|

1 解 AB?0CD?1010101021?00?102000100?2010?400201

0?1010?101而 ||CA||||DB||?1111?0 故 CADB? ||CA||||DB||

?3 28

设A???4?43O?? 求|A8|及A4

??O2?202?? 解 令A341?????4?3?? A??202??22???

则 A???A1O??OA

2??8故 A8???A1O??A18O??OA 2?????OA28?? |A8|?|A18||A28|?|A1|8|A2|8?1016 ?540 A4??4O?A1O??054???OA4??2???240?

?O2624?? 29 设n阶矩阵A及s阶矩阵B都可逆

?1 (1)??OBOA????

解 设??OA??1?BO?????C1C2??C3C4?? 则

求

OA??C1C2???AC3AC4???EnO? ??BO??CC??BCBC??OE????34??1s?2???AC3?En?由此得 ?AC4?OBC1?O?BC?E?2s?1

?C3?A?1?C4?O?C?O?C1?B?1?2

?1OAOB???????1所以 ????BO??AO?

AO (2)??CB?????1

?1D1D2?AO??? 解 设????DD?CB???34? 则

AD2??EnO?AO??D1D2???AD1 ??CB??DD??CD?BDCD?BD???OE????34??1324??s??AD1?En?由此得 ?AD2?OCD1?BD3?O?CD?BD?E?24s?1?D1?A?1?D2?O?D??B?1CA?1?D3?B?1?4

?1AOA???????1?1O所以 ???1??CB???BCAB? 30 求下列矩阵的逆阵

?5?2 (1)?0?0?210000850?0?3?2??

5 解 设A???2?5 A?1???2?2?1???18 B???5?3?2?? 则 8 B?1???5?3???2?3???58?2?????12???1?2???25?1????