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?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)

22 ?2diag(1? ?2? 1)?

?1?0 22? 已知矩阵A的伴随阵A*??1?0?010?300100?0?? 0?8??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?010300100??600???060??60??6???03?100600?0?? 0??1???1?4? ????1 23? 设P?1AP??? 其中P???11???0???0?? 求A11?

2?? 解 由P?1AP??? 得A?P?P?1? 所以A11? A=P?11P?1.

1 |P|?3? P*????1??1而 ?11???0?114?? P?1?1?14??

??1?3??1?1??0????10 ??

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

???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?210010200??11??01??0?3???0031?12?1?? 0?23?00?3??1 解 设A1???0?2?? A??21?? B??31?? B???23??

2?03?1?2?1?2?0?3?1????????A1E??EB1??A1A1B1?B2?则 ??O??OB???OAB?? A??2??2?22?1?而 AB?B?112?0?2 A2B2???0?2??31????23???52?? ??????1???2?1??0?3??2?4?1???23????43?? ????3???0?3??0?9?252?12?4?? 0?43?00?9???1A1E??EB1??A1A1B1?B2??0?所以 ???OB???OAB???0OA??2??2?22??0??1?0即 ?0?0?210010200??11??01??0?03???031??112?1???00?23??0?000?3???252?12?4?? 0?43?00?9??1 27? 取A?B??C?D???0?1 解 AB?0CD?10010?110100?? 验证AB? |A||B|?

CD|C||D|1??020?1001000?2010?4? 002011021?00?110而 |A||B|?11?0?

|C||D|11AB? |A||B|故 CD|C||D|?

?34O??4?3?84

28? 设A??? 求|A|及A? ?20?O22???34? A??2 解 令A1???4?3??2?2???A1O?则 A???OA??

?2?0??

2??A1O??A18O??故 A?????OA8?? OA?2??2?88888816 |A8|?|A|A1||A2|?1||A2|?10?

?540O?4?O??0544?A1 A???? 4???4OA20?2??O64?22?? 29? 设n阶矩阵A及s阶矩阵B都可逆? 求 OA?? (1)??BO????1OA??C1C2?? 则 解 设??BO???CC????34?C1C2??AC3AC4??EnO?OA??? ???CC???BCBC???OE?? BO???34??1s?2???1?AC3?En?C3?A?1?AC4?O?C4?O由此得 ????

BC1?OC1?O?C?B?1?BC?E?2s?2?1OAOB??? ?????1所以 ???BOAO?????1AO? (2)??CB?????1AO???D1D2?? 则 解 设??CB??DD????34??1

AD2??EnO?AO??D1D2???AD1??CB??DD??CD?BDCD?BD???OE?? ???34??1324??s??D1?A?1?AD1?En?D?O?由此得 ?AD2?O??2?

CD1?BD3?OD3??B?1CA?1?D?B?1?CD?BD?E?24s?4?1AOA????所以 ?CB????1?1O?1?? ????BCAB??1 30? 求下列矩阵的逆阵?