??233??033??033? B?(A?2E)?1A??1?10??110????123??
??121???123??110????????1?101? 20? 设A??020?? 且AB?E?A2?B? 求B?
?101??? 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)
22 ?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?
010?300100?0?? 0?8?? 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?03?6????100600?0?? 0??1???1?4? ????10?? 求A11? 23? 设P?1AP??? 其中P???11???02????? 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?1?1?3???0????10 ??
?11?2???02??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??22?01???1?? B??31?? B???23??
1?2?1?2?0?3?3??????A1E??EB1??A1A1B1?B2?则 ??OA??OB???OAB??
??2??2?22?1而 A1B1?B2???0?2 A2B2???0?2??31????23???52?? ?2?1??0?3??2?4?1????????1???23????43?? ?0?3??0?9?3???????1A1E??EB1??A1A1B1?B2??0???所以 ???????0OAOBOAB??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??252?12?4?? 0?43?00?9??1 27? 取A?B??C?D???0?0?? 验证AB? |A||B|?
CD|C||D|1??1 解 AB?0CD?10010?11010021?00?110020?1001000?2010?4? 002011|A||B|11而 ??0?
|C||D|11|A||B|故 AB? ?
CD|C||D|?34O??4?3? 28? 设A??? 求|A8|及A4? ?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|?|A1||A2|?|A1||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????OA??C1C2?? 则 解 设??BO???CC????34??1?1OA??C1C2???AC3AC4???EnO?? ??BO??CC??BCBC??OE????34??1s?2???AC3?En?C3?A?1??由此得 ?AC4?O??C4?O?
C1?OBC1?O??BC?E?1?2s?C2?B?1OAOB??? ?????1所以 ????BO??AO??1AO? (2)??CB?????1