(2)若A2?A? 则A?0或A?E?
1 解 取A???0? 解 取
1?? 则A2?A? 但A?0且A?E? 0?? (3)若AX?AY? 且A?0? 则X?Y ?
1 A???0?0?? X??11?? Y??1??11??00?????1??
1??则AX?AY? 且A?0? 但X?Y ?
10? 求A2? A3? ? ? ?? Ak? 7? 设A????1????10?10???10?? 解 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Ak ?
?00???? 解 首先观察
??10???10???22?1? A2??0?1??0?1???0?22???
?00???00???00?2?????????33?23?? A3?A2?A??0?33?2??
?00?3?????44?36?2? A4?A3?A??0?44?3??
?00?4?????55?410?3? A5?A4?A??0?55?4??
?00?5??? ? ? ? ? ? ??
??kk?k?1k(k?1)?k?2?2k A??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，
??kk?k?1k(k?1)?k?2?????10?2 Ak?1?Ak?A??0?kk?k?1??0?1?
?00??00???k????????k?1(k?1)?k?1(k?1)k?k?1???2 ??0?k?1(k?1)?k?1??
??k?100?????由数学归纳法原理知?
??kk?k?1k(k?1)?k?2???2k?k?1?? Ak??0?k?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??12?? |A|?1? 故A?1存在? 因为 解 A???25???A11A21??5?2? A*???AA????21??
??1222??5?2? 故 A?1?1A*????21??|A|??cos??sin?? (2)??sin?cos?????co?s?si?n?? |A|?1?0? 故A?1存在? 因为 解 A???si??nco?s??A11A21??cos?sin?? A*???AA????sin?cos???
??1222??cos?sin??? 所以 A?1?1A*????sin?cos??|A|???12?1? (3)?34?2??
?5?41????12?1? 解 A??34?2?? |A|?2?0? 故A?1存在? 因为
?5?41????A11A21A31???420? A*??A12A22A32????136?1??
????3214?2?AAA??132333????210??13?11?1所以 A?A*???3???
22?|A|??167?1???a1a?0??2 (4)??(a1a2? ? ?an ?0) ?
???0an???a10??a?2 解 A???? 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?
2 (1)??1?5?X??4?6??
?21?3????5??4?6??3?5??4?6??2?23?? ?21????12??21???08?3???????????12 解 X???1??21?1?1?13 (2)X?210????432???
?1?11??????21?1?1?13???210? 解 X????432??1?11????101?1?131 ???432????23?2?
3????330?????221??? 82 ????5??3??3?11 (3)???1?4?X?2?2????10???31??
??1???0?1??11 解 X????1?4??31??2?0?1???12?????0?
1???12?4?31??10? ?1??11???0?1??12?12??????