?x1?∴?x2?x?3??1?????1??2??315?2???1??3???1?2?1?????2????1?4??3????111?1??1??4???????1??2???1?
?????2???4??3?∴ 方程组的解为x1?4,x2?1,x3?3, (2) ∵方程组用矩阵形式可表示为： ?1? ?2?3??x1?∴?x2?x?3??1?????2??3??1141141???1?5??1??x1???1??x2?5???x3?1??1???????3? ??1?????12?1?2??3??1???1??4???????43? ????1??1??????1??9????3????13?1??5???∴ 方程组的解为x1?4,x2??4,x3?1,
12．证明：
(1)∵A2?5A?7E?0 ∴A2?5A?7E
∴A(17A?57E)?E ∴A可逆，且A?1?17A?57E
(2)∵A2?5A?7E?0 ∴(A?3E)(A?8E)?7E?24E??17E ∴(A?3E)(?13．证明：
∵
117)(A?8E)?E ∴A?3E可逆，且(A?3E)?1??117(A?8E)
A?0
2k?1k∴(E?A)(E?A?A???A2k?1)
2k?1k?E?A?A???A?A?A???A?A
k?E?A?E
?12k?1∴E?A可逆，且(E?A)?(E?A?A???A)
14．解：∵AA?1?AA?1?E?1 ∴A又∵A?2且A为四阶行列式 ∴(14A)?1?1?1A?12
?3A*?4A?1?3AA?1??2A?1?(?2)A4?1?8
15．证明：∵ A和B为同阶的可逆方阵，且AB?BA，
∴ 两边左乘A?1得 B?A?1BA
两边右乘A?1得 BA?1?A?1B
16．证明：
?a11??a21 (1) 设A?????a?n1?a11??a12?????a?1nA11A21?An1a12a22?an2???a1n??A11??a2n??A12* ， 则A????????Aann???1nan1??A11??an2??A21T*， (A)????????Aann???n1A1n??A2n?*T?(A) ???Ann???1A21A22?A2n???An1??An2? ， ???Ann??A1n??A2n?， ???Ann??其中Aji是aij代数余子式
a21a22?a2nA12A22?An2??????A12A22?An2???∴AT???T*∴(A)?????(2)若A?0，则A可逆，且A ∴A?AA ∴(A)*T*?1?1AA ，
*?(AA)?1T?A(A)
?1T又∵(AT)*?AT(AT)?1?A(AT)?1，(A*)T?(AT)* ∴(A?1)T?(AT)?1 ?a11??a21(3) 设A?????a?n1?????则(?A)??????a11a21?an1a12a22?an2???a1n??A11??a2n??A12*A?，则????????Aann??1n?a1n???a2n?， ????ann??n?1A21A22?A2n???An1??An2? ， ???Ann??其中Aji是aij代数余子式
?a12?a22??an2???由于矩阵(?A)中任一元素(?aij)的代数余子式为(?1)?(?1)n?1A11?n?1(?1)A12?*∴ (?A)?????(?1)n?1A1n???n?1??(?1)????A11A12?A1n(?1)(?1)n?1n?1Aji
A21A22A2n???(?1)A21A22?A2n?n?1?An1??n?1(?1)An2?? ??n?1(?1)Ann??(?1)n?1??An1??An2?n?1*?(?1)A ???Ann??(4) ∵AA*????AE?????A0?0n0A?0????0??0????A??
n?n∴AA*?AA*?A ∴A*?A*n?1
(5) ∵AB(AB)?ABE，且A、B可逆， ∴两边左乘B?1A?1得，
(AB)?B*?1A?1ABE?B?1A?1ABE?BBEAAE?BA
12?1?1?0
?1?1**
?1?17．解：∵A??0?2?1101?1?2?，A?0?1?2?110∴ A可逆.
又∵ A2?AB?E， ∴A2?E?AB，两边左乘A?1得
A?1(A2?E)???1*?A?1A??分别计算?4A??2?B
1?321??2??2?2?，A?E??4?01???04?20??4? ?2??2022??0? 2???2??12B?A(A?E)?∴ ??4?4?
?1?18．解：∵A?diag(1,1,2)??0?0?010?0??1??1?0?， ∴A?0??2??0?*010?0?0?， 1??2?∵ABA**?6AB?20E，∴AB(A?6E)?20E
?1∴B(A?6E)?20A
又∵A?6E?AA*?1??4??6E??0?0?0?140?400??0? ?5??∴ (A?6E)*?1?1???4??0???0?0?0??0? ?1???5?B?20A(A?6E)?1*?1??1?20?0??0?010?1???0??40??01????2??0?0?140?0????5??0??0??1??0??5?0?500??0??2??
?219．∵P???2??1??1?1?， ∴P??2????10??19?? ，∴ ???0?2????1??2? 1??? 9(?2)??0 又∵ ????0??1又∵P?1AP??， ∴A?P?P?1， ∴A?P?P
?2??220．解：∵A??0??0?340000720??0??C???0?6??1??0??2?C??，其中?2?D??3??7??D?，?24???6?? 1??99?1?2???2?1??1???2???0???1?9??(?2)??1?0?1?514????2??1??1026?513? ???1025??C7∴A???0?70??， ∴ A?C7D7?C7D??A??0???17D7?2(?5)??10
777
?021．解：(1) 设??B??D1???D?3D2??0??，则?B?D4??A??D1???0???D3D2??E????D4???00?? ?E??AD3??AD4∴??BD1?BD2??0∴??B??E?0?0?E?1 ??D3?A?1??D4?0 ??D1?0?D?B?1?2B?? 0???1A??0???0???1?A??1?A(2) 设??C?0??B???D1???D?3D2??A??，则?CD4????????D3??0??B???1?D1??D?3D2??E????D4???00?? E??AD1?E??AD2?0?∴? ?CD?BD?013??CD?BD?E24?D1?A?1D2?0??BCAD4?B?1?1
?A??C?0??B???1?1?A????B?1CA??10?? ?1?B?