《线性代数》第二章习题解答
??5?1． 解：（1）2AB?3B??10?6???3??2E???1?3?3???9???6????6?697?34?35???4? 7???2???2? 4???2?0?? （2）?10???3?1?232334（2）ABT2．解：（1）????2??46??0????4???04213??1???4??????1???3????3? ????4???2??8??1????2（3）???3?3????7?（5）?5?6?6741?3??1????6?9??1?5????2?4??1??2????2?1??2? （4）?3?3??067?5??4? 7??1?1????1???2???2
?3????4?3?33???54????8?5??7?6??（6）?x1x2?a11?x3??a21?a?31a12a22a32a13??x1??a23??x2?a33???x3??? ???x1???a13x1?a23x2?a33x3??x2?
?x??3?2??a11x1?a21x2?a31x3a12x1?a22x2?a32x3?(a11x1?a21x2x1?a31x3x1)?(a12x1x2?a22x2?a32x3x2)?(a13x1x3?a23x2x3?a33x3
3．解：AT22?1????3?0?4113?2413???2T1?， B???4???2?3??2??1??1??2???14??3??3????1?2????1?341131?? 2???1?TAB? (1)??3?0??2T（2）BA???4??24??2? ?1???12?1?? ??1??11?????2?2???30???31????4???2?(3)(AB)T?BATT?2???4?13?11?????3?2???0?2413????11?????5???2?165?? ?11??x1?4．解：从y1,y2,y3到x1,x2,x3的线性变换可表示为：?x2?x?3?1?A??3?2??2?B??1?1???y1?????Ay??2?，其中??y???3??3?4?5543?x1??x2?x?32??y1??z1??????1?；从z1,z2,z3到y1,y2,y3的线性变换可表示为：?y2??B?z2?，其中
?y??z?3???3??3?6??3?，所以从z1,z2,z3到x1,x2,x3的线性变换可表示为： 2????z1??1?????AB??z2???3??z??2??3???3?4?52??1?3???2??1?1?5436??3?2???z1??z2?z?3??1?????3??2???12?11??8?3???z1???z?2? ?z??3?所以，从z1,z2,z3到x1,x2,x3的线性变换为： ?x1?z1?z2?z3 ? ?x2?3z1?2z2?8z3
?x?2z?z?3z 123?35．解：（1）f(A)?A2?3A?4E????3?111?xx1?1?2??2???1???3??2??1??－3???32???2??8??＝?4E??02??0?? ?8? (2) f(x)?0x?123?x?2x?2 0?11??0???1??2?22 f(A)?A?2A?2E???2?21??1???2?20???1???1????2E??10???2?2?? 0??6．（1）∵(A?B)?(A?B)(A?B)?A?AB?BA?B ∴要使(A?B)?A?2AB?B，则必须BA?AB (2) ∵(A?B)(A?B)?A?AB?BA?B
22 ∴要使(A?B)(A?B)?A?B，则必须?AB?BA?0 ，即BA?AB
22222 (3) 当BA?AB时，用数学归纳法证明(AB)?AB
kkk①k?1时，显然(AB)?AB
k222 k?2时，(AB)?(AB)?ABAB?A(AB)B?A(AB)B?AB，
kkk 所以(AB)?AB
kkk ②设k?n时，有(AB)?AB，则k?n?1时
kkk (AB)K?(AB)nn?!n?1?(AB)(AB)?ABAB?A(BA)B?A(BAB)B?A(Bnn?1nnnnnnn?1BA)B
n?1 ?A(BA)B???A(AB)B2nn?An?1B
可见，k?n?1时，也有(AB)k?AkBk
所以，当BA?AB时，对一切正整数k都有 (AB)k?AkBk
?11?n7．解：(1) ??2n?12n?1??11????????2n?12n?1? ??2 (2) ∵??2?1?0???3?2??1??????01?? ???10??? ∴??2?1?n?1?? n为偶数???3?2?????0?????1?
??2???3?2?? n为奇数??110?221?(3) ∵??011???1???012?? ， ??001????001????110?3?21??10?33??011???112??1??0??011???1???013?? ??001????001????001????001???3?110?410???110??33???011???1???011??011???1??01?1?3???0??001????001????001????001????0?146????014?? ??001???10?n??1nn(n?1)?∴?12??011????1???0n?
?001????001????8．证明：∵A、B为对称矩阵，∴AT?A，BT?B (1) ∵ (CTAC)T?CTAT(CT)T?CTAC
∴ CTAC是对称矩阵
(2) ∵ (ABABA)T?ATBTATBTAT?ABABA
∴ ABABA是对称矩阵
(3) ∵(AA?1)T?ET?E，AT?A
∴(AA?1)T?(A?1)TAT?(A?1)TA?E?A?1A
∴ (A?1)T?A?1
∴ A?1是对称矩阵
10?11??01?? 9．解：(1) ∵
2347?2?0
?1?2∴??3?4??7???123?7?4???37?4??2???2??3?4??7???1?1?7??2??3?4? ??2? (2) ∵
sin??cos?cos?sin??sin??cos??1?0
?122?sin?∴ ???cos??cos???sin????1?sin???1?cos??cos???sin??????sin???cos??cos??? sin???2324536 (3) ∵13r1?2r2r3?3r2365653?1010?12?2?2?3??1?0 ∴?1?3?3?133656?3， A13???3， A23???13245??3?可逆 6??又∵A11?2434?0， A12???2， A22?13232434??2
?1
A21?? A31??2?∴?1?3??3?(4) ??2?1??1?(5) ?3??3?3223??1， A32??2153??1， A33??23?12132?1
324?21?15??3?6???A11????A12?A?13?1A21A22A23A31??0??A32????3?A33????2A21A22A231??1? ?1???1?3?1?3???3? 1??3???3?2??2???4?0???A111???A122??A13???4???2??3??22A31??1?1?A32????12?A33????133?6?1?112??3?5?? 3??1???123??012(6) 把D??001??000??12?其中A???01??，B??4???A3??分块为?02???1???12????01??，C???C??， B???3??2?4??， 3??则D?AB?1?0，∴矩阵D可逆。根据课本P44的例2可知，
?11D?1???A?A?1CB?????0B?1?? 又∵A?1???1?2???2???01??， B?1??1???01?? ?∴ ?A?1CB?1????1?2??4????2?0???01???3??23??1??1?????1?0????21?? ????1?210?∴ D?1??01?21????001?2? ???0001??10．解：(1) 由X??34???42???12???4???3?得， ??62???42??42?X???4?34??1??3?5?4????3??2????4??1?2?62??3????0???1?3??62?2????1???1? ??5?9???11??11? (2) 由?1?234???3?X?201??得， ??368????121??
?11??1??311?X??1??02?1???311??2?234???201????4?52???201???3???48??368????121?????33?1????121?????4?5?01??010?06? (3)由?0?100???X?001????3???5?24??得， ??010????100????213???001??1???306???010??1X???100???5?24???001??
??010????213????100???010?06??45????001????3?001????5?24???100????2??132?? ??100????213????010????06?3??11．解：(1) ∵方程组用矩阵形式可表示为： ?13?2??x1? ??11?1??????x2??1????2? ??25?3????x?3???4??1?1???1??