线性代数课后习题答案
习题一
1.2.3(答案略)
4. (1) ∵ ?(127435689)?4?1?5 (奇数) ∴ ?(127485639)为偶数
故所求为127485639
(2) ∵?(397281564)?2?5?1?1?9 (奇数) ∴所求为397281564 5.(1)∵??????? (偶数)
∴项前的符号位??1?6??1 (正号)
(2)∵a32a53a26a11a44a65?a11a26a32a44a53a65 ?(162435)?4?1?5
∴ 项前的符号位(?1)5??1 (负号) 6. (1) 原式=(?1)?(234?n1)1?2?n?(?1)(n?1)n! (2)原式=??1??((n?1)(n?2)?2?1?n)(n?1)(n?2)(n?1)(n?2)?2?1?n?(?1)2n!
n(n?1)(3)原式=(?1)?(n(n?1)?2?1)a21na2(n?1)?an1 ?(?1)a1na2(n?1)?an1
7.8(答案略)
9. ∵D?1?6?2?x?0?19?(?4?2)?0
∴x?7
10. (1)从第2列开始,以后各列加到第一列的对应元素之上,得
x1?1x?(n?1)1?110?1x?xx?(n?1)x?1???????????x?(n?1)?1x?1????11?xx?(n?1)1?x10???x?(n?1)?(x?1)n?1
(2)按第一列展开:
y0?00D?x?xn?1?(?1)n?1yxy?00nn??????x?(?1)n?1y
00?xy - 1 -
00?x?1
11234?n110211?11?n11?1?n1345?12311?1?n1?????????????????11?n?111?n1n?1n1?n?3n?2n?111?n?111?n1?1111?n?11?????n12?n?2n?1n1?n1?11(3)D?n(n?1)12?11
?n(n?1)02?0011n(n?1)?211?n
?
?????11?1?n10011?1?n111?11?n11?11?n?n(n?1)(n?2)?(n?3)???2?1?(?1)?211?1(n?1)(n?2)2
11?n?110?n?00??????(?1)?1n(n?1)??2?1?1?1
00??n0?(?1)(n?1)(n?2)2?1n(n?1)??2?1?1??(?1)0?nn(n?1)2n?n?2nn?1
- 2 -
习题二
1.2.3.4.5(答案略) 6. 设 B???x11?x21?1?1x12??为与A可交换的矩阵,则有AB?BA x22?1??x1??1??x2x12??x???x2?2?xx1??21??x2??211? ?1?即 ?111121解之得 x11?a,x12?b,x21?b,x22?a
?x1??3???7. (1)?x2???2?x??0?3???y1??1???y?1?2???y??0?3??031?1??y1????1y , 记为X=AY ??2????2???y3?1???z1??1?? ,记为Y=BZ ?z?2?1??2???z1?0?? ?z?2??3???x1??3???(2)X=A?BZ?=?AB?Z 即 ?x2???5?x??1?3??8(答案略)
?3?9.f(A)?A2?3A?2E??18?3??41045??10 ?1??10.(1)(A?B)(A?B)?A2?BA?AB?B2?A2?B2
(2) (A?B)2?(A?B)(A?B)
?A?BA?AB?B22
=A2?2AB?B2
111. ∵A2?A,A?(B?E)
2 ∴ B?2A?E,B2?4A2?4A?E?E 反之 若 B2?E,
则 4A2?4A?O,即 A2?A
12. (1) 设A?(aij),A2?(bij) ∵AT?A ∴aij?aji
又∵ A2?O ∴bii?0
又 bij?ai1a1j?ai2a2j???ainanj?ai12?ai22???ain2 (i,j?1,?2,n ,
- 3 -
当 i?j?1,2时,有a11?a12???a1n?0,a21?a22???a2n?0,an1?an2???ann?0 ?,n,∴ A?0
(2)设 A?(aij),AAT?(bij) 则bij?ai1aj1?ai2aj2???ainajn
∵ ATA?0 ∴ bij?0(i,j?1,2,?,n)
当 i?j 时,有 ai12?ai22???ain2?0(i?1,2,?,n) 故 ai1?ai2???ain?0(i?1,2,?,n) 即 A?0 13.(1) ∵ (ATA)T?ATA ∴ATA为对称矩阵
同理 AAT也为对称矩阵
(2) ∵ (A?AT)T?AT?A?A?AT ∴ A?AT 为对称矩阵
又 ∵(A?AT)T?AT?A??(A?AT) ∴ A?AT为反对称矩阵
(3)∵A?1(A?AT?A?AT)?1(A?AT)?1(A?AT)
222 由(2)知,1(A?AT)为对称矩阵,1(A?AT)为反对称矩阵
22故 A可表示成一个对称矩阵与一个反对称矩阵的和。
14. (1)必要性:∵AT?A,BT?B,(AB)T?AB ∴AB?(AB)T?BTAT?BA 充分性: ∵ AT?A,BT?B,AB?BA ∴ (AB)T?(BA)T?ATBT?AB (2) 必要性: ∵A2?E,B2?E,(AB)2?E
∴ BA?EBAE?A2BAB2?A(AB)2B?AB 充分性:∵A2?E,B2?E,AB?BA
∴ (AB)2?(AB)(AB)?A(BA)B?A2B2?E
(3) 必要性 :∵A2?A,B2?B,(A?B)2?A?B
∴(A?B)2?A2?AB?BA?B2?A?BA?AB?B?A?B
- 4 -
即 AB??BA
充分性: ∵A2?A,B2?B,AB??BA ∴ (A?B)2?A?B
15(答案略)
16. ∵ (E?A)(E?A?A???Ak?1)?E ∴ E?A 可逆。
且 (E?A)?1?E?A?A2???Ak?1 17. ∵ Ak(A?1)k?AkA?k?A?AA?1?A?1?E
∴ Ak可逆,且 (Ak)?1?(A?1)k 18.(答案略)
19. ∵AA*?AE,若 A可逆,则A?0
??∴ ?1A?A*?E 故 A*可逆,且(A*)?1?A
A?A?20.设 A?(aij),∵A是对称矩阵 ∴aij?aji 记 A*?(Nij),则
* Nij?Nji,即A*为对称矩阵,又∵ A?1?A, ∴ A?1为对称矩阵。
A21.(1)设 A*?(Nij),则 (?A)*?(?1)n?1Nij?(?1)n?1A* (2) ∵ AA*?AE ∴A*?AA?1 又 ∵(A?1)(A?1)*?A?1E ∴ (A?1)*?A?1(A?1)?1?A-1A
于是 A*(A?1)*?AA?1A?1A?E 即 (A?1)*?(A*)?1
(3)∵ AA*?AE ∴A*?AA?1
*T?1TT?1TT?1T* 于是 (A)?A(A)?A(A)?A(A)?(A)
(4) (注意加条件:A可逆) ∵ A可逆 ∴ A*?AA?1 ∴ (A*)*?(AA?1)*?An?1(A)?A?1*n?1(A)*?1
- 5 -