线性代数习题1-4作业与解答（同济五版）

第一章 行列式

1? 利用对角线法则计算下列三阶行列式?

201(1)1?4?1? ?183

解

2011?4?1 ?183 ?2?(?4)?3?0?(?1)?(?1)?1?1?8 ?0?1?3?2?(?1)?8?1?(?4)?(?1) ??24?8?16?4??4?

111(3)abca2b2c2?

解

111abca2b2c2 ?bc2?ca2?ab2?ac2?ba2?cb2 ?(a?b)(b?c)(c?a)?

4? 计算下列各行列式?

41 (1)10012512021125142? 0720214c?c4?123122??????0c?7c103300742?104?1?1002?122?(?1)4?3 2?14103?141041 解 100

23 (2)154?110c2?c39910?12?2??????00?2?0? 10314c1?1c17171423

1?120423611? 2223 解 151?12042361c?c421?????22423023151?12042360r?r422?????0223121?12142340200

r4?r121?????3?1120002?0? 00

?abacae(3)bd?cddebfcf?ef?

解

?abacae?bcebd?cdde?adfb?cebc?ebfcf?ef?111?adfbc1e?11?4abcde? f11?1

a1(4)?001b?1001c?1001d?

0r?ar120?????1d01?aba0?1b100?1c100?1d 解

a?1001b?1001c?1

1?aba0c3?dc21?abaad2?1?(?1)(?1)?1c1??????1c1?cd0?1d0?10

6. 证明:

?(?1)(?1)3?21?abad?abcd?ab?cd?ad?1? ?11?cda2abb2 (1)2aa?b2b?(a?b)3;

111 证明

a22a1abb2c2?c1a2ab?a2b2?a2a?b2b?????2ab?a2b?2a0011c3?c11

?(?1)3?1ab?a2b?a3b2?a2?(b?a)(b?a)ab?a?(a?b) ? 122b?2aax?byay?bzaz?bxxyz (2)ay?bzaz?bxax?by?(a3?b3)yzx;

az?bxax?byay?bzzxy 证明

ax?byay?bzaz?bx ay?bzaz?bxax?by

az?bxax?byay?bzxay?bzaz?bxyay?bzaz?bx ?ayaz?bxax?by?bzaz?bxax?by

zax?byay?bzxax?byay?bzxay?bzzyzaz?bx ?a2yaz?bxx?b2zxax?by

zax?byyxyay?bzxyzyzx ?a3yzx?b3zxy

zxyxyzxyzxyz ?a3yzx?b3yzx

zxyzxyxyz ?(a3?b3)yzx?

zxy8. 计算下列各行列式(Dk为k阶行列式)?

a (1)Dn?1? ? ?1, 其中对角线上元素都是a? 未写出的元素都

a是0? 解

a000a000aDn?? ? ?? ? ?? ? ?0001000an?1?(?1)0? ? ?000a? ? ?0? ? ?01? ? ?00? ? ?00(按第n

? ? ?? ? ?? ? ?? ? ?a0? ? ?0a? ? ?? ? ?? ? ?? ? ?? ? ?000? ? ?a行展开)

000? ? ?01a0 ?(?1)2n?a? ? ?0? ? ?a(n?1)?(n?1)0(n?1)?(n?1)a

?(?1)n?1?(?1)n? ? ??an?a

a(n?2)(n?2)n

?an?2?an?2(a2?1)?

x (2)Dn?? a? ?aax? ? ?a? ? ?? ? ?? ? ?? ? ?aa; ? ? ?x 解 将第一行乘(?1)分别加到其余各行? 得

xaaa?xx?a0Dn?a?x0x?a? ? ?? ? ?? ? ?a?x00? ? ?a? ? ?0? ? ?0? ? ?? ? ?0x?a?

再将各列都加到第一列上? 得

x?(n?1)aaa0x?a0 Dn?00x?a? ? ?? ? ?? ? ?000? ? ?a? ? ?0n ? ? ?0?[x?(n?1)a](x?a)? ? ?? ? ?0x?a第二章 矩阵及其运算 1. 计算下列乘积? (5)(x1x2?a11a12a13??x1?x3)?a12a22a23??x2?? ????aaa?132333??x3?

解

(x1x2?a11a12a13??x1?x3)?a12a22a23??x2? ????aaa?132333??x3?

?x1? ?(a11x1?a12x2?a13x3 a12x1?a22x2?a23x3 a13x1?a23x2?a33x3)?x2?

???x3? ?a11x12?a22x22?a33x32?2a12x1x2?2a13x1x3?2a23x2x3?

?111??123?2. 设A??11?1?? B???1?24??

?1?11??051?????求3AB?2A及ATB?

解

?111??123??111?3AB?2A?3?11?1???1?24??2?11?1?

?1?11??051??1?11????????058??111???21322??3?0?56??2?11?1????2?1720?? ?290??1?11??429?2????????111??123??058?AB??11?1???1?24???0?56??

?1?11??051??290???????T

3.

已知两个线性变换

x?2y1?y3??1?x2??2y1?3y2?2y3? ??x3?4y1?y2?5y3y??3z1?z2??1?y2?2z1?z3??y3??z2?3z3 ?

求从z1? z2? z3到x1? x2? x3的线性变换? 解 由已知

?x1??201??y1??201???310??z1??x????232??y????232??201??z? ?x2??415??2??415??0?13??2???y2??????z3??3????613??z1???12?49??z2?? ??10?116??????z3?

x??6z1?z2?3z3??1所以有?x2?12z1?4z2?9z3?

??x3??10z1?z2?16z34.

12??10?设A???13?? B??12?? 问?

???? (1)AB?BA吗? 解 AB?BA?

34??12? 因为AB????? BA??38?? 所以AB?BA?

???46?

(3)(A?B)(A?B)?A2?B2吗? 解 (A?B)(A?B)?A2?B2?

22??02?? 因为A?B????? A?B????25??01?22??02??06? (A?B)(A?B)?????????? 250109??????38??10??28?而 A2?B2???????????

?411??34??17?故(A?B)(A?B)?A2?B2?

5. 举反列说明下列命题是错误的? (1)若A2?0? 则A?0?

01?2 解 取A????? 则A?0? 但A?0?

?00? (2)若A2?A? 则A?0或A?E?

11?2 解 取A????? 则A?A? 但A?0且A?E? 00?? (3)若AX?AY? 且A?0? 则X?Y ? 解 取

10??1 A??? X?????00???11??11?? ? Y????1??01?则AX?AY? 且A?0? 但X?Y ? 7.

??10?设A??0?1??

?00????求Ak ?

解 首先观察

2??10???10???2?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??

?005???? ? ? ? ? ? ??

??kk?k?1k(k?1)?k?2?2kA??k0?k?k?1?0k0????? ? ??

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?1k(k?1)k?2??k??????10?2???0?1? Ak?1?Ak?A?0?kk?k?1????00?k00????????k?1?k?1(k?1)k?(k?1)??k?1??2?k?1?0?(k?1)?k?1?? ??k?100?????

由数学归纳法原理知?

?kk?1k(k?1)k?2??k????2?? Ak??0?kk?k?1??00?k????

8. 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? 11? 求下列矩阵的逆矩阵?

12? (1)??25??

??12??1 解 A????? |A|?1? 故A存在? 因为

?25? A*???A11A21??5?2?????21?? AA??1222??5?2?故 A?1?1A*????21??

|A|???12?1?(3)?34?2?? ?5?41???

解

?12?1?A??34?2?? |A|?2?0? ?5?41???故A?1存在? 因为

?A11A21A31???420?A*??A12A22A32????136?1??

????3214?2?AAA??132333??所以

??210??1311??1A?A*???3???

|A|22??167?1????a1a0???2(4)??(a1a2? ? ?an ?0) ?

??0?a?n?

解

?a1?0??a2A??? ???0?a?n?由对角矩阵的性质知

?1??0?a11??a2?? A?1????1?0???an???

12. 利用逆矩阵解下列线性方程组?

?x1?2x2?3x3?1?(1)?2x1?2x2?5x3?2? ??3x1?5x2?x3?3

解 方程组可表示为

?123??x1??1??225??x???2?? ?351??2??3????x3????1

故

?x1??123??1??1??x???225??2???0?? ?2??351??3??0???????x3??

从而有

?x1?1??x2?0? ??x3?0?1?4???119.设P?1AP??? 其中P????? ????11??00?11

?? 求A? 2? 解 由P?1AP??? 得A?P?P?1? 所以A11? A=P?11P?1.

14?1?14??1 |P|?3? P*????? P????

??11?113??1?1??10???10 ?而 ?11???02???0211??

????故 A11?14??1?4???10??33??27312732??????0211??????683?684?? 1111???????????33???1??111?其中P??10?2?? ???1??

?1?11???5????20. 设AP?P??

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???

21. 设Ak?O (k为正整数)? 证明(E?A)?1?E?A?A2?? ? ??Ak?1? 证明 因为Ak?O ? 所以E?Ak?E? 又因为 E?Ak?(E?A)(E?A?A2?? ? ??Ak?1)? 所以 (E?A)(E?A?A2?? ? ??Ak?1)?E? 由定理2推论知(E?A)可逆? 且 (E?A)?1?E?A?A2?? ? ??Ak?1?

证明 一方面? 有E?(E?A)?1(E?A)?

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