线性代数 - 北京邮电大学出版社(戴斌祥 - 主编)习题答案(3、4、5)
更新时间:2023-12-20 04:18:02 阅读量: 教育文库 文档下载
- 线性代数知识点总结推荐度:
- 相关推荐
习题 三
(A类)
1. 设α1=(1,1,0),α2=(0,1,1),α3=(3,4,0).求α1-α2及3α1+2α2-α3. 解:α1-α2=(1,1,0)-(0,1,1)=(1,0,-1),3α1+2α2-α3=(3,3,0)+(0,2,2)-(3,4,0)=(0,1,2)
2. 设3(α1-α)+2(α2+α)=5(α3+α),其中α1=(2,5,1,3),α2=(10,1,5,10),α=(4,1,-1,1).求α.
解:由3(α1-α)+2(α2+α)=5(α3+α) 整理得:α=
3
11(3α1+2α2-5α3),即α= (6,12,18,24) 66=(1,2,3,4)
3.(1)× (2)× (3)√ (4)× (5)×
4. 判别下列向量组的线性相关性.
(1)α1=(2,5), α2=(-1,3);
(2) α1=(1,2), α2=(2,3), α3=(4,3);
(3) α1=(1,1,3,1),α2=(4,1,-3,2),α3=(1,0,-1,2);
(4) α1=(1,1,2,2,1),α2=(0,2,1,5,-1),α3=(2,0,3,-1,3),α4=(1,1,0,4,-1). 解:(1)线性无关;(2)线性相关;(3)线性无关;(4)线性相关.
5. 设α1,α2,α3线性无关,证明:α1,α1+α2,α1+α2+α3也线性无关. 证明:设
k1?1?k2(?1??2)?k3(?1??2??3)?0,
即
(k1?k2?k3)?1?(k2?k3)?2?k3?3?0.
由?1,?2,?3线性无关,有
?k1?k2?k3?0,? ?k2?k3?0,?k?0.?3所以k1?k2?k3?0,即?1,?1??2,?1??2??3线性无关.
6.问a为何值时,向量组
?1?(1,2,3)',?2?(3,?1,2)',?3?(2,3,a)'
线性相关,并将?3用?1,?2线性表示.
1解:A?2111?13?7(5?a),当a=5时,?3??1??2.
7732a32
7. 作一个以(1,0,1,0)和(1,-1,0,0)为行向量的秩为4的方阵. 解:因向量(1,0,0,0)与(1,0,1,0)和(1,-1,0,0)线性无关,
所以(1,0,0,0)可作为方阵的一个行向量,因(1,0,0,1)与(1,0,1,0),(1,-1,0,0),(1,0,0,
?10?1?10)线性无关,所以(1,0,0,1)可作为方阵的一个行向量.所以方阵可为??10??10
10??00?. 00??01?8. 设?1,?2,?,?s的秩为r且其中每个向量都可经?1,?2,?,?r线性表出.证明:
?1,?2,?,?r为?1,?2,?,?s的一个极大线性无关组.
【证明】若 ?1,?2,?,?r (1) 线性相关,且不妨设
?1,?2,?,?t (t 是(1)的一个极大无关组,则显然(2)是?1,?2,?,?s的一个极大无关组,这与?1,?2,?,?s的秩为r矛盾,故?1,?2,?,?r必线性无关且为?1,?2,?,?s的一个极大无关组. 9. 求向量组?1=(1,1,1,k),?2=(1,1,k,1),?3=(1,2,1,1)的秩和一个极大无关组. 【解】把?1,?2,?3按列排成矩阵A,并对其施行初等变换. ?1?1A???1??k11??111??111??11?0??0??0k?112?0101?????????k1??0k?10??0k?10??00??????11??01?k1?k??001?k??001?0?? 1??0?当k=1时,?1,?2,?3的秩为2,?1,?3为其一极大无关组. 当k≠1时,?1,?2,?3线性无关,秩为3,极大无关组为其本身. 10. 确定向量?3?(2,a,b),使向量组?1?(1,1,0),?2?(1,1,1),?3与向量组?1=(0,1,1), ?2=(1,2,1),?3=(1,0,?1)的秩相同,且?3可由?1,?2,?3线性表出. 【解】由于 ?0A?(?1,?2,?3)??1???1?1B?(?1,?2,?3)??1???01211111??1??00????1????02??1??0a???b????00?;?1?1??00?? 12?,1b??0a?2??2而R(A)=2,要使R(A)=R(B)=2,需a?2=0,即a=2,又 a??0112??120?, c?(?1,?2,?3,?3)??120a???0112??????11?1b????000b?a?2??要使?3可由?1,?2,?3线性表出,需b?a+2=0,故a=2,b=0时满足题设要求,即?3=(2,2,0). 11. 求下列向量组的秩与一个极大线性无关组. (1) α1=(1,2,1,3),α2=(4,-1,-5,-6),α3=(1,-3,-4,-7); (2) α1=(6,4,1,-1,2),α2=(1,0,2,3,-4),α3=(1,4,-9,-6,22),α4=(7,1,0,-1,3); (3) α1=(1,-1,2,4),α2=(0,3,1,2),α3=(3,0,7,14),α4=(1,-1,2,0),α5=(2,1,5,6). 解:(1)把向量组作为列向量组成矩阵Α,应用初等行变换将Α化为最简形矩阵B,则 11???1 4 1??1 0 ???1 4 1??1 4 1??9???5??????2 ?1 ?30 ?9 ?550 1 ????????A??9???0 1 ??B ?1 ?5 ?4??0 ?9 ?5??9??0 0 0??????0 0 0???3 ?6 ?7??0 ?18 ?10??0 0 0?????0 0 0???可知:R(Α)=R(B)=2,B的第1,2列线性无关,由于Α的列向量组与B的对应的列向 量有相同的线性组合关系,故与B对应的Α的第1,2列线性无关,即α1,α2是该向量组的一个极大无关组. (2)同理, ? 6 1 1 7??0 -11 55 7??1 2 -9 0??????? 4 0 4 10 ?8 40 10 -11 55 7??????? 1 2 -9 0???1 2 -9 0???0 -8 40 1?????????1 3 -6 ?10 5 -15 -10 5 -15 -1??????? 2 ?4 22 3??0 ?8 40 1??0 0 0 0????????1 2 -9 0? ??7?0 1 -5 -??1 2 -9 0??1 0 0 0??11??0 1 -5 0??0 1 0 0?????45???0 0 0 -???0 0 10 0???0 0 1 0??B11??????0 0 0 10 0 0 1????24??????0 0 10 ???0 0 0 0??0 0 0 0?11???0 0 0 0???可知R(Α)=R(B)=4,Α的4个列向量线性无关,即α1,α2,α3,α4是该向量组的极大无关组. (3)同理, ?1 0 3 1 2??1 0 3 1 2??1 0 3 1 2??1 0 3 1 2?????????-1 3 0 -1 10 3 3 0 30 1 1 0 10 1 1 0 1??????????, A???2 1 7 2 5??0 1 1 0 1??0 0 0 -4 -4??0 0 0 1 1?????????4 2 14 0 60 2 2 -4 -20 0 0 0 00 0 0 0????????可知R(Α)=R(B)=3,取线性无关组α1,α3,α5为该向量组的一个极大无关组. 12.求下列向量组的一个极大无关组,并将其余向量用此极大无关组线性表示. (1) α1=(1,1,3,1),α2=(-1,1,-1,3),α3=(5,-2,8,-9),α4=(-1,3,1,7); (2) α1=(1,1,2,3),α2=(1,-1,1,1),α3=(1,3,3,5),α4=(4,-2,5,6),α5=(-3,-1,-5,-7). 解:(1)以向量组为列向量组成Α,应用初等行变换化为最简形式. 3???1 -1 5 -1??1 0 1??1 -1 5 -1??1 -1 5 -1??2??7???????1 1 -2 3??0 2 -7 4?7?0 1 - 2?A?????2???0 1 - 2??B, ?3 -1 8 1??0 2 -7 4??2??0 0 0 0??????0 0 0 0???1 3 -9 7??0 4 -14 8 ??0 0 0 0?????0 0 0 0???可知,α1,α2为向量组的一个极大无关组. ?x1?x2?5?x?x??237?12设α3=x1α1+x2α2,即?解得,x1?,x2?? 22?3x1?x2?8??x1?3x2??9 ?x1?x2??1?x?x?3?12设α4=x3α1+x4α2,即?解得,x1?1,x2?2 ?3x1?x2?1??x1?3x2?737所以a3?a1?a2,a4?a1?2a2. 22?1 1 1 4 -3??1 1 1 4 -3??1 0 2 1 -2???????1 -1 3 -2 -10 -2 2 -6 20 1 -1 3 -1????????B (2)同理, A???2 1 3 5 -5??0 -1 1 -3 1??0 0 0 0 0???????3 1 5 6 -70 -2 2 -6 20 0 0 0 0??????可知, α1、α2可作为Α的一个极大线性无关组,令α3=x1α1+x2α可得:?2 ?x1?x2?1即x1=2,x2=-1,令α4=x3α1+x4α2, ?x1?x2?3可得:??x1?x2?4即x1=1,x2=3,令α5=x5α1+x6α2, x?x??2?122 ?x1?x2??3可得:?即x1=-2,x2=-1,所以α3=2α1-α x?x??1?12α4=α1+3α2,α5=-2α1-α 2 13. 设向量组?1,?2,?,?m与?1,?2,?,?s秩相同且?1,?2,?,?m能经?1,?2,?,?s线性表出.证明?1,?2,?,?m与?1,?2,?,?s等价. 【解】设向量组 ?1,?2,?,?m (1) 与向量组 ?1,?2,?,?s (2) 的极大线性无关组分别为 ?1,?2,?,?r (3) 和 ?1,?2,?,?r (4) 由于(1)可由(2)线性表出,那么(1)也可由(4)线性表出,从而(3)可以由(4)线性表出,即 ?i??aij?jj?1r(i?1,2,?,r). 因(4)线性无关,故(3)线性无关的充分必要条件是|aij|≠0,可由(*)解出?j(j?1,2,?,r),即(4)可由(3)线性表出,从而它们等价,再由它们分别同(1),(2)等价,所以(1)和(2)等价. 14. 设向量组α1,α2,…,αs的秩为r1,向量组β1,β2,…,βt的秩为r2,向量组α1,α2,…,αs,β1,β2,…,βt的秩为r3,试证: max{r1,r2}≤r3≤r1+r2. 证明:设α s1,…, ?S为α1,α2,…,αs的一个极大线性无关组, βt1,βt2,…,?t为β1, r1r2β2,…,βt的一个极大线性无关组. μ1,…,?r3为α1, α2,…,αs,β1,β2,…,βt的一个极大线性无关组,则α s1, …,?Sr1和βt1,…,β tr2 可分别由μ1,…,?r3线性表示,所 s1, 以,r1≤r3,r2≤r3即max{r1,r2}≤r3,又μ1,…,?r3可由α …,αsr1,βt1,…,βtr2线性 表示及线性无关性可知:r3≤r1+r2. 15. 已知向量组α1=(1,a,a,a)′,α2=(a,1,a,a)′,α3=(a,a,1,a)′,α4=(a,a,a,1)′的秩为3,试确定a的值. 解:以向量组为列向量,组成矩阵A,用行初等变换化为最简形式: ?1 a a a??1 a a a??1?3a a a a???????a 1 a aa-1 1?a 0 00 1-a 0 0???????? ?a a 1 a??a-1 0 1-a 0??0 0 1-a 0???????a a a 1a-1 0 0 1-a0 0 0 1-a??????由秩A=3.可知a≠1,从而1+3a=0,即a=- 1. 3 16. 求下列矩阵的行向量组的一个极大线性无关组. ?25?75(1)??75??2543??1?09453132??; (2)??29454134???322048??13117121?215?1??. 03?13??104?1?2??1????2【解】(1) 矩阵的行向量组??的一个极大无关组为?1,?2,?3; ??3?????4? ??1????2(2) 矩阵的行向量组??的一个极大无关组为?1,?2,?4. ??3?????4?17. 集合V1={(x1,x2,?,xn)|x1,x2,?,xn∈R且x1?x2???xn=0}是否构成向量空间?为什么? 【解】由(0,0,…,0)∈V1知V1非空,设??(x1,x2,?,xn)?V1,??(y1,y2,?,yn)?V2,k?R)则 ????(x1?y1,x2?y2,?,xn?yn) k??(kx1,kx2,?,kxn).因为 (x1?y1)?(x2?y2)???(xn?yn)?(x1?x2???xn)?(y1?y2???yn)?0, kx1?kx2???kxn?k(x1?x2???xn)?0,所以????V1,k??V1,故V1是向量空间. 18. 试证:由?1?(1,1,0),?2?(1,0,1),?3?(0,1,1),生成的向量空间恰为R3. 【证明】把?1,?2,?3排成矩阵A=(?1,?2,?3),则 110A?101??2?0, 011所以?1,?2,?3线性无关,故?1,?2,?3是R3的一个基,因而?1,?2,?3生成的向量空间恰为R3. ?2?(1,1,1,2),?3?19. 求由向量?1?(1,2,1,0),生的向量空间的一组基及其维数. 【解】因为矩阵 (3,4,3,4),?4?(1,1,2,1),?5?(4,5,6,4)所 A?(?1,?2,?3,?4,?5)?1?2???1??01314??11314??11314??0?1?2?1?3??0?1?2?1?3? 1415????????,1326??00012??00012??????2414??02414??00000?∴?1,?2,?4是一组基,其维数是3维的. 20. 设?1?(1,1,0,0),?2?(1,0,1,1),?1?(2,?1,3,3),?2?(0,1,?1,?1),证明: L(?1,?2)?L(?1,?2). 【解】因为矩阵 A?(?1,?2,?1,?2)?1?1???0??010??112?0?1?30?11????13?1??000??13?1??00020? 1??,0??0?由此知向量组?1,?2与向量组?1,?2的秩都是2,并且向量组?1,?2可由向量组?1,?2线性表出.由习题15知这两向量组等价,从而?1,?2也可由?1,?2线性表出.所以 L(?1,?2)?L(?1,?2). 21. 在R3中求一个向量?,使它在下面两个基 (1)?1?(1,0,1),?2?(?1,0,0)?3?(0,1,1)(2)?1?(0,?1,1),?2?(1,?1,0)?3?(1,0,1)下有相同的坐标. 【解】设?在两组基下的坐标均为(x1,x2,x3),即 ?x1??x1???(?,?,?)?x?,??(?1,?2,?3)?x2123?2??????x3???x3?? ?1?10??x1??011??x1??001??x????1?10??x????2????2???101????101????x3????x3??即 ?1?2?1??x1??111??x??0, ???2???000????x3??求该齐次线性方程组得通解 x1?k,x2?2k,x3??3k (k为任意实数) 故 ??x1?1?x2?2?x3?3?(k,2k,?3k). 22. 验证?1?(1,?1,0),?2?(2,1,3),?3?(3,1,2)为R3的一个基,并把?1?(5,0,7), ?2?(?9,?8,?13)用这个基线性表示. 【解】设 A?(?1,?2,?3),B?(?1,?2), 又设 ?1?x11?1?x21?2?x31?3,?2?x12?1?x22?2?x32?3, 即 ?x11(?1,?2)?(?1,?2,?3)??x21??x31记作 B=AX. 则 x12?x22??, x32???1235(A?B)???1110???0327?1235?0327???002?2?9??12r2?r1?????03?8?????13???03?9??1作初等行变换???????0?13?????4???0354527001001?9?r2?r3?????17??r2?r3?13?? 23?3?3???1?2??因有A?E,故?1,?2,?3为R3的一个基,且 ?23?(?1,?2)?(?1,?2,?3)?3?3?, ?????1?2??即 ?1?2?1?3?2??3,?2?3?1?3?2?2?3. (B类) 1.A 2.B 3.C 4.D 5.a=2,b=4 6.abc≠0 7.设向量组α1,α2,α3线性相关,向量组α2,α3,α4线性无关,问: (1) α1能否由α2,α3线性表示?证明你的结论. (2) α4能否由α1,α2,α3线性表示?证明你的结论. 解:(1)由向量组α1,α2,α3线性相关,知向量组α1, α2, α3的秩小于等于2,而α2, α3, α4线性无关,所以α2, α3线性无关,故α2, α3是α1, α2, α3的极大线性无关组,所以α1能由α2, α3线性表示. (2)不能.若α4可由α1,α2,α3线性表示,而α2,α3是α1,α2,α3的极大线性无关组,所以α4可由α2,α3线性表示.与α2,α3,α4线性无关矛盾. 8.若α1,α2,…,αn,αn+1线性相关,但其中任意n个向量都线性无关,证明:必存在n+1个全不为零的数k1,k2,…,kn,kn+1,使 k1α1+k2α2+…+kn+1αn+1=0. 证明:因为α1,α2,…,αn,αn+1线性相关,所以存在不全为零的k1,k2,…,kn,kn+1使k1α1+k2α2+…+kn+1αn+1=0 若k1=0,则k2α2+…+kn+1αn+1=0,由任意n个向量都性线无关,则k2=…=kn+1=0,矛盾.从k1≠0,同理可知ki≠0,i=2, …,n+1,所以存在n+1个全不为零的数k1,k2,…,kn,kn+1,使k1a1+k2a2+…+kn+1an+1=0. 9. 设A是n×m矩阵,B是m×n矩阵,其中n<m,E为n阶单位矩阵.若AB=E,证明:B的列向量组线性无关. 证明:由第2章知识知,秩A≤n,秩B≤n,可由第2章小结所给矩阵秩的性质,n=秩E≤min{秩A,秩B}≤n,所以秩B=n,所以B的列向量的秩为n,即线性无关. 习题四 (A类) 1. 用消元法解下列方程组. ?x1?4x2?2x3?3x4?6,?x1?2x2?2x3?2,?2x?2x?4x?2,??124(1) ? (2) ?2x1?5x2?2x3?4, ?x?2x?4x?6;?3x1?2x2?2x3?3x4?1,23?1?x?2x?3x?3x?8;234?1【解】(1) ?1?2(A?b)??4?2206??14?111r2?42?2?????3?206?r4?r121?r2?r1?????r?3r3得 所以 (2) 解②?①×2得 ③?① 得 得同解方程组 ??322?31??322?31?32?123?38????123?38????14?236??0?32?1?5?(?1)?r2?r3??0?12?9?2????????0?25?62????14?236?4?23?01?2??16?921?292?r4?r3??0?32?1?5??????r3?3r20r4?2r2???00?4261???????0?25?62????001126????14?236???236??01?292??1401?292???001126?????r4?4r3???001126??,?00?4261????0007425????x1?4x2?2x3?3x4?6?? x2?2x3?9x4?2 x12x ?3?4?6?? 74x4?25??x1871??,?74?x?211?2,?74 ??x3?144?74,??x254?74.?① ?x1?2x2?2x3?2?2x1?5x2?2x3?4 ?② ?x1?2x2?4x3?6③ x2?2x3=0 2x3=4 ?x1?④ ?2x2?2x3?2? x2?2x3?0?? 2x3?4 ⑤ ⑥ 由⑥得 x3=2, 由⑤得 x2=2x3=4, 由④得 x1=2?2x3 ?2x2 = ?10, 得 (x1,x2,x3)T=(?10,4,2)T. 2. 求下列齐次线性方程组的基础解系. ? x1?3x2?2x3?0,?(1) ? x1?5x2? x3?0, (2) ?3x?5x?8x?0;23?1? x1? x2?5x3? x4?0,? x? x?2x?3x?0,?1234 ??3x1? x2?8x3? x4?0,?? x1?3x2?9x3?7x4?0;? x1?2x2?2x3?2x4? x5?0,?? x1?2x2? x3?3x4?2x5?0, ?2x?4x?7x? x? x?0.2345?1? x1? x2?2x3?2x4?7x5?0,?(3) ?2x1?3x2?4x3?5x4 ?0, (4) ?3x?5x?6x?8x ?0;234?1【解】(1) ?x1?3x2?2x3?0,??x1?5x2?x3?0, ?3x?5x?8x?0.23?1?132??132??132?r3?2r2r2?r1?02?1?????02?1? A??151??????r?3r??31????????358???0?42???000??得同解方程组 7?x??2x?3x??x3,32?12?x1?3x2?2x3?0???1 ?x?x,23?2x2?x3?0?2?x3?x3,?得基础解系为 T?7???2(2) 系数矩阵为 1?1?. 2? ?1?1A???3??1?1?0??0??0?11?13?12005?28?95?700?1??1?15?1??02?74?3?r3?r2r?r21???????????r4?2r23?3r1??1?r02?74r4?r1???7??04?148? ?1?4??r(A)?2.0??0?∴ 其基础解系含有4?R(A)?2个解向量. 3?x1???x3?x??2?2?x1?x2?5x3?x4?0???7?????x3?2???2x2?7x3?4x4?0?x3??x3???x?4???基础解系为 ??3??x4???2???1??????2?7????2x4?x3?x4?? ??2??0???1???????1??x4??0????3???2????7?,?2??1?????0?? (3) ??1???2???. ?0????1??1122A??2345???3568?112r3?2r2?????010???000得同解方程组 7??11227?r2?2r1?0101?14?????0??r3?3r1???0???0202?21?? 27?1?14??07???x1?x2?2x3?2x4?7x5?0,?x2?x4?14x5?0, ??7x5?0?x5?0.?取? ?x3??1??0????0?,?1?得基础解系为 x?4????? (?2,0,1,0,0)T,(?1,?1,0,1,0). (4) 方程的系数矩阵为 ?12?2A??12?1???24?7?12r3?3r2?????00???002?1??12?22?1?r2?r1?0011?1?????3?2??r3?2r1???11???00?3?33?? ?22?1?R(A)?2,11?1??000???x2??0??1??0??x???0?,?0?,?1?, ?4?????????1????0????0???x5???∴ 基础解系所含解向量为n?R(A)=5?2=3个 ?x2???取x4为自由未知量 ????x5???3???2???4??0??1??0???????得基础解系 ?1?,?0?,??1?. ???????0??0??1???1????0????0??3. 解下列非齐次线性方程组. ?x1?x2?2x3?1,?2x1?x2?x3?x4?1,?2x?x?2x?4,??123(1) ? (2) ?4x1?2x2?2x3?x4?2, ?2x?x?x?x?1;?x1?2x2?3,1234???4x1?x2?4x3?2;?x1?x2?x3?x4?x5?7,x?2x?x?x?1,?1234?3x?2x?x?x?3x??2,??12345(3) ?x1?2x2?x3?x4??1, (4) ? ?x?2x?x?x?5;?x2?2x3?2x4?6x5?23,234?1??5x1?4x2?3x3?3x4?x5?12.【解】 (1) 方程组的增广矩阵为 ?11?2?1(A?b)???1?2??41?11?0?3??00??00 21??11?0?324?r?2r21??????r3?r103?r4?4r1?0?3??42??0?321??1?01?r4?r3?22?2????????000????2?4??02?2?2?41?3001?2?r3?r2?????r4?r22???2? 21??22??12??00? 得同解方程组 x3?2,??x1?x2?2x3?1?2?2x3???3x?2x?2?x???2, ??223?3??x?23???x1?1?x2?2x3??1.(2) 方程组的增广矩阵为 ?21?111??21?111?r3?r1?000?10? (A?b)??42?212???????r2?2r1?????21?1?11???000?20??得同解方程组 ?2x1?x2?x3?x4?1,??x4?0 ?x4?0,???2x4?0,?即 ?2x1?x2?x3?1, ?x?0.?4令x1?x3?0得非齐次线性方程组的特解 xT=(0,1,0,0)T. 又分别取 ?x2??1??0??x???0?,?1? ?3?????得其导出组的基础解系为 TT1??1???,1,0,0?;??2?∴ 方程组的解为 1??2???,0,1,0?, ?2??1??1???0??2??2??1?????x????k1?1??k2?0?.?0??0??1????????0?0??????0??k1,k2?R 1??1?2111??1?211??r2?r1??000?2?2? (3) 1?21?1?1?????r3?r1????4??1?2115???0000?R(A)?R(A)∴ 方程组无解. ?1?0(A?b)???0??3?1?0??0??0?1?0??0??011?12110?111000?221?r3?r2?????r4?3r1(a?3)?2b??1a?1?110?221?r4?r2????? a?10b?1???2a?3?1?110?221??.a?10b?1??0a?10?11(i) 当a?1≠0时,R(A)=R(A)=4,方程组有惟一解. ?b?a?2??a?1?x?1???a?2b?3?x????2???a?1?. ?x3???b?1?????x4??a?1???0??(ii) 当a?1=0时,b≠?1时,方程组R(A)=2 (iii) 当a=1,b= ?1时,方程组有无穷解. 得同解方程组 ?x1?x2?x3?x4?0, ?x?2x?2x?1.34?2取 ?x1?x3?x4?1,?x??2x?2x?1,?234 ?x?x,33??x4?x4,?∴ 得方程组的解为 ?x1??1??1???1??x???2???2??1?2???k???k?????.?x3?1?1?2?0??0?????????x0???1??0??4??112???8. 设A?224,求一秩为2的3阶方阵B使AB=0. ????336??【解】设B=(b1 b2 b3),其中bi(i=1,2,3)为列向量, 由 k1,k2?R AB?0?A(b1?Abi?0?b1为Ax=0的解. b2b3)?0 (i?1,2,3)b3b2?112??x1?????求224x2=0的解.由 ??????336????x3???112??112?r2?2r1?000? A??224???????r3?3r1?????336???000??得同解方程组 ?x1??x2?2x3,? ?x2?x2,?x?x,33?∴ 其解为 ?x1???1???2??x??k?1??k?0?.?2?1??2??????0???1???x3??取 k1,k2?R ??1???2??0?b1??1?;b2??0?;b3??0?, ??????????0???1???0??则 ??1?20?B??100? ????010?? 9.已知?1,?2,?3是三元非齐次线性方程组Ax=b的解,且R(A)=1及 ?1??1??1??,?????1?,?????1?, ?1??2??03??2??13??????0???0???1??求方程组Ax=b的通解. 【解】Ax=b为三元非齐次线性方程组 R(A)=1?Ax=0的基础解系中含有3?R(A)=3?1=2个解向量. ?1?1??0?????1?,?1??3?(?1??2)?(?2??3)??0?1??????0????0?? ?1?1??0????0?,?1??2?(?1??3)?(?2??3)??1?1??????1?0????1??由?1,?2,?3为Ax=b的解??1??3,?1??2为Ax=0的解, 且(?1??3),(?1??2)线性无关??1??3,?1??2为Ax=0的基础解系. 又 ?1?1?(?1??2)?(?2??3)?(?1??3)?2?1??1??1??1??2? 1??1??1?????0?1?1??0?,2??2??2????????0???0???1???1??2?∴ 方程组Ax=b的解为 x??1?k1(?1??3)?k2(?1??2)?1??2??0??0?????0??k1??1??k2?0?.?????1???0?1????????2???2??3?????(1) ξ1=1,ξ2=0; ???????0???1?? k1,k2?R10. 求出一个齐次线性方程组,使它的基础解系由下列向量组成. ?1??2??1???2???3???2???????(2) ξ1=?0?,ξ2=?2?,ξ3=?1?. ??????35?????2??????1????3????2??【解】 ??2??3???ξ=?0?设齐次线性方程组为Ax=0 (1) ξ1=1??2?????0???1??由?1,?2为Ax=0的基础解系,可知 ??2??3??x1??x1???2k1?3k2????x???? x?k1?1??k2?0???xk221?????????????k2?0???1??????x3????x3???令 k1=x2 , k2=x3 ?Ax=0即为x1+2x2?3x3=0. ?121???2?3?2???(2) A(?1?2?3)=0?A的行向量为方程组为(x1x2x3x4x5)?021??0的解. ??352?????1?3?2??x1?2x2?3x4?x5?0??即?2x1?3x2?2x3?5x4?3x5?0的解为 ?x?2x?x?2x?2x?02345?1?1?203?1??1?203?1?r3?r1?2?325?3?????012?1?1? ?r?2r??21?????1?212?2???001?1?1??得基础解系为?1=(?5 ?1 1 1 0)T ?2=(?1 ?1 1 0 1)T A=?方程为 ??5?1110? ???1?1101???5x1?x2?x3?x4?0, ??x?x?x?x?0.?1235 ?x1?x2?a1?x?x?a2325??11. 证明:线性方程组?x3?x4?a3有解的充要条件是?ai?0. i?1?x?x?a4?45??x5?x1?a5【解】 ?1?0?A??0??0???1?1?0??0??0??0?1?0??0??0??0?1?0??0??0??0??11000?1100?1?11000?110000?11000?11000?110?10?110000?11000?11000?11000?110000?11000?11000?11000?11a1?a2??r2?r1a3??????a4?a5??a1?a2??r5?r2a3??????a4?a1?a5??a1??a2????a3??a4?a1?a2?a5??a1?a2?? a3??a4?5?ai??i?1?方程组有解的充要条件,即R(A)=4=R(A) ??ai?0得证. i?1512. 设?是非齐次线性方程组Ax=b的一个解,ξ1,ξ2,?,ξn?r是对应的齐次线性方程组的一个基础解系.证明 (1)?,ξ1,?,ξn?r线性无关; (2)?,?+ξ1,?,?+ξn?r线性无关. 【 证明】 ***** (1) ?,ξ1,?,ξn?r线性无关? *k?*?k1ξ1???kn?rξn?r?0成立, 当且仅当ki=0(i=1,2,…,n?r),k=0 A(k?*?k1ξ1???kn?rξn?r)?0?kA??k1Aξ1???kn?rAξn?r?0∵ξ1,ξ2,?,ξn?r为Ax=0的基础解系 * ?A?i?0(i?1,2,?,n?r) ?kA?*?0由于A?*?b?0 ?k?b?0?k?0.. 由于ξ1,ξ2,?,ξn?r为线性无关 k1ξ1?ξ2?k2???kn?r?ξn?r?0?ki?0∴?,ξ1,ξ2,?,ξn?1线性无关. (2) 证?,?+ξ1,?,?+ξn?r线性无关. ****(i?1,2,?,n?r) ?k?*?k1(?*?ξ1)???kn?r(?*?ξn?r)?0成立 当且仅当ki=0(i=1,2,…,n?r),且k=0 k?*?k1(?*?ξ1)???kn?r(?*?ξn?r)?0 即 (k?k1???kn?r)?*?k1ξ1???kn?rξn?r?0 由(1)可知,?,ξ1,?,ξn?1线性无关. 即有ki=0(i=1,2,…,n?r),且 *k?k1?kn?r?0?k?0 ∴?,?+ξ1,?,?+ξn?r线性无关. (B类) 1.B 2. C 3. D 4. C 5. t=?3 *** 6. R(A)=2;2;2 7. 设η1,η2,…,ηs是非齐次线性方程组Ax=b的一组解向量,如果c1η1+c2η2+…+csηs也是该方程组的一个解向量,则c1+c2+…+cs= . 解:因为η1, η2,…, ηs是Ax=b的一组解向量,则Aη1=b, Aη2=b,…, Aηs=b,又 C1η1+ C2η2+…+ Csηs也是Ax=b的一解向量,所以A(C1η1+…+ Csηs)=b,即C1Aη1+ CAη2+…+ CsAηs=b,即C1b+ C2b+…+ Csb=b,(C1+…+Cs)b=b,所以C1+…+ Cs=1. 8. 设向量组?1=(1,0,2,3),?2=(1,1,3,5),?3=(1,?1,a+2,1),?4=(1,2,4,a+8),?=(1,1,b+3,5) 问:(1) a,b为何值时,?不能由?1,?2,?3,?4线性表出? (2) a,b为何值时,?可由?1,?2,?3, ?4惟一地线性表出?并写出该表出式. (3) a,b为何值时,?可由?1,?2,?3,?4线性表出,且该表出不惟一?并写出该表出式. 【解】 ??x1?1?x2?2?x3?3?x4?4 (*) ?1?0A?(A?b)???2??3?1?0??0??0113511121??121?r3?2r1?????a?24b?3?r4?3r1?1a?85?111??1?0?121?r?r32??????r4?2r2?0a2b?1????2a?52??0111?1?121??0a?10b??00a?10?111 (1) ?不能由?1,?2,?3,?4线性表出?方程组(*)无解,即a+1=0,且b≠0.即a=?1,且b≠0. (2) ?可由?1,?2,?3,?4惟一地线性表出?方程组(*)有惟一解,即a+1≠0,即a≠?1. (*) 等价于方程组 ?x1?x2?x3?x4?1?x?x?2x?1?234??(a?1)x3?b??(a?1)x4?0bba?b?1 ?x4?0x3?x2?x3?1??1?a?1a?1a?1b2b?b?x1?1???0???1??a?1?a?1?a?12ba?b?1b?????1??2??3a?1a?1a?1(3) ?可由?1,?2,?3,?4线性表出,且表出不惟一?方程组(*)有无数解,即有 a+1=0,b=0?a=?1,b=0. ?x1?k2?2k1??x1?x2?x3?x4?1?x2?k1?2k2?1方程组(*)?? ??x?x?2x?1x3?k14?23??x4?k2?k1,k2,k3,k4为常数. ∴??(k2?2k1)?1?(k1?2k2?1)?2?k1?3?k2?4 9. 设有下列线性方程组(Ⅰ)和(Ⅱ) ?x1?x2?2x4??6?x1?mx2?x3?x4??5??(Ⅰ)?4x1?x2?x3?x4?1 (Ⅱ) ?nx2?x3?2x4??11 ?3x?x?x?3?x3?2x4?1?t123??(1) 求方程组(Ⅰ)的通解; (2) 当方程组(Ⅱ)中的参数m,n,t为何值时,(Ⅰ)与(Ⅱ)同解? 解:(1)对方程组(Ⅰ)的增广矩阵进行行初等变换 ?110?2?6??1?4?1?1?11???0????3??3?1?10???0?1 ???0??010?5?1?4?100?110?101?2?2?6?725???621???2??4???5???110?2?6??00?125?????010?1?4?? 由此可知系数矩阵和增广矩阵的秩都为3,故有解.由方程组 ?x1?x4?0??x2?x4?0 (*) ??x3?2x4?0得方程组(*)的基础解系 ??1???1??1?? ?2??1?????2??4?令x4?0,得方程组(Ⅰ)的特解 ????? ??5??0??于是方程组(Ⅰ)的通解为x???k?,k为任意常数。 (2) 方程组(Ⅱ)的增广矩阵为 ??1m?1?1?5???44m?3n0012?t??0n?1?2?11??21?t???0n0?4?10?t? ?001?????001?21?t???系数矩阵与增广矩阵的秩均为3,令 ??4x1?(4m?3n)x2?0?nx2?4x4?0 ??x3?2x4?0方程组(**)的基础解系为 ??3?m??4n?当n?0时,??????m??4??1??1?n?,当n?0时,?2??? ?2????0???0??1???方程组(Ⅱ)与方程组(Ⅰ)同解,则n?0,故有 ??3??m?1?4n ? ??4?m?2 ??n?4?n?1把m,n代入方程组,同时有 1?t??5,即t = 6. 也就是说当m=2,n=4,t=6时,方程组(Ⅱ)与方程组(Ⅰ)同解. (**) 10. 设四元齐次线性方程组(Ⅰ)为??x1?x2?0,又已知某齐次线性方程组(Ⅱ)的通解为 ?x2?x4?0,k1(0,1,1,0)′+k2(-1,2,2,1)′. (1) 求齐次线性方程组(Ⅰ)的基础解系; (2) 问方程组(Ⅰ)和(Ⅱ)是否有非零公共解?若有,则求出所有的非零公共解;若没有,则说明理由. ??1??? 1?x1?x2?0?x1??x2解:(1)由?,所以?,以x2,x3为自由未知数可得基础解系?1???, ? 0??x2?x4?0?x4?x2??? 1??0???0?2???. ?1????0??0???1???1??0?????????1 2 10?????????k?k?k(2)令k1,则可得: ?1?2? 2?3? 0?4?1?????????0 1 1???????0???k2??k3?k3?k2???k1?2k2?k3,即?k1??k2 ??k?k?k1?2k2?k42?4?k?k3?2?0???1???1???????1 2 1所以有公共解k1???k2???k?? ?k?k2??k1? ?1?? 2?? 1???????0 1????? 1? 习题五 (A类) 1. 计算??,??. (1)??(?1,0,3,?5),??(4,?2,0,1);?313??3? 2(2)???,?,,?1?,????,?2,3,?.343??2??2【解】 (1)??,???(?1)?4?0?(?2)?3?0?(?5)?1??9 32?????1?(2)??,?????3???3??(?2)?????3???(?1)?043?3??2??2?2. 把下列向量单位化. (1) ?=(3,0,-1,4); (2) 【解】 ?=(5,1,-2,0). (1)e?aaa??a,a??32?02?(?1)2?42?26 ?e?(2)a?e?aa?14?1?3,0,,?(3,0,?1,4)???;26262626???a,a??25?1?4?0?30?1?2?1?5,,,0?.(5,1,?2,0)??30?303030? 3. 在R4中求一个单位向量,使它与以下三个向量都正交:α1=(1,1,-1,1), α2=(1,-1,-1,1), α3=(2,1,1,3). 解:设向量a=(x1,x2,x3,x4)与a1,a2,a3都正交,则 ?x1?x2?x3?x4?0?x1?4x3??令x3=1得a=(4,0,1,-3) ?x1?x2?x3?x4?0 得:?x2?0?2x?x?x?3x?0?x??3x34?123?4单位化可得单位向量为?1(4,0,1,?3). 26 4. 利用施密特正交化方法把下列向量组正交化. (1) ?1 =(0,1,1)′, ?2 =(1,1,0)′, ?3 =(1,0,1)′; (2) ?1 =(1,0,?1,1), ?2 =(1,?1,0,1), ?3 =(?1,1,1,0) 【解】 (1)?1??1?(0,1,1)?,?2,?1??111????2??2??1?(1,1,0)??(0,1,1)???1,,??,2?22???1,?1??3,?1??3,?2????222???3??3??1??2??,?,?;?333???1,?1???2,?2?(2)?1??1?(1,0,?1,1)?, ?2,?1??221???1???2??2??1?(1,?1,0,1)?(1,0,?1,1)??,?1,,?,3?,?33??3?11??3,?1??3,?2????1334???3??3??1??2???,,,?.?5555???1,?1???2,?2?5. 试证,若n维向量?与?正交,则对于任意实数k,l,有k?与l?正交. 【证】?与?正交???,???0. ?k,l?R.∴ k?与l?正交. 6. 下列矩阵是否为正交矩阵. (k?,l?)?kl??,???0 ??1?1(1)???2??1??3?121121??1013???2?10?11?;(2)?2?0102???0?10??1??0?0??. 1??1?【解】 (1) A′A≠E, ∴A不是正交矩阵 (2) A′A=E?A为正交矩阵 7. 设x为n维列向量,x′x=1,令H=E-2xx′.求证H是对称的正交矩阵. 【证】 H?E?2xx?H?(E?2xx?)??E??2(xx?)??E?2(xx?)?H∴ H为对称矩阵. H?H?(E?2xx?)(E?2xx?)?E2?2E(xx?)?2(xx?)E?4(xx?)(xx?) ?E2?4(xx?)?4x(xx?)x??E∴ H是对称正交矩阵. 8. 设A与B都是n阶正交矩阵,证明AB也是正交矩阵. 【证】A与B为n阶正交矩阵?A′A=EB′B=E (AB)(AB)′=AB·(B′A′)=A(BB′)A′=AEA′=AA′=E ∴ AB也是正交矩阵. 9. 判断下列命题是否正确. (1) 满足Ax=?x的x一定是A的特征向量; (2) 如果x1,…,xr是矩阵A对应于特征值?的特征向量.则k1x1+k2x2+…+krxr也是A对应于?的特征向量; (3) 实矩阵的特征值一定是实数. 【解】 (1) ╳.Ax=?x,其中当x=0时成立,但x=0不是A的特征向量. ?1???1?????(2) ╳.例如:E3×3x=?x特征值?=1, ?的特征向量有2,?2 ??????3?????3???1???1??0??0?????????则2??2?0,0不是E3×3的特征向量. ??????????3?????3????0????0??(3) ╳.不一定.实对称矩阵的特征值一定是实数. 10. 求下列矩阵的特征值和特征向量. ?2?3?(1)?,???31??2?20?(3)??21?2?,????0?20??【解】(1) ?6(2)?2???4?2?0(4)??0??024?,32??26??3?1?4? ?1?21??.12?2??112?3?E?A???当????23??1?(??2)(??1)?9?0??2?3??7?0 3?37.23?37时, 2??1?37?2(?E?A)x?0为??3????37?1?3??x1????x1?? ????0得解???6?1?37??x2??x2??1?????2?对应的特征向量为 ?37?1???,k??6??1????k?R且k?0. ??1?37?3?372当??时, ?2?3???3??x???1??0 1?37??x2??2??37?1??37?1????其基础解系为?1??,1?,对应的特征向量为k??6?,k?R且k?0. ?6???1??6?? 23??2426???6??2023??42 (2)A??E?24?4?2?2???(??2)2(??11)?0,∴ 特征值为 ?1??2?2,(i) 当?1??2?2时, ?3?11. ?424??x1??212??x??0?2x?x?2x?0, 123???2???424????x3??其基础解系为 ?1???2???,?1??0???∴ 对应于?=2的特征向量为 ??1??0?. ????1???1???1???2?k1???k2?0????1???1???0???(ii)当?3?11时, k1,k2?R且使得特征向量不为0. ??524??x1??2?82??x??0, ???2???42?5????x3??解得方程组的基础解系为 1???(x1,x2,x3)??1,,1?. ?2?T?1?∴ 对应于?3?11的特征向量为k?1,,1?,k?R且k?0. ?2?2??(3)A??E??20?21???20?2??(??2)(??4)(??1)?0 0??T?特征值为?1??2,?2?4,?3?1. (i) 当?1??2时, ?4?20??x1???23?2??x??0 ???2???0?22????x3???1??.得基础解系为?,11,? ?2??1??2??1??2对应的特征向量为k????1??1???(ii) 当?2?4时, k?R且k?0. ??2?20??x1???2?3?2??x??0 ???2???0?2?4????x3??其基础解系为(2,?2,1)′, ?2???所以与?2?4对应的特征向量为k??2,????1??(iii) 当?3?1时, k?R且k?0. ?1?20??x1??0???20?2??x???0? ???2?????0?2?1????0???x3???其基础解系为(2,1,?2)′ ?2???∴ 与?3?1对应的特征向量为k?1,?????2??k?R且k?0. 2??(4)A??E?000∴ A的特征值为1,2. (i) 当?1??2??3?1时, 3?1??11?1?22??1?41?22???(2??)??1??11?22??11?22?? ??(??1)3?(2??)?0??1??2??3?1,?4?2?13?1?4??x1??0??0?2?21??x??0????2???? ?011?2??x3??0???????0111???x4??0?其基础解系为(4,?1,1,0)′. ∴ 其对应的特征向量为k·(4,?1,1,0)T,k∈R且k≠0. (ii) 当?4?2时, ?03?1?4??x1??0??0?3?21??x??0????2????, ?010?2??x3??0???????0110???x4??0?其基础解系为:(1,0,0,0)′. ∴ 其对应的特征向量为 ?1??0?k???,?0????0??1?x1??2?,????2?? k?R且k?0. 11.设3阶方阵A的特征值为λ1=1,λ2=0,λ3=-1,对应的特征向量依次为 ?2?x2???2?,????1????2?x3???1?, ????2?? 求矩阵A. 【解】 Ax1??1x1,Ax2??2x2,Ax3??3x30? ?0??3????10?A(x1,x2,x3)?(?1x1,?2x2,?3x3)?(x1,x2,x3)??0?2??00由于?1?1,?2?0,?3??1为不同的特征值?x1,x2,x3线性无关,则有 ?12?2?(x1,x2,x3)??2?2?1?可逆 ???2??21??12?2??100??12?2???102?1?A??2?2?1??000??2?2?1???012?. ???????3???2?2??21???00?1????21??220??12. 设3阶实对称矩阵A的特征值为-1,1,1,与特征值-1对应的特征向量x=(-1, 1,1)′,求A. 【解】?1??1对应的特征向量为x1=(?1,1,1)T,设?2?1对应的特征向量为x2=(x1,x2,x3)T,A为实对称矩阵,所以(x1,x2)=0,即有?x1+x2+x3=0. 得方程组的基础解系为 ?1?1??,?1??1????0??可知?1,?2为?2?1对应的特征向量. 将x1,?1,?2正交化得 ?1??, ?2??0????1?????111?,,?1?x1=(?1,1,1)T, 单位化:e1?1??; ?1?333????11?,,0?; ?2??1 =(1,1,0)T, e2?2???2?22?T?3,?1??3,?2????11???3??3??1??2??,?,1?,e3???22???1,?1???2,?2??TT112?,?,?. 666?T ??1?3??1P??3??1??3??1?3??1?A??3??1??3121201?6????100?1??1?010?. ?PAP? 则有 ??6????001??2?6??121201???1?36????100??1????1?010??36?????001???2??1?6???3121201??1??36??1??2???36???2??2??36???123132?32?3??2??. 3??1?3??13. 若n阶方阵满足A2=A,则称A为幂等矩阵,试证,幂等矩阵的特征值只可能是1或者 是零. 【证明】设幂等矩阵的特征值为?,其对应的特征向量为x. Ax??x;A(Ax)?A(?x)?A2x??Ax??2x;由A2=A可知?Ax??x??x??x; 所以有??????0或者?=1. 222 14. 若A2=E,则A的特征值只可能是±1. 【证明】设?是A的特征值,x是对应的特征向量. 则Ax=?x A2x=?(Ax)=?2x 由A2=E可知 x=Ex=A2x=?2x ?(?2?1)x=0, 由于x为?的特征向量,∴ x≠0??2?1=0??=±1. 15. 设λ1,λ2是n阶矩阵A的两个不同的特征根,?1,?2分别是A的属于λ1, λ2的特征向量,证明?1+?2不是A的特征向量. 证明:假设?1+?2是A的属于特征根λ的特征向量,则 A(?1+?2)=λ(?1+?2)=λ?1+λ?2. 又 A(?1+?2)= A?1+ A ?2=λ1?1+λ2?2 于是有 (λ?λ1)?1+(λ?λ2)?2 =0 由于?1??2,?1与?2线性无关,故λ?λ1=λ?λ2=0. 从而?1??2与?1??2矛盾,故?1+?2不是A的特征向量. ??200???100?????16. 设矩阵A?2x2与B?020相似. ???????211???00y?? (1) 求x与y; - (2) 求可逆矩阵P,使P1AP=B. 【解】(1)由A~B可知,A有特征值为?1,2,y. ??2???A??E??2??2由于?1为A的特征值,可知 0???(2??)??(x??)(1??)?2??0 x??2??11????0A+E??(2?1)?(x?1)2?2??0?x?0. 将x=0代入|A??E|中可得 A??E??(2??)?(??)(1??)?2??0 ??(2??)(??2)(??1)?0,??1??2,?2?2,?3??1可知y= ?2. (2) (i) 当?1=?1时, ??100??x1??212??x??0 ???2???212????x3??其基础解系为 ?1=(0,?2,1)T, ?1= ?1对应的特征向量为 ?1=(0,?2,1)T. (ii) 当?2=2时, ??400??x1??0??2?22??x???0? ???2?????21?1????0???x3???其基础解系为 ?2=(0,1,1)T 所以?2=2对应的特征向量为 ?2=(0,1,1)T (ⅲ) 当?3=?2时, ?000??x1??0??222??x???0?, ???2?????213????0???x3???其基础解系为 ?3=(?2,1,1)T, 取可逆矩阵 ?00?2?p?(?1,?2,?3)???211? ????111??则 p?1AP?B. ?11?1???, 求A100. 17. 设A?001????0?23??【解】 1??A??E?001???11?(1??)(??1)(??2)?0 ?23???特征值为?1??2?1,?3?2. (i) 当?1??2?1时, ?01?1??x1??0??0?11??x???0? ???2?????0?22????0???x3???其基础解系为 ?0??1?,????1??(ii) 当?3?2时, ?1??1?. ????1????11?1??x1??0?21??x??0 ???2???0?21????x3??其基础解系为(?1,1,2)T. ?01?1??1????1?1? 令p?111,则pAP???????2??112???? (p?1AP)100?1??p?1A100P??1????2????A100100?01?1??1???111??1??????2??112?????100100?01?1??111?????112???1 ?01?1??1???111??1??????2??112???????13?2??1?11?????0?11???12100?11?2100?????02?21002100?1?.100?2100?1??02?2?18. 求正交矩阵T,使T -1 AT为对角矩阵. ?0?22?(1)A???2?34?,????24?3???410?1??14?10??,(3)A???0?141?????1014?【解】 ?124?(2)A??2?22?,????421???3?20?(4)A???22?2?.????0?21?? ???224?3????(??1)2(??8)?0 (1)A??E??2?3??24??1??2?1,?3??8.(i)当?1??2?1时, ??1?22??x1??0???2?44??x???0? ???2?????24?4????0???x3???方程组的基础解系为 (?2,1,0)T, (2,0,1)T. (ii) 当?3??8时, ?8?22??x1???254??x??0 ???2??45??2???x3???1?其基础解系为??,?1,1?. ?2?T??122??1?取?1??,?1,1?,单位化为,p1?1???,?,? ?1?333??2?取?2??2,1,0?,取?3??2,0,1?,使a2,a3正交化. T?,???24??32T???,,1?, 令?2??2?(?2,1,0),?3??3??55???2,?2?2?TTTT单位化 ??????21?,,0?,p3?3??25,45,5? p2?2???3?15153??2?55??1???3?T???2?3??2?3?1??(2)A??E?24得?1??2??3,?3?6.. (i) 当?1??2??3时, TT?2515025??15?45?. ?15??5??3?2?2??2421????(??3)2(??6) ?424??x1??0??212??x???0?, ???2?????424????0???x3???其基础解系为 ??1???1??2????0?? ??1??. ?2??0????1?? 正交化得 ?1?(?1,2,0)单位化得 TT?2,?1???42?,?2??2??1???,?,1?, ?55???1,?1???????12?,,0?,p2?2???45,?25,5?. p1?1???1?55??2?15153?TT (ii) 当?3?6时, 其基础解系为 单位化得 4??1(3)A??E?14??0?1?12?1??2?4,?3?2,?4?6.(i) 当?1??2?4时, 其基础解系为 ???524??x1??0??2?82??x???0?,?5???2??? ?42????x3????0?? ?3=(2,1,2)T. p33????1?2,1,2?T, 33??2?1?35?45?15??T???12?3?5?25?15?. ??2??305?3???0?1?1024??1?(4??)?(?2?8??12)?0 14????010?1??x1??10?10?????x?2??0?0?101??x 3???1010????x?4? ?1??0??1???,?1????0?由于(?1,?2)=0,所以?1,?2正交. 将它们单位化得 ?0??1??2???. ?0????1?????p1?????? (ii) 当?3?2时, 1?2??0?,1??2?0??????p2??????0?1??2?. 0??1??2??210?1??x1??0??12?10??x??0????2????, ?0?121??x3??0????????1012???x4??0?其基础解系为?3=(1,?1,?1,1)T, 单位化得 p3??3?3?1??2?????1??2????. ??1??2??1????2?(iii) 当?4?6时, ??210?1??x1??0??1?2?10??x??0????2????, ?0?1?21??x3??0????????101?2???x4??0?其基础解系为?4=(?1,?1,1,1)T, 单位化为 ?1111?p4?4????,?,,?, ?4?2222?T??101?1??222???01?1??T???2?122??4?11??0?1?,T?1AT??4??222??2??1????011222??3???20(4)A??E??22???2?(??2)(??5)(??1)?0,0?21?? ?1?2,?2?5,?3??1,(i) 当?1=2时, ??1?20??x1???20?2??x???0?2?1???2?0 ????x3??其基础解系为?T1=(2,1,?2), 单位化得 ??212?Tp1?1???3,3,?3??, 1?(ii) 当?2=5时, ???2?20??x1???2?3?2??x??0?2?4???2??0 ????x3??其基础解系为?T2=(2,?2,1). 单位化得 Tp?????23,?23,1?2?23??. 2?(iii) 当?3=?1时, ????. 6?? ?4?20??x1??0???23?2??x???0?, ???2?????0?22????0???x3???其基础解系为?3=(1,2,2)T, 单位化得 ??122?p3?3??,,?, ?3?333?得正交阵 T?2?3?1T???3???2?3?232?3131?3??2?,?3?2??3? ?2??. T?1AT??5????1???19.将下列二次型用矩阵形式表示. (1) f(x1,x2,x3)?x1?2x2?5x3?2x1x2?6x2x3?2x3x1; (2) f(x1,x2,x3,x4)?x1x2?x2x3?x3x4?x4x1; (3) f(x1,x2,x3,x4)?6x1?3x1x2?2x1x3?5x1x4?2x2?x2x4. 【解】 22222?111??x1?????(1)f(x1,x2,x3)?(x1,x2,x3)1?23x2; ??????135????x3????0??1?2(2) f(x1,x2,x3,x4)?(x1,x2,x3,x4)??0??1??21201200120121?2???x?1?0????x2?; ???1?x3??2??x4??0?? 3?6?2??32(3) f(x1,x2,x3,x4)?(x1,x2,x3,x4)?2??10??51???22 20.写出下列各二次型的矩阵. ?10005?2??x??11?????x2?. 2???x30???x???4?0??(1)f(x1,x2,x3)?x1?4x2?x3?4x1x2?2x1x3?4x2x3; (2) f(x1,x2,x3)?x1?x2?7x3?2x1x2?4x2x3. 222222?1 2 1??x1?????解:(1)由f(x1,x2,x3)?(x1,x2,x3)?2 4 2??x2? ?1 2 1??x????3??1 2 1???所以二次型的矩阵为?2 4 2? ?1 2 1?????1 ?1 0??x1?????(2)由f(x1,x2,x3)?(x1,x2,x3)??1 1 ?2??x2? ? 0 ?2 ?7??x????3???1 ?1 0???所以二次型的矩阵为??1 1 ?2? ? 0 ?2 ?7??? 21. 当t为何值时,二次型f(x1,x2,x3)?x1?6x1x2?4x1x3?x2?2x2x3?tx3的秩为2. 222?132??x1?????【解】 f(x1,x2,x3)?(x1,x2,x3)311x2 ??????21t????x3??2??132??131?r2?????r3?2r1?0?8?5?????8?A??311?????? r?3r??21?????21t???0?5t?4?? 2?13?2??13??5???5?r3?5r2?01?01?????8? 8???25??0?5t?4??00t?4????8??7?R(A)?2?t?. 8 22.用正交变换把下列二次型化为标准形,并写出所作的变换. (1) f(x1,x2,x3,x4)=2x1x2-2x3x4; (2) f(x1,x2,x3)?x1?2x2?3x3?4x1x2?4x2x3. 222?0 1 0 0???1 0 0 0?,A的特征多项式 解:(1)f的正交矩阵A???0 0 0 ?1????0 0 ?1 0??? 1 0 0A??E? 1 ?? 0 0 0 0 ?? ?1 0 0 ?1 ??于是A的全部特征值为?1?1(二重),?2??1(二重). ?(?2?1)2?0, ?1?? 0?????1 0??????,???1?1,解(A-E)x=0,得基础解系1 ?0?2??1?????0??? 1??1?? 0??2????1?? 0? 0????????1? 0??1??1???,?2?,??正交化得?1?,单位化得?1???? ?0???1?2?2?2?????????0??1??0?? 1??0?? ?2??????1??0????? 10????,???2??1解(A+E)x=0,得基础解系?3? ? 0?4?1????? 0???1? ?1??0????????1??0?20???????? 10?1??1?,??正交化 ?3???,?4???,单位化得:?3?? ? 0??1?2?4?2????????? 0???1?? 0??1?? 0??????2?1?1? 0 ? 0?2?2??1?1? 0 0?2?2? 取正交矩阵T?(?1,?2,?3,?4)??11??0 ? 0 ?22????0 1 0 1???22??令x=Ty,得f?x?Ax?y?(T?AT)y?y1?y2?y3?y3 2222? 1 ?2 0???(2)f的矩阵A???2 2 ?2? ? 0 ?2 3???特征多项式 A??E?0,得特征值?1??1,?2?2,?3?5 ?2???当?1??1,解(A+E)x=0,得基础解系?1??2? ?1?????2???当?2?2,解(A-2E)x=0,得基础解系?2?? 1? ? 2????1?? 2???当?3?5,解(A-5E)x=0,得基础解系?3???1? ? 1??????2??2??1???3??3?? 3???????212单位化得?1???,?2?? ?,?3???? ?3??3??3???????122??????? ??? ???3??3??3? 取正交矩阵T?(?1,?2,?3),令x=Ty,得 22f?x?Ax?y?(T?AT)y??y12?2y2?5y3 23. 用配方法把下列二次型化为标准型,并求所作变换. 22(1)f(x1,x2,x3)?2x1x2?4x1x3?x2?8x3;(2)f(x1,x2,x3)?2x1x2?4x1x3.【解】 22(1)f(x1,x2,x3)?2x1x2?4x1x3?x2?8x3; 22 ??(x2?2x1x2?x12)?x12?4x1x3?8x32??(x2?x1)2?(x1?2x3)2?12x3令 ?y1??x1?x2?x1?y2?2y3??y?x?2x??2?x2?y1?y2?2y3 13?y?x?x3?y333???110由于102?0 010∴ 上面交换为可逆变换. 得f(x1,x2,x3)??y1?y2?12y3. 222(2)f(x1,x2,x3)?2x1x2?4x1x3. x1x2?y??12?2,?x1?y1?y2??x1x2 令?x2?y1?y2为可逆线性变换??y??,?x?y?2223?3??y3?x3.22f(x1,x2,x3)?2(y12?y2)?4(y1?y2)y3?2y12?2y2?4y1y3?4y2y3222?2(y12?2y1y3?y3)?2y2?4y2y3?2y3 ?2(y1?y3)2?2(y2?y3)2??1?y1?y3?22令??2?y2?y3为可逆线性交换f(?1,?2,?3)?2?1?2?2 ???y33?所作线性交换为
- exercise2
- 铅锌矿详查地质设计 - 图文
- 厨余垃圾、餐厨垃圾堆肥系统设计方案
- 陈明珠开题报告
- 化工原理精选例题
- 政府形象宣传册营销案例
- 小学一至三年级语文阅读专项练习题
- 2014.民诉 期末考试 复习题
- 巅峰智业 - 做好顶层设计对建设城市的重要意义
- (三起)冀教版三年级英语上册Unit4 Lesson24练习题及答案
- 2017年实心轮胎现状及发展趋势分析(目录)
- 基于GIS的农用地定级技术研究定稿
- 2017-2022年中国医疗保健市场调查与市场前景预测报告(目录) - 图文
- 作业
- OFDM技术仿真(MATLAB代码) - 图文
- Android工程师笔试题及答案
- 生命密码联合密码
- 空间地上权若干法律问题探究
- 江苏学业水平测试《机械基础》模拟试题
- 选课走班实施方案
- 北京邮电大学出版社
- 线性代数
- 习题
- 主编
- 答案
- 戴斌祥
- 管理模拟题1
- 导游综合知识测试
- 让诗香满校园
- 哈尔滨工业大学2012matlab考查题答案
- 数值分析教案 - 图文
- 假如我是一线社工
- 初中英语课堂提问的有效方法
- 2018年供应商大会欢迎晚宴主持人串词-word范文(2页)
- 八年级下册历史中国人民站起来了说课稿
- 人生只如初见青年教师发言稿
- 借题发挥,让阅读教学更有效
- 马克思主义基本原理概论读书笔记
- 噶米高等数学函数、极限、无穷小、连续性
- 修改病句练习及答案
- 嘉力丰学院:糯米胶发展进化史 - 图文
- 建筑工程管理中的成本控制分析
- 2019年中考数学模拟试题及答案(3)
- U8界面按钮二次开发手册
- ANSYS单元生死总结2
- 山东省高等教育评估所成功举办2012年全国大学排行榜新闻发布会doc - 图文