简明线性代数(邓小成)第一章习题答案

-1-习题1-11．已知两个线性变换112212331232,232,45,xyyxyyyxyyy1122133233,2,3,yzzyzzyzz求从123,,zzz到123,,xxx的线性变换.解:112233210232415xyxyxy123210310232201415013zzz123421124910116zzz,所求为11232123312342,1249,1016.xzzzxzzzxzzz2．设1357A,3479B,32CAB,求21nC.解:133432325779CAB3968152114183113,23131100101313010CI,21231()(10)101013nnnnnCCCICC.3．设(1,2,3)a,TAaa,求nA.解一:T121233Aaa123246369,21231231428422462462856843693694284126A1231424614369A,3222(14)1414AAAAAAA,……,111231414246369nnnAA.解二:T121233Aaa123246369,2TTTTTT()()()(14)1414AaaaaaaaaaaaaA,-2-3222(14)1414AAAAAAA,……,111231414246369nnnAA.4．设100010001000aaaaA,求nA.解:将矩阵A写成aAIB,其中0100001000010000B.对于矩阵B,有20010000100000000B,30001000000000000B,4BO.显然B与aI可交换,所以0()()nnnrnrrnraCaAIBIB11222333nnnnnnnaCaCaCaIBBB12312111(1)(1)(2)2610(1)200000nnnnnnnnnnanannannnaanannaanaa.5．求与1101A可交换的所有矩阵.解:若矩阵B与A可交换,即ABBA,则B为2阶方阵.设abcdB,则1101abacbdcdcdAB,1101abaabcdccdBA.由ABBA得acbdaabcdccd,解得0,cda.因此-3-0abaB,其中,ab为任意数.6．设2112A,2()43fxxx,98()gxxx,计算()fA和()gA.解:2()43fAAAI221211043121201548430454803O,988()(1)gxxxxx888(1)(3)3(1)xxx7780()33(1)nnnfxxx,7780()()33()nnngfAAAAI83()AI811311.7．设,AB为n阶对称矩阵,证明:AB为对称矩阵的充分必要条件是ABBA.证明:因,AB为对称阵,所以TAA,TBB,于是TTT()ABBABA.因此,T()ABAB的充要条件是ABBA.也就是说,AB为对称阵的充要条件是ABBA.8．举反例说明下列命题是错误的:(1)若2AO,则AO;(2)若2AA,则AO或AI;(3)若AXAY,且AO,则XY.解:(1)对于矩阵0100A,有2AO,但AO.(2)对于矩阵1000A,有2AA,但AO且AI.(3)对于矩阵0100A,0100X,YO,有AXAY,且AO,但XY.习题1-21．计算下列行列式:(1)abacaebdcddebfcfef;解:原式111111111abcdef111002020abcdef111020002abcdef4abcdef.(2)xyxyyxyxxyxy.-4-解:原式222222xyyxyxyxyxxyxy2200xyyxyxyxyx(22)xyxyxyx2233(22)()22xyxxyyxy.2．设1111222233334444xabcxabcxabcxabcA,1111222233334444yabcyabcyabcyabcB,且已知||4,||1AB,试求||AB.解:11111222223333344444222222||222222xyabcxyabcxyabcxyabcAB111112222233333444448xyabcxyabcxyabcxyabc8(||||)40AB.3．计算下列行列式:(1)1357153323851389;解:1357022403290032D1357011220329003213570112200515003213570112100013003213570112100013000770.(2)2240413531232051.解:原式602104135701220516210712251806712330986339270.4．解方程1111110111111xxxx.-5-解:111111111111xxxx1111111111111xxxxxxx1111012202120001xxxx122(1)212001xxxx212(1)21xxx22(1)(214)xxx3(1)(3)xx,原方程的解为12343,1xxxx.5．已知100120143A,求T|(4)(4)|IAIA.解:300|4|1206141IA,TT|(4)(4)||(4)||(4)|IAIAIAIA2|(4)|36IA.6．已知三阶矩阵A的行列式||2A,123211110B,求||ABA.解:|||()|||||ABAABIABI023220111102320231112322423.7．计算下列行列式:(1)11200000001111nnaaaaa;解:将原行列式D的第1列加于第2列,第2列加于第3列,…,第n列加于第1n列,得12000000000121naaDann1(1)(1)nnnaa.(2)121100000001nnxxxaaaxa.-6-解:原行列式nD按第1列展开,得nD1111000100(1)000001nnnnxxDaxx1nnxDa,由此递推公式并注意11Dxa,就可推知121100000001nnxxxaaaxa12121nnnnnxaxaxaxa.8．证明:(1)121211111111(1)111nniinaaaaaaa,其中120naaa;证明:左边行列式nD的第i列提取ia(1,2,,)in,得nD1211212111111111111nnnnaaaaaaaaaaa121121211111111111111niinnininniinaaaaaaaaaaa1211111010001niinnaaaaa111(1)nniiaaa.4(2)222244441111abcdabcdabcd()()()()()()()abacadbcbdcdabcd.证一:左边行列式222222244444441000abacadaDabacadaabacada1()()()bacadaD,其中-7-1322332233223111Dbacadabbabaaccacaaddadaa3223322322322322100bacbdbbbabaaccacabbabaddadabbaba22222211()()cbdbcbcbacabadbdbadaba22()()()cbdbdbdadcbcac()()()()cbdbdcdcba,于是1()()()DbacadaD()()()()()()()abacadbcbdcdabcd.证二:不妨设,,,abcd各不相同,否则等式两边都等于零.考虑22222432333334444411111()abcdxabcdxfxAxBxCxDxEabcdxabcdx,其中222233331111abcdAabcdabcd,222244441111abcdBabcdabcd.显然,,,abcd为()0fx的四个根,由韦达定理知BabcdA.注意A为范德蒙行列式,所以222244441111abcdabcdabcd()BabcdA()()()()()()()abacadbcbdcdabcd.习题1-31．已知A是三阶方阵,且||2A,求1|3|A,*||A和1*1|(3)|2AA.解:1|3|A112727||27||2AA,*||A2||4A,-8-1*1|(3)|2AA1*11||32AA**11||62AA*1||3A*14||2727A.2．求线性变换11232123312322,35,323xyyyxyyyxyyy的逆变换.解:线性变换的系数矩阵221315323A,其伴随矩阵为*749637324A.将||A按第1行展开,得||2(7)26131A,因此A可逆,其逆矩阵为1*7491637||324AAA.所求逆变换为112321233123749,637,324.yxxxyxxxyxxx3．解矩阵方程XAB,其中211210111A,113432B.解:矩阵A的伴随矩阵为*101232330A.将||A按第1行展开,得||211(2)(1)(3)3A,因此A可逆,其逆矩阵为1*10111232||3330AAA.所求解为110122111132328243253330