研究生矩阵论课后习题答案(全)习题四

习题四1．求下列微分方程组的通解（1）;34,2212211xxdtdxxxdtdx（2）.,3233212321,xxdtdxxxxdtdxxxdtdx解：(1)设,3421A21xxx,则原方程组可写为Axdtdx，矩阵A的特征方程为0)1)(5(3421AI，则矩阵A的特征值为51,12,求得矩阵A的特征向量分别为11,21,令1211P,则1211311P,有10051APP,1PPA,则tttttttttttAteeeeeeeeeePPee55555122231121100121131.故该方程组的通解为ttttttttttttAteccecceccecccceeeeeeeecex)2()22()2()(31222312152121521215555其中21,cc为任意常数.(2)设,110111110A321xxxx,则原方程可写为Axdtdx，矩阵A的特征方程为0)1(2AI，则矩阵A的特征值为01,132.A的属于特征值01的特征向量为1121,由方程组32322AA解得A的属于特征值132的广义特征向量为111,10132.令111101112,,321P,则1113121011P,有11,100110000PJPAJAPP,由于tttJteteee000001,则1113121010000011111011121tttJtAteteePPeetttttttttttttteeteteeeeeteetetee21111222,故方程组的通解为32121111222cccteeteteeeeeteeteteecextttttttttttttAt,其中321,,ccc为任意常数.2．求微分方程组Axdtdx满足初始条件)0(x的解：（1）33,3421A，（2）001,102111121A.解:(1)由第1题知ttttttttAteeeeeeeee555522231,故微分方程组Axdtdx满足初始条件)0(x的解为ttttttttttttAteeeeeeeeeeeeex555555423322231.(2)矩阵A的特征方程为0)1)(3(2AI,故矩阵A的特征值为31,132.A的属于特征值31的特征向量为2121,由方程组32322AA解得A的属于特征值132的广义特征向量为021,21232,令022211122,,321P,则24025122312811P,有11,100110003PJPAJAPP,又ttttJteteeee000003,故微分方程组Axdtdx满足初始条件)0(x的解为1PPeexJtAt0012402512231200000022211122813tttteteeetttttteeeeee44224481333.3．求)(tBuAxdtdx满足条件)0(x的解：（1）21,)(,41,3421ccetuBAt（2）101,1)(,262,0061011016tuBA解:(1)由第1题知ttttttttAteeeeeeeee555522231,则ttttttttttttAteccecceccecccceeeeeeeee)2()22()2()(31222312152121521215555,vttvttvvvtvtvtvtvtvtvtvtvtAeeeeeeeeeeeeeevBue6565)()(5)()(5)()(5)()(5)(6636314222231)(故tttttttvtAeeteeetedvvBue550)(62121631)(则该方程组的解为tttttttvtAAtteecceccteecceccdvvBueetx2])12()122[(312])212()21[(31)()(21521215210)((2)矩阵A的特征方程为0)3)(2)(1(AI,则A的特征值为11,3,232,求得其特征向量分别为231,341,651321.令13924811121,2363451111PP,有11,300020001PJPAJAPP,又tttJteeee32000000,则1PPeeJtAt1011392481110000002363451112132ttteeetttttteeeeee32323289121243,)(3)(2)()(3)(2)()(3)(2)(1)()(2663852262)(vtvtvtvtvtvtvtvtvtvtJvtAeeeeeeeeePPevBue故373236453131)(3232320)(ttttttttttvtAeeeeeeeeedvvBue则该方程组的解为37323645313189121243)()(3232323232320)(ttttttttttttttttvtAAteeeeeeeeeeeeeeedvvBueetx3732212611165313114323232ttttttttteeeeeeeee.4．求方程teyyyy6116满足0)0()0()0(yyy的解.解:令yxyxyx321,,,则,6116,,32133221texxxyxxxxx写成向量方程组为tBeAxx,其中100,6116100010BA.对于矩阵A,有JPAP1,其中321,13228615621,9413211111JPP于是tttJteeee32,1PPeeJtAtttttttttttttttttttttttttttteeeeeeeeeeeeeeeeeeeeeeeeeee3232323232323232329827325182463491656126238526621由于000)0(x,则tvvtAtvvtAAtdvBeedvBeexetx0)(0)()0()()1(29)1(8)1(23)1(4)1(21)1(221232232232ttttttttttttttteeeeteeeeeteeeeete故原方程的解为ttttttttteeeteeeeetexy322321414321)]1(21)1(2[215．试证明：若A为2阶方阵，其特征值为21,，特征向量为21,PP，则方程Axdtdx的解一定能表示成221121PecPecxtt，其中21,cc由下式确定:2211)0(PcPcx，然后利用这一结论求解定解问题：11)0(,651021xxxdtdx的解,并将这一结论推广到n阶方阵情形.(1)证明:令],[21PPP,则,,121211PPAAPP于是xPPdtdx121,xPdtdxP1211令,1xPy则dtdxPdtdy1,微分方程化为ydtdy21其解为2121cceeytt,故方程Axdtdx的解一定能表示成221121212121],[cecPeccceePPPyxtttt若是定解问题,则21,cc由2211)0(PcPcx确定.(2)解:矩阵6510的特征值为5,121,特征向量分别为51,1121PP,则方程组216510xxdtdx的通解为ttttttecececececec52152152155111,由于11)0(x,则1512121cccc,解之,得212321cc,故原方程组的解为tttteeeexx552125232123.(3)n阶方阵的情形:设微分方程组Axdtdx,其中系数矩阵A为n阶可对角化矩阵，其特征值为n,,,21，特征向量分别为nPPP,,,21，则该方程组的通解为ntntPecPecPecxnt221121,其中nc