线性代数期中试卷

1Linearalgebraquestions2010/12Papercode：10043CClassTimes：45ClassesCourseTitle：LinearAlgebraClass：ElectiveclassesQ.1Q.2Q.3Q.4Q.5Q.6TotalScores【Note：Writeyouranswersandthequestionnumbersonyouranswersheet!Otherwise,invalid!】Q.1/Scores1.Fillintheblank（2*10=20scores）(1)LetA=(1，–1，0)andB=(1，1，1，1，2),soRank(ATB)=(1).Solution:1)(so000002111121111)2,1,1,1,1(011BARBATT(2)LetAisa(3×3)matrix,|A|=−3，so|−3A|=(81)(3)IfAisan(n×n)matrix，X1andX2arethesolutionstothelinearequationsAX=B，(X1≠X2)，so|A|=(0).(4)LetAbealinearlydependentthevectorsset,A={β1,β2,β3},hereβ1=(1,2,3),β2=(4,t,3),β3=(0,0,1),sot=(8).Solution:.8isThat,2)(sosetdependentlinearlyaisFor100080021~100980321~10034321tARtttA(5)Therearendimensionvectors:α1，α2，α3，if(αi，αj)=i+j，sowehave(α12+α2，α1–α3)=(4).Solution:For(αi，αj)=i+j，sowehave4)32()12()31()11(),(321231113121(6)IfX1=(1,0,2)TandX2=(3,4,5)Taretwosolutionstothethreevariablesnon-homogeneouslinearequationsAX=B,sothecorrespondinghomogeneouslinearequationsAX=0hassolution((2,4,3)T).Solution:X1–X2=(1,0,2)T–(3,4,5)T=(2,4,3)TisasolutiontotheequationAX=0.(7)Letβ=(1，–1，0，2)Tandγ=(a，1，–1，1)Tareorthogonalvectors,soa=(–1).Solution:βandγareorthogonalvectors,soβ·γ=0,thatis1so01)1,1,1,)(2,0,1,1(aaaTT(8)A，Bare(4×4)matrices，forA=(α，α1，α2，α3)andB=(β，α1，α2，α3)，if|A|=1and|B|=2,sowecanfindthat|A+B|=(24).Solution:A+B=(α，α1，α2，α3)+(β，α1，α2，α3)=(α+β，2α1，2α2，2α3),so|A+B|=det(α+β，2α1，2α2，2α3)=det(α，2α1，2α2，2α3)+det(β，2α1，2α2，2α3)=23det(α，α1，α2，α3)+23det(β，α1，α2，α3)=8+16=24(9)LetAis(3×3)matrices,for|A|=2，so|A*－3A－1|=(–1/2).3Solution:for|A|=2,|A－1|=1/2A*=A－1|A||A*－3A－1|=|A－1|A|－3A－1|=|A－1(|A|－3)|=|(－1)A－1|=–1/2(10)LetdcbaA，|A|≠0,soA−1=(acbdcbad1).Solution:A－1=A*/|A|Q.2/Scores2.MultipleChoice(1*20=20scores)(1)LetA，B，C，andDare(n×n)matrices，Iisan(n×n)identitymatrix，thefollowingstatementiscorrect(D).A．IfA2=0，thatisA=0B．IfA2=A，sowehaveA=0orA=IC．IfAB=AC，andA≠0，soB=CD．IfAB=BA，wehave(A+B)2=A2+2AB+B2(2)Ifthevectorsgroupβ1，β2，andβ3arelinearlyindependent,andβ1，β2，andβ4arelinearlydependent,so(C).(A)β1canbeexpressedasalinearcombinationbythevectorsβ2，β3andβ4.(B)β2can’tbeexpressedasalinearcombinationbythevectorsβ1，β3andβ4.(C)β4canbeexpressedasalinearcombinationbythevectorsβ1，β2andβ3.(D)β4can’tbeexpressedasalinearcombinationbythevectorsβ1，β2andβ3.4(3)LetA,Bareinvertiblematriceswiththesameorders，thefollowingconclusionsiscorrect(C).(A)111)(BABA(B)111)(BAAB(C)0000111ABBA(D)1111100BCABABCA(4)Let,333231232221131211aaaaaaaaaA131312113333323123232221222aaaaaaaaaaaaB,0011000101P,1200100012PsoB=(A).(A)P1AP2(B)AP1P2(C)AP2P1(D)P2AP1(5)LetAisan(m×n)matrix,Bisan(n×m)matrix,sothat(A).(A)ifm＞n,wehave∣AB∣＝0(B)ifm＞n,wehave∣AB∣≠0(C)ifn＞m,wehave∣AB∣≠0(D)ifn＞m,wehave∣AB∣＝0.(6)Letβ1，β2，β3，andβ4arethreedimensionvectors,so(B)iscorrect.Note：an+1’sndimensionsvectorssetlinearlydependent.(A)β1，β2，β3，andβ4linearlyindependent.(B)β1，β2，β3，andβ4linearlydependent.5(C)β1canbeexpressedasalinearcombinationbythevectorsβ2，β3，andβ4.(D)β1can’tbeexpressedasalinearcombinationbythevectorsβ2，β3，andβ4.(7)Ifan(3×3)matrixAhaseigenvalues2,1,0sowehave|A+2E|=(D).|A|=所有特征值的积=0，A+2E的特征值为2+2,1+2,0+2,即4,3,2，|A+2E|=4*3*2=24.(A)0(B)2(C)3(D)24(8)IfAandBareequivalentmatrices,sowehave(A).(A)r(A)=r(B)(B)|λE−A|=|λE−B|(C)|A|=|B|(D)P−1AP=B(9)IfAandBaresimilarmatrices,sowehave(C).(A)r(A+1)=r(B)(B)|λE−A|=|λE−B|(C)|A|=|B|(D)P′TAP=B(forPisainvertiblematrix)(10)Ifan(3×3)realsymmetricmatrixAhaseigenvalues2,1,0sothat(B).(A)Aisapositivedefinitematrix.(B)Aisapositivesemidefinitematrix.(C)Aisanegativesemidefinitematrix.(D)Aisanegativedefinitematrix.Positivelydefinitematrix(正定矩阵)：Alleigenvaluesaregreaterthenzero;Negativedefinitematrix(负定矩阵)：Alltheeigenvaluesarelessthanzero;Positivesemidefinitematrix(半正定矩阵)：Alltheeigenvaluesarenotlessthanzero(greaterthanorequaltozero);所有特征值都大于等于0，半正定；Negativesemidefinitematrix(半负定矩阵)Alltheeigenvaluesarenotgreaterthanzero6(arelessthanorequaltozero);Indefinitematrix(不定矩阵);Positivedefinitequadraticform(正定二次型);Negativedefinitequadraticform(负定二次型);(11)Letvectorsβ1，β2，andβ3arefundermentalsetofsolutionstothehomogeneouslinearequationssystemAX=0，sothefundermentalsetofsolutionoftheequationssystemAX=0canbeexpressedinotherwayas(C).(A)β1–β2，β2–β3，β3–β1.(B)avectorssetthathasthesamerankasvectorsset{β1，β2，β3}.(C)β1，β1+β2，β1+β2+β3.(D)beequivalenttothevectorsset{β1，β2，β3}.(12)LetAisan(m×n)matrix，AX=0isthecorrespondinghomogeneouslinearequationsfortheequationsAX=b,sofollowingconclusions(D)iscorrect.(A)IfAX=0hasthetrivialsolution,sothattheequationsAX=bhasonlysolution.(B)IfAX=bhasinfinitesolutions,sothattheequationsAX=0hasthetrivialsolution.(C)IfAX=0hasthenontrivialsolutions,sothattheequationsAX=bhasinfinitesolutions.(D)IfAX=bhasinfinitesolutions,sothattheequationsAX=0hasthenontrivialsolutions.(13)Let,sothefollowing(B)matrixissimilartothematrixA.320230002A7(A)(B)(C)(D)(14)Let0isaneigenvalueforth