Answers, Solution Outlines and Comments to Exercises

1 AAnswers,SolutionOutlinesandCommentstoEx ercisesChapter1 PreliminaryTest(page3) [c2=a2+b2 2abcosC.](5 marks) 4=3+y16= 1.[Verifythatthepoint is onthecurve. Findslopedydx= 12(atthatpoint)andthetangenty+ 8 = 12(x+ 2).(5 marks)Rearrangetheequationto getit in interceptform,or solvey= 0 forx-interceptandx= 0fory-intercept.](5 marks)3. One.[Show thatg0(t)<08t2R, hencegis ](5 marks)Comment:Graphicalsolutionis alsoacceptable,butarguments mustbe 4 (9 2p6) or 16:4 or [V=R2 0 RRh02pR2 r2rdrd ,R= 3; Rh= domain.]

2 (5 marks)V= 4 RRh0pR2 r2rdr, substituteR2 r2=t2.(5 marks)EvaluateintegralV= 4 RpR2 R2hRt2dt.](5 marks)5. (a)(2,1,8).[Consider~p ~q=~s ~r.](5 marks)(b)p3=5.[cosQ=~QP ~QRk~QPk k~QRk.](5 marks)(c)95(j+ 2k).[Vectorprojection= (~QP ^QR)^QR.](5 marks)(d)6p6 squareunits.[Vectorarea=~QP ~QR.](5 marks)(e)7x+ 2y z= 8.[Take normalnin thedirectionof vectorareaandn (x q).](5 marks)Comment:Parametricequationq+ (p q) + (r q) is alsoacceptable.(f) 14,4, 2 onyz,xzandxyplanes.[Take components of vectorarea.](5 marks)6. [A= ab)dAA=daa+dbb.

3 (5 marks)Puta= 10,b= 16andda=db= 0:1.](5 marks)7. 9/2.[Sketch domain:regioninsidetheellipsex2+ 4y2= 9 andabove thex-axis.(5 marks)Bychangeof order,I=R3 3R12p9 x20ydydx.(5 marks)435436 AppliedMathematicalMethodsThen,evaluateI =R3 39 x28dx.](5 marks)8. Connect(1;1);(0;1);( 1=2;0);(0; 1);(1; 1) by straight segments.[Split:fory 0; y= 1 +x jxjandfory <0; y= 1 x+jxj.(5 marks)Next,splitinxto describe thesetwo functions.](5 marks)Comment:yis unde nedforx < (page13) :Theyshow threedi erent `cases'.In onecase,bothproductsareof thesamesize,buttheelements aredi erent.

4 In thesecond,theproductsareof di erent thethird,oneof theproductsis noteven de 2,b= 1,c= 2,d= 1,e= 0,f=i.[Equateelements of bothsidesin anappropriatesequenceto have oneequationin oneunknownat everystep.]Comment:In thecaseof a successfulsplitwithallrealentries,we concludethatthematrixis positive de positive de nitematriceshasgreatutility laterchapters,thesenotionswillbe thepresentinstance,thegiven matrixisnotpositive de (a)Domain:R2; Co-domain:R3.(b)Range:linearcombinationo f thetwo mapped :f0g.

5 [Inthiscase,an inspectionsu ces;becausetwo linearlyindependent vectorsareallthatwe arelookingfor.](c)24 7=239 6 5=2235.[WriteasA[x1x2] = [y1y2] anddetermineA.](d)[ 4 12 3]T.(e)ThematrixrepresentationchangestoA P, wherethe2 2 matrixPrepresents therotation( and10).Chapter3 (page19)1. (a)Thenullspaceis non-trivial,certainlycontaining(x1 x0) anditsscalarmultiples.(b)Thesetof pre-imagesis anin niteset,certainlycontaining(possiblyasa subset)allvectorsof theformx0+ (x1 x0).2. [LetRank(A) =r. Identifyrlinearlyindependent columnsofAin basisB.]

6 Expressother(n r) dependent columns(togetherreferredtoasD) in thisbasis,notether (n r)coe cientscjkandestablish[DB] In rC =0. TheleftoperandherecontainscolumnsofA, hencethecolumnsof theright operandgives (n r) linearlyindependent membersofNull(A).(Coordinatespossiblynee dto be permuted.)So,nullity is at leastn r. Inthe rstn rcoordinates, ,allowingforadditionalpossibility in thelower coordinates,proposex= yCy+p andshow that[DB]x=0)p=0. Thisestablishesthenullity to ben r.]Comment:Alternative (a)No.(b) Answers, SolutionOutlinesandCommentsto Exercises4374.

7 (a)v1= [0:820 0:410:41]T.(b)v2= [ 0:210:640:270:69]T.(c)v3= [0:20 0:590:720:33]T.(d)v4= [ 0:50 0:50 0:500:50]T.(e)Settingv1=u1=ku1kandl= 1; fork= 2;3; ;m, vk=uk Plj=1(vTjuk)vj;if vk6=0, thenvl+1= vk=k vkkandl l+ (a)C= cos10 5 15 sin5 sin10 515 cos5 .(b)C 1=13 cos15 15 cos5 15 sin5 sin10 5cos10 5 .6. (a)241000100a135.(b)2664100001000 (page27)1. Rank= 3,Nullity = 2,Range=<q1;q2;q3>,Nullspace=<n1;n2>, wheren1= [ 0:7 0:1 2:1 1 0]T;n2= [0:1 0:7 2:7 0 1]T;Particularsolution: x= [2:3 2:9 13:9 0 0]T, Generalsolution:x= x+ 1n1+ 5,b= 2,c= 1,d= 2,e= 2,f= 3,g= 1,h= 1,i= 3.

8 [Equateelements todeterminetheunknownsin thissequence,to getoneequationin oneunknownat everystep.]3. Hecanifnis odd,butcannotifnis even.[Forthecoe cient matrix,determinant =1 ( 1)n2n.]4. Realcomponent = (Ar+AiA 1rAi) 1; Imaginarycomponent = A 1rAi(Ar+AiA 1rAi) 20=17,x4= 1.[Partitionx= y1y2 andthensolve fory2= [x3x4]T.]6. =1aTa aTATPAa,b= PAa ,Q=P(Im AabT).[Equateblocks in ( A AT) P=Im+1.]In thesecondcase,withP= QqqT andA= AaT , P=Q qqT= . :50:50:5 0:50:50:5 0:50:50:5 0:50:50:50:5 0:5 0:5 0:53775andR=26641260404 22000 :Asa diagonalentryofRturnsoutas zero,thecorrespondingcolumnofqbecomesjus tunde nedor indeterminate(notnon-existent) andtheonlyconstraint thatremainsis theorthogonality ofQ.

9 Any vectororthonormalto previouscolumnsis ,however,resultsof thenextstepsmay di a way, theQRdecompositionisanextensionof Theseelements areusedto constructthebasisvectorsforthenullspace, by appending` 1' atthecoordinatecorrespondingto serve as thecoe cientsin theexpressionof non-basiscoordinatesin termsof thebasiscoordinates.[For theproof,438 AppliedMathematicalMethodsconsideronlyth osecolumns(towhich thepremisemakes reference)in theRREFas well as of elementarymatricesto establishequality of theirranks.]

10 ]9. [Identifymlinearlyindependent vectorsorthogonaltoM, fromde thatbasis,determineit to satisfyPv=v w2 M. Thiswouldyieldw=AT(AAT) 1 Avandleadto theresult.]Chapter5 (page34)1. 0:090:73 , 2:27 0:18 , 0:73 0:18 , 0:36 0:270:090:18 , 1:45 0:091:360:73 . 1=241a00 bad1d0be cdadf For 1< j n,pj=ajqj 1andqj=bj pjcj 1. FLOP:3(n 1).Forwardsubstitution:2(n 2).Back-substitution:3n ,w1=c1=b1, 1=r1= < j < n,qj=bj ajwj 1,wj=cj=qjand j=rj aj j 1qj. Finally,qn=bn anwn 1, n=rn an n 1qn. FLOP:6n (a)Rank= 1.(b)Theformulais usedto make smallupdatesonanalreadycomputedinverse.