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.(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.

2 [Take components of vectorarea.](5 marks)6. [A= ab)dAA=daa+dbb.(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. 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.

3 (b)Range:linearcombinationof thetwo mapped :f0g. [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. 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.]

4 Thisestablishesthenullity to ben r.]Comment:Alternative (a)No.(b) Answers, SolutionOutlinesandCommentsto Exercises4374. (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.[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.

5 [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. 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.]9. [Identifymlinearlyindependent vectorsorthogonaltoM, fromde thatbasis,determineit to satisfyPv=v w2 M.]

6 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.[ For derivation,expandleft-hand-sidein binomialseriesandsandwich thescalarquantity (1 vTA 1u) in everyterm.]5. A 1ddt(A)A 1.[Di erentiateAA 1=I.]Chapter6 (page41) ,Pmj=1ajixj;Ay,ATx;Pnj=1(aij+aji)xj; (A+AT) Yes.[Therearevariousapproachestothisprob lem, 'scriterion,Choleskydecomposition,usingS turmsequenceproperty of characteristicpolynomials( )oractualevaluationof theeigenvalues!]3. MatrixIn+cATAis of sizen n, largerthanthesizeofIm+cAAT, henceinversionin termsof thelattergives ,theformermatrixis ill-conditioned,withmeigenvaluesof largemagnitudesandtherestequalto :Such matricesappearin thesolutionstepswitha :Canyourelatethisproblemwiththepreviouso ne?

7 5. 3, 4, 4, 3, 3, 3, 2, 2.[Thisis acyclictridiagonalsystem, thatdi ersfromanordinarytridiagonalcasein thattheelements at (1;n) and(n;1) locationsin thecoe cient reduceit to a tridiagonalsystem.]A. Answers, SolutionOutlinesandCommentsto 0:840:340:410:39 0:080:69 0:19 0:34 0:06 0:13 0:38 0:120:190:110:260:55 0:05 0:32 0:270:09 0:020:020:14 0:07 0:01377775,266664 0:890:370:440:40 0:080:72 0:20 0:36 0:06 0:13 0:39 0:110:190:120:260:57 0:06 0:33 0:280:09 0:020:020:15 0:07 0 (a) dTkekdTkAdk.(b) 1dT0Ad1, 1dT0Ad1+ 2dT0Ad2and 2dT1Ad2.(c)Fori < j 1 and2 j k,dTiej=Pj 1l=i+1 ldTiAdl.(d)dTiAdj= 0 fori < j (page50)1. (a)[1 +u22u11 u22u1]T.(b)q2 +u21+p4 +u41and2+u21+p4+u412u1.(c)[11]T, (d)Thesolutionis insensitive to smallchangesinu1andwithu2it changesas x= 50[1 1]T :Thegivensystemexhibitstherudimentaryfea turesof ill-conditioning,as dis-cussedin thechapteranddepictedin :Atw= 1, notethesingularity andhow theTikhonov (a)fora singularzone,it givessmoothsolutionsthatareusableas workingsolutionsformany ,evenaway fromsingularity, theresultingsolutionhasanerror, a trade-o theparameter resultsinbetteraccuracyof solutionsaway froma singularity, at thecostof smoothnessaroundit!

8 (b). 10 50510wSolution with = solutionTikhonov Solution (a) 5 4 3 2 1012345wError in Solution = = (b) :Solutionof anill-conditionedsystem:Tikhonov regularization440 AppliedMathematicalMethods3. Firstiteration:[17:832:0210:92]T, seconditeration:[15:684:6811:79] thethirditeration,theerrormoves to secondplaceof decimal!However,withthechangedorder,thea lgorithmdiverges,becausethecoe cientmatrixis (a)ATA= 10000 7500 75005625 ;Eigenvalues: 1= 15625, 2= 0;Eigenvectors:v1= 0:8 0:6 ,v2= 0:60:8 .(b) =241250000035.(c)U=240:800:000:600:360:8 0 0:48 0:480:600:6435.[Keepingthe rstcolumnu1 xed,any orthogonalmatrixwillbe valid.](d)<v2>.(e)<u1>.(f) y=c, wherey=VTxandc= (page57)1. (a)Eigenvalues:-2,3, 3;Eigenvectors:[1 1 1]T, [34 0]T, [10 2] :Linearcombinationsof thelasttwo arealsoeigenvectors.(b)Eigenvalues:1, 2, 2;Eigenvectors:[11 1]T, [12 1] :Thethirdeigenvectordoes matrixis n+a1 n 1+a2 n 2+ +an 1 +an. Signi cance:For everyn-thdegreepolynomial,wecanconstruct ann nmatrix,calleditscompanionmatrix, whoseeigenvaluesaretherootsof thegiven :Thisresultis thebasisof oneof thecommonlyusedmethods forsolvingpolyno-mialequations.

9 ( )3. Thesolutionis discussion,seethetextof thenextchapter( ).4. (a)A1= 1bT10A1 ,bT1=qT1AQ1,A1=QT1AQ1. (b) Q2= [q2Q2],Q2="100 Q2#.(c)For a realmatrixA, havingrealeigenvalues,thereexistsanortho gonalmatrixQsuchthatQTAQis upper triangular,withtheeigenvaluesat [For the rstpart,useAT(Im+cAL 1AT) =AT+cATAL 1AT= (In+cATAL 1)AT= (L+cATA)L thesecondpart,operateAL 1 ATonitseigenvectorandnotethat(Im+cAL 1AT)v=v+c v= (1 +c )v.]6. Eigenvalue = 28 andcorrespondingeigenvectorv= [ 0:410:41 0:82]Tconvergeswithinoneplaceof decimal,in seven iterations(inthiscase).Chapter9 (page64)1. (a)Jordanform:2421002000135.[Inthesecond columnof thesimilarity transformationmatrix,usegeneralized eigenvectorw2, givenby (A 2I)w2=v1.](b)Diagonalform:A. Answers, SolutionOutlinesandCommentsto Exercises441diag(3;1;1).(c)Jordanform:24 11001100135.[S= [v1w2w3].](d)DiagonalforminC: diag( 1;i; i).[Inthiscase,eigenvectorsarealsocomple xand,hence,so is thesimilarity transformation.]

10 ] :25 0:20 0:150:100:90 0:100:25 0:200:8535andA6=242:63 1:24 0:850:470:53 0:471:63 1:240:1535.[UseA=S S 1andA6=S 6S 1.]3. [If a diagonalizablematrixhasa fullsetoforthogonaleigenvectors,thenthea ssociatedorthog-onalsimilarity transformationwillpreserve thesymmetryof a diagonalmatrixandthegivenmatrixturnsoutt o be thenon-diagonalizablecase,theresultingJo rdanformisnotsymmetric,so thematrixcanhave allitseigenvectors(lessthanthefullset)or thogonal.]4. De atedmatrix:240:640:4800:480:36000035; othereigenvalues:1, 0; andcorrespondingeigenvectors:[0:8 0:6 0]T, [ 0:360:480:8] (page71)1. Eigenvaluesandeigenvectors:0:620+ 0:785i, [ 0:115 0:500i 0:115+ 0:500i0:688]T;0:620 0:785i, [ 0:115+ 0:500i 0:115 0:500i0:688]T;1,[0:6880:6880:229] (Q) = ,Qrepresents a rotation.[Check orthogonality.]Planeorthogonalto thethirdeigenvector,or theplaneof the rsttwo eigenvectors,of whichtherealplanecanbe obtainedwiththebasisf[ 1 1 6]T;[1 1 0] ,2, (aftertwo sweeps, , , ).