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INSTRUCTOR’S SOLUTIONS MANUAL L INEAR A LGEBRA WITH A PPLICATIONS NINTH EDITION Steven J. Leon University of Massachusetts, Dartmouth Boston Columbus Indianapolis New York San Francisco Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Linear Algebra with Applications 9th Edition Leon Solutions Manual Full Download: http://testbanktip.com/download/linear-algebra-with-applications-9th-edition-leon-solutions-manual/ Download all pages and all chapters at: TestBankTip.com

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2 3 4 5 6 OPM 17 16 15 14

Contents Preface v 1 MatricesandSystemsofEquations 1 1SystemsofLinearEquations1 2RowEchelonForm2 3MatrixArithmetic3 4MatrixAlgebra6 5ElementaryMatrices12 6PartitionedMatrices17 MATLABExercises20 ChapterTestA22 ChapterTestB24 2 Determinants 27 1TheDeterminantofaMatrix27 2PropertiesofDeterminants30 3AdditionalTopicsandApplications33 MATLABExercises35 ChapterTestA35 ChapterTestB36 3 VectorSpaces 38 1DeﬁnitionandExamples38 2Subspaces 42 3LinearIndependence47 4BasisandDimension50 5ChangeofBasis52 6RowSpaceandColumnSpace52 MATLABExercises59 ChapterTestA60 ChapterTestB62 4 LinearTransformations 66 1DeﬁnitionandExamples66 2MatrixRepresentationsofLinearTransformations69 3Similarity 71 MATLABExercise72 iii Copyright © 2015 Pearson Education, Inc.

iv Contents ChapterTestA73 ChapterTestB74 5 Orthogonality 76 1TheScalarproductin Rn 76 2OrthogonalSubspaces78 3LeastSquaresProblems81 4InnerProductSpaces85 5OrthonormalSets90 6TheGram-SchmidtProcess98 7OrthogonalPolynomials100 MATLABExercises103 ChapterTestA104 ChapterTestB105 6 Eigenvalues 109 1EigenvaluesandEigenvectors109 2SystemsofLinearDiﬀerentialEquations114 3Diagonalization115 4HermitianMatrices123 5SingularValueDecomposition130 6QuadraticForms132 7PositiveDeﬁniteMatrices135 8NonnegativeMatrices138 MATLABExercises140 ChapterTestA144 ChapterTestB145 7 NumericalLinearAlgebra 149 1Floating-PointNumbers149 2GaussianElimination150 3PivotingStrategies151 4MatrixNormsandConditionNumbers152 5OrthogonalTransformations162 6TheEigenvalueProblem164 7LeastSquaresProblems168 MATLABExercises171 ChapterTestA172 ChapterTestB173 Copyright © 2015 Pearson Education, Inc.

Preface

Thissolutionsmanualisdesignedtoaccompanythenintheditionof LinearAlgebrawithApplications byStevenJ.Leon.Theanswersinthismanualsupplementthosegivenintheanswerkeyofthe textbook.Inaddition,thismanualcontainsthecompletesolutionstoallofthenonroutineexercises inthebook.

Attheendofeachchapterofthetextbooktherearetwochaptertests(AandB)andasection ofcomputerexercisestobesolvedusingMATLAB.ThequestionsineachChapterTestAaretobe answeredaseither true or false.Althoughthetrue-falseanswersaregivenintheAnswerSectionofthe textbook,studentsarerequiredtoexplainorprovetheiranswers.Thismanualincludesexplanations, proofs,andcounterexamplesforallChapterTestAquestions.ThechaptertestslabeledBcontain problemssimilartotheexercisesinthechapter.Theanswerstotheseproblemsarenotgiveninthe AnswerstoSelectedExercisesSectionofthetextbook;however,theyareprovidedinthismanual. CompletesolutionsaregivenforallofthenonroutineChapterTestBexercises.

IntheMATLABexercises.mostofthecomputationsarestraightforward.Consequently,they havenotbeenincludedinthissolutionsmanual.Ontheotherhand,thetextalsoincludesquestions relatedtothecomputations.Thepurposeofthequestionsistoemphasizethesigniﬁcanceofthe computations.Thesolutionsmanualdoesprovidetheanswerstomostofthesequestions.Thereare somequestionsforwhichitisnotpossibletoprovideasingleanswer.Forexample,someexercises involverandomlygeneratedmatrices.Inthesecases,theanswersmaydependontheparticular randommatricesthatweregenerated.

StevenJ.Leon sleon@umassd.edu

Chapter1 Matricesand Systems ofEquations 1 SYSTEMSOFLINEAREQUATIONS 2. (d)                11111 021 21 0041 2 0001 3 00002                5. (a)3x1 +2x2 =8 x1 +5x2 =7 (b)5x1 2x2 + x3 =3 2x1 +3x2 4x3 =0 (c)2x1 + x2 +4x3 = 1 4x1 2x2 +3x3 =4 5x1 +2x2 +6x2 = 1 1 Copyright © 2015 Pearson Education, Inc.

9. Giventhesystem

onecaneliminatethevariable x2 bysubtractingtheﬁrstrowfromthesecond.Onethen obtainstheequivalentsystem

(a)If m1 = m2,thenonecansolvethesecondequationfor x1

Onecanthenplugthisvalueof x1 intotheﬁrstequationandsolvefor x2.Thus,if m1 = m2,therewillbeauniqueorderedpair(x1,x2)thatsatisﬁesthetwoequations.

(b)If m1 = m2,thenthe x1 termdropsoutinthesecondequation

b2 b1

Thisispossibleifandonlyif b1 = b2.

(c)If m1 = m2,thenthetwoequationsrepresentlinesintheplanewithdiﬀerentslopes. Twononparallellinesintersectinapoint.Thatpointwillbetheuniquesolutionto thesystem.If m1 = m2 and b1 = b2,thenbothequationsrepresentthesamelineand consequentlyeverypointonthatlinewillsatisfybothequations.If m1 = m2 and b1 = b2, thentheequationsrepresentparallellines.Sinceparallellinesdonotintersect,thereis nopointonbothlinesandhencenosolutiontothesystem.

10. Thesystemmustbeconsistentsince(0, 0)isasolution.

11. Alinearequationin3unknownsrepresentsaplaneinthreespace.Thesolutionsettoa3 × 3 linearsystemwouldbethesetofallpointsthatlieonallthreeplanes.Iftheplanesare paralleloroneplaneisparalleltothelineofintersectionoftheothertwo,thenthesolution setwillbeempty.Thethreeequationscouldrepresentthesameplaneorthethreeplanes couldallintersectinaline.Ineithercasethesolutionsetwillcontaininﬁnitelymanypoints. Ifthethreeplanesintersectinapoint,thenthesolutionsetwillcontainonlythatpoint.

2 ROWECHELONFORM

2. (b)Thesystemisconsistentwithauniquesolution(4, 1).

4. (b) x1 and x3 areleadvariablesand x2 isafreevariable.

(d) x1 and x3 areleadvariablesand x2 and x4 arefreevariables.

(f) x2 and x3 areleadvariablesand x1 isafreevariable.

5. (l)Thesolutionis(0, 1 5, 3 5).

6. (c)Thesolutionsetconsistsofallorderedtriplesoftheform(0, α,α).

7. Ahomogeneouslinearequationin3unknownscorrespondstoaplanethatpassesthrough theoriginin3-space.Twosuchequationswouldcorrespondtotwoplanesthroughtheorigin. Ifoneequationisamultipleoftheother,thenbothrepresentthesameplanethroughthe originandeverypointonthatplanewillbeasolutiontothesystem.Ifoneequationisnot amultipleoftheother,thenwehavetwodistinctplanesthatintersectinalinethroughthe

2 Chapter1 •

MatricesandSystemsofEquations

(d)4x1 3x2 + x3 +2x4 =4

x1 + x2 5x3 +6x4 =5

1 + x2 +2x3 +4x4 =8

x1 + x2 +3x3 2x4 =7

m1x1 + x2 = b1 (m1 m2)x1 = b2 b1

x1 + x2 =

x1 + x

= b

b2 b1

origin.Everypointonthelineofintersectionwillbeasolutiontothelinearsystem.Soin eithercasethesystemmusthaveinﬁnitelymanysolutions.

Inthecaseofanonhomogeneous2 × 3linearsystem,theequationscorrespondtoplanes thatdonotbothpassthroughtheorigin.Ifoneequationisamultipleoftheother,thenboth representthesameplaneandthereareinfinitelymanysolutions.Iftheequationsrepresent planesthatareparallel,thentheydonotintersectandhencethesystemwillnothaveany solutions.Iftheequationsrepresentdistinctplanesthatarenotparallel,thentheymust intersectinalineandhencetherewillbeinfinitelymanysolutions.Sotheonlypossibilities foranonhomogeneous2 × 3linearsystemare0orinfinitelymanysolutions.

9. (a)Sincethesystemishomogeneousitmustbeconsistent.

13. Ahomogeneoussystemisalwaysconsistentsinceithasthetrivialsolution(0,..., 0).Ifthe reducedrowechelonformofthecoeﬃcientmatrixinvolvesfreevariables,thentherewillbe inﬁnitelymanysolutions.Iftherearenofreevariables,thenthetrivialsolutionwillbethe onlysolution.

14. Anonhomogeneoussystemcouldbeinconsistentinwhichcasetherewouldbenosolutions. Ifthesystemisconsistentandunderdetermined,thentherewillbefreevariablesandthis wouldimplythatwewillhaveinﬁnitelymanysolutions.

16. Ateachintersection,thenumberofvehiclesenteringmustequalthenumberofvehiclesleaving inorderforthetraﬃctoﬂow.Thisconditionleadstothefollowingsystemofequations

17. If(c1,c2)isasolution,then

1,αc2)isalsoasolution.

18. (a)If x4 =0,then x1, x2,and x3 willallbe0.Thusifnoglucoseisproduced,thenthere isnoreaction.(0, 0, 0, 0)isthetrivialsolutioninthesensethatiftherearenomoleculesof carbondioxideandwater,thentherewillbenoreaction.

(b)Ifwechooseanothervalueof x4,say x4 =2,thenweendupwithsolution x1 =12, x2 =12, x3 =12, x4 =2.Notetheratiosarestill6:6:6:1.

3 MATRIXARITHMETIC

Section3 • MatrixArithmetic 3

x1 + a1 = x2 + b1 x2 + a2 = x3 + b2 x3 + a3 = x4 + b3 x4 + a4 = x1 + b4 Ifweaddallfourequations,weget x1 + x2 + x3 + x4 + a1 + a2 + a3 + a4 = x1 + x2 + x3 + x4 + b1 + b2 + b3 + b4 andhence a1 + a2 + a3 + a4 = b1 + b2 + b3 + b4

a11c1 + a12c2 =0 a21c1 + a22c2 =0 Multiplyingbothequationsthroughby α

a11(αc1)+ a12(αc2)= α 0=0 a21(αc1)+ a22(αc2)= α 0=0 Thus(αc

,oneobtains

(g)        5 1015 5 14 8 96        2. (d)    361056 10316     5. (a)5A =        1520 55 1035        2A +3A =        68 22 414        +        912 33 621        =        1520 55 1035        (b)6A =        1824 66 1242        3(2A)=3        68 22 414        =        1824 66 1242        (c) AT =     312 417     (AT )T =     312 417     T =        34 11 27        = A 6. (a) A + B =     546 051     = B + A (b)3(A + B)=3     546 051     =     151218 0153     3A +3B =     12318 6915     +     390 66 12     =     151218 0153     (c)(A + B)T =     546 051     T =        50 45 61        AT + BT =        42 13 65        +        1 2 32 0 4        =        50 45 61        7. (a)3(AB)=3        514 1542 016        =        1542 45126 048        (3A)B =        63 189 612            24 16     =        1542 45126 048        A(3B)=        21 63 24            612 318     =        1542 45126 048        Copyright © 2015 Pearson Education, Inc.

4 Chapter1 • MatricesandSystemsofEquations

11. Thegiveninformationimpliesthat

arebothsolutionstothesystem.Sothesystemisconsistentandsincethereismorethanone solution,therowechelonformof A mustinvolveafreevariable.Aconsistentsystemwitha freevariablehasinﬁnitelymanysolutions.

12. Thesystemisconsistentsince x =(1, 1, 1, 1)T isasolution.Thesystemcanhaveatmost3 leadvariablessince A onlyhas3rows.Therefore,theremustbeatleastonefreevariable.A consistentsystemwithafreevariablehasinﬁnitelymanysolutions.

13. (a)Itfollowsfromthereducedrowechelonformthatthefreevariablesare x2, x4, x5.Ifwe set

(b)Ifwesetthefreevariablesequalto0,then

Section3 • MatrixArithmetic 5 (b)(AB)T =        514 1542 016        T =     5150 144216     BT AT =     21 46         26 2 134     =     5150 144216     8. (a)(A + B)+ C =     05 17     +     31 21     =     36 38     A +(B + C)=     24 13     +     12 25     =     36 38     (b)(AB)C =     418 213         31 21     =     2414 2011     A(BC)=     24 13         4 1 84     =     2414 2011     (c) A(B + C)=     24 13         12 25     =     1024 717     AB + AC =     418 213     +     146 94     =     1024 717     (d)(A + B)C =     05 17         31 21     =     105 178     AC + BC =     146 94     +     4 1 84     =     105 178    

(b) x =(2, 1)T isasolutionsince b =2a1 + a2.Therearenoothersolutionssincetheechelon formof A isstrictlytriangular. (c)Thesolutionto Ax = c is x =( 5 2 , 1 4 )T .Therefore c = 5 2 a1 1 4 a2.

x1 =        1 1 0        and x2 =        0 1 1       

x2 = a, x4 = b, x5 = c,then x1 = 2 2a 3b c x3 =5 2b 4c andhencethesolutionconsistsofallvectorsoftheform x =( 2 2a 3b c,a, 5 2b 4c,b,c)T

x0 =( 2, 0, 5, 0, 0)T

isasolutionto

x =

14. If w3 istheweightgiventoprofessionalactivities,thentheweightsforresearchandteaching shouldbe w1 =3w3 and w2 =2w3.Notethat 1 5w2 =3w3 = w1, sotheweightgiventoresearchis1.5timestheweightgiventoteaching.Sincetheweights mustalladdupto1,wehave

C isthematrixintheexampleproblem fromtheAnalyticHierarchyProcessApplication,thentheratingvector

15. AT isan n × m matrix.Since AT has m columnsand A has m rows,themultiplication AT A ispossible.Themultiplication AAT ispossiblesince A has n columnsand AT has n rows.

16. If A isskew-symmetric,then AT = A.Sincethe(j,j)entryof AT is ajj andthe(j,j)entry of A is ajj ,itfollowsthat ajj = ajj foreach j andhencethediagonalentriesof A must allbe0.

17. Thesearchvectoris x =(1, 0, 1, 0, 1, 0)T .Thesearchresultisgivenbythevector

The ithentryof y isequaltothenumberofsearchwordsinthetitleofthe ithbook.

18. If α = a21/a11,then

Theproductwillequal A provided

4 MATRIXALGEBRA

6 Chapter1 •

MatricesandSystemsofEquations

1= w1 + w2 + w3 =3w3 +2w3 + w3 =6w3 andhenceitfollowsthat w3 = 1 6 , w2 = 1 3 , w1 = 1 2 .If

r iscomputedby

timestheweightvector w r = Cw =            1 2 1 5 1 4 1 4 1 2 1 2 1 4 3 10 1 4                       1 2 1 3 1 6            =            43 120 45 120 32 120           

multiplying C

y = AT x

, 2, 2, 1, 1, 2, 1)T

=(1

    10 α 1         a11 a12 0 b     =     a11 a12 αa11 αa12 + b     =     a11 a12 a21 αa12 + b    

αa12 + b = a22 Thuswemustchoose b = a22 αa12 = a22 a21a12 a11

(a)(A + B)2 =(A + B)(A + B)=(A + B)A +(A + B)B = A2 + BA + AB + B2 Forrealnumbers, ab + ba =2ab;however,withmatrices AB + BA isgenerallynotequal to2AB. (b) (A + B)(A B)=(A + B)(A B) =(A + B)A (A + B)B = A2 + BA AB B2

Forrealnumbers, ab ba =0;however,withmatrices AB BA isgenerallynotequal to O

2. Ifwereplace a by A and b bytheidentitymatrix, I,thenbothruleswillwork,since

3. Therearemanypossiblechoicesfor A and B.Forexample,onecouldchoose

4. Toconstructnonzeromatrices A, B, C withthedesiredproperties,ﬁrstﬁndnonzeromatrices C and D suchthat DC = O (seeExercise3).Next,foranynonzeromatrix

Section4 • MatrixAlgebra 7

(A + I)2 = A2 + IA + AI + B2 = A2 + AI + AI + B2 = A2 +2AI + B2 and (A + I)(A I)= A2 + IA AI I 2 = A2 + A A I 2 = A2 I 2

A =     01 00     and B =     11 00     Moregenerallyif A =     ab cacb     B =     dbeb da ea     then AB

O foranychoiceofthescalars a, b, c, d, e.

A,set B = A + D Itfollowsthat BC =(A + D)C = AC + DC = AC + O = AC 5. A2 × 2symmetricmatrixisoneoftheform A =     ab bc     Thus A2 =     a2 + b2 ab + bc ab + bcb2 + c2     If A2 = O,thenitsdiagonalentriesmustbe0. a 2 + b2 =0and b2 + c 2 =0 Thus a = b = c =0andhence A = O. 6. Let D =(AB)C =     a11b11 + a12b21 a11b12 + a12b22 a21b11 + a22b21 a21b12 + a22b22         c11 c12 c21 c22     Itfollowsthat d11 =(a11b11 + a12b21)c11 +(a11b12 + a12b22)c21 = a11b11c11 + a12b21c11 + a11b12c21 + a12b22c21 d12 =(a11b11 + a12b21)c12 +(a11b12 + a12b22)c22 = a11b11c12 + a12b21c12 + a11b12c22 + a12b22c22 d21 =(a21b11 + a22b21)c11 +(a21b12 + a22b22)c21 = a21b11c11 + a22b21c11 + a21b12c21 + a22b22c21 d22 =(a21b11 + a22b21)c12 +(a21b12 + a22b22)c22 = a21b11c12 + a22b21c12 + a21b12c22 + a22b22c22 Ifweset E = A(BC)=     a11 a12 a21 a22         b11c11 + b12c21 b11c12 + b12c22 b21c11 + b22c21 b21c12 + b22c22     thenitfollowsthat e11 = a11(b11c11 + b12c21)+ a12(b21c11 + b22c21) = a11b11c11 + a11b12c21 + a12b21c11 + a12b22c21 Copyright © 2015 Pearson Education, Inc.

8 Chapter1 • MatricesandSystemsofEquations e12 = a11(b11c12 + b12c22)+ a12(b21c12 + b22c22) = a11b11c12 + a11b12c22 + a12b21c12 + a12b22c22 e21 = a21(b11c11 + b12c21)+ a22(b21c11 + b22c21) = a21b11c11 + a21b12c21 + a22b21c11 + a22b22c21 e22 = a21(b11c12 + b12c22)+ a22(b21c12 + b22c22) = a21b11c12 + a21b12c22 + a22b21c12 + a22b22c22 Thus d11 = e11 d12 = e12 d21 = e21 d22 = e22 andhence (AB)C = D = E = A(BC) 9. A2 =           0010 0001 0000 0000           A3 =           0001 0000 0000 0000           and A4 = O.If n> 4,then An = An 4A4 = An 4O = O 10. (a)Thematrix C issymmetricsince CT =(A + B)T = AT + BT = A + B = C (b)Thematrix D issymmetricsince DT =(AA)T = AT AT = A2 = D (c)Thematrix E = AB isnotsymmetricsince ET =(AB)T = BT AT = BA andingeneral, AB = BA. (d)Thematrix F issymmetricsince F T =(ABA)T = AT BT AT = ABA = F (e)Thematrix G issymmetricsince GT =(AB + BA)T =(AB)T +(BA)T = BT AT + AT BT = BA + AB = G (f)Thematrix H isnotsymmetricsince H T =(AB BA)T =(AB)T (BA)T = BT AT AT BT = BA AB = H 11. (a)Thematrix A issymmetricsince AT =(C + CT )T = CT +(CT )T = CT + C = A (b)Thematrix B isnotsymmetricsince BT =(C CT )T = CT (CT )T = CT C = B (c)Thematrix D issymmetricsince AT =(CT C)T = CT (CT )T = CT C = D (d)Thematrix E issymmetricsince ET =(CT C CCT )T =(CT C)T (CCT )T = CT (CT )T (CT )T CT = CT C CCT = E Copyright © 2015 Pearson Education, Inc.

14. If A werenonsingularand

that B = I.Soif B = I,then A mustbesingular.

15. Since

itfollowsfromthedeﬁnitionthat

16. Since

17. If Ax = Ay and x = y,then

Section4 • MatrixAlgebra

(e)Thematrix F issymmetricsince F T =((I + C)(I + CT ))T =(I + CT )T (I + C)T =(I + C)(I + CT )= F (e)Thematrix G isnotsymmetric. F =(I + C)(I CT )= I + C CT CCT F T =((I + C)(I CT ))T =(I CT )T (I + C)T =(I C)(I + CT )= I C + CT CCT F and F T arenotthesame.Thetwomiddleterms C CT and C + CT donotagree. 12. If d = a11a22 a21a12 =0,then 1 d     a22 a12 a21 a11         a11 a12 a21 a22     =          a11a22 a12a21 d 0 0 a11a22 a12a21 d          = I     a11 a12 a21 a22     1 d     a22 a12 a21 a11     =          a11a22 a12a21 d 0 0 a11a22 a12a21 d          = I Therefore 1 d     a22 a12 a21 a11     = A 1 13.

    35 2 3    

(b)

= A,thenitwouldfollowthat A 1AB = A 1A andhence

A 1A = AA 1 = I

A 1 isnonsingularanditsinverseis A

AT (A 1)T =(A 1A)T = I (A 1)T AT =(AA 1)T = I itfollowsthat (A 1)T =(AT ) 1

A mustbesingular,forif A werenonsingular,thenwecould

A 1Ax = A 1Ay x = y

(A1) 1 = A 1 =(A 1)1 Assumetheresultholdsinthecase m = k,thatis, (Ak) 1 =(A 1)k Itfollowsthat (A 1)k+1Ak+1 = A 1(A 1)kAkA = A 1A = I and Ak+1(A 1)k+1 = AAk(A 1)kA 1 = AA 1 = I Copyright © 2015 Pearson Education, Inc.

multiplyby A 1 andget

18. For m =1,

19.

10 Chapter1 •

(A 1)k+1 =(Ak+1) 1

MatricesandSystemsofEquations Therefore

andtheresultfollowsbymathematicalinduction.

If A2 = O,then (I + A)(I A)= I + A A + A2 = I and (I A)(I + A)= I A + A + A2 = I Therefore I A isnonsingularand(I A) 1 = I + A

If Ak+1 = O,then (I + A + + Ak)(I A)=(I + A + + Ak) (A + A2 + + Ak+1) = I Ak+1 = I and (I A)(I + A + + Ak)=(I + A + + Ak) (A + A2 + + Ak+1) = I Ak+1 = I Therefore I A isnonsingularand(I A) 1 = I + A + A2 + + Ak . 21. Since RT R =     cos θ sin θ sin θ cos θ         cos θ sin θ sin θ cos θ     =     10 01     and RRT =     cos θ sin θ sin θ cos θ         cos θ sin θ sin θ cos θ     =     10 01     itfollowsthat R isnonsingularand R 1 = RT 22. G2 =     cos2 θ +sin2 θ 0 0cos2 θ +sin2 θ     = I 23. H 2 =(I 2uu T )2 = I 4uu T +4uu T uu T = I 4uu T +4u(u T u)u T = I 4uu T +4uu T = I (since u T u =1) 24. Ineachcase,ifyousquarethegivenmatrix,youwillendupwiththesamematrix. 25. (a)If A2 = A,then (I A)2 = I 2A + A2 = I 2A + A = I A (b)If A2 = A,then (I 1 2 A)(I + A)= I 1 2 A + A 1 2 A2 = I 1 2 A + A 1 2 A = I and (I + A)(I 1 2 A)= I + A 1 2 A 1 2 A2 = I + A 1 2 A 1 2 A = I Therefore I + A isnonsingularand(I + A) 1 = I 1 2 A 26. (a) D2 =              d2 11 0 0 0 d2 22 ··· 0 . 00 d2 nn              Copyright © 2015 Pearson Education, Inc.

20.

34. False.Forexample,if

35. False.Forexample,if

36. True.If A and B arenonsingular,thentheirproduct AB mustalsobenonsingular.Usingthe resultfromExercise23,wehavethat(AB)T isnonsingularand((AB)

followsthenthat

Section4 • MatrixAlgebra 11 Sinceeachdiagonalentryof D isequaltoeither0or1,itfollowsthat d2 jj = djj ,for j =1,...,n andhence D2 = D (b)If A = XDX 1,then A2 =(XDX 1)(XDX 1)= XD(X 1X)DX 1 = XDX 1 = A 27. If A isaninvolution,then A2 = I anditfollowsthat B2 = 1 4 (I + A)2 = 1 4 (I +2A + A2)= 1 4 (2I +2A)= 1 2 (I + A)= B C 2 = 1 4 (I A)2 = 1 4 (I 2A + A2)= 1 4 (2I 2A)= 1 2 (I A)= C So B and C arebothidempotent. BC = 1 4 (I + A)(I A)= 1 4 (I + A A A2)= 1 4 (I + A A I)= O 28. (AT A)T = AT (AT )T = AT A (AAT )T =(AT )T AT = AAT 29. Let A and B besymmetric n × n matrices.If(AB)T = AB,then BA = BT AT =(AB)T = AB Conversely,if BA = AB,then (AB)T = BT AT = BA = AB 30. (a) BT =(A + AT )T = AT +(AT )T = AT + A = B CT =(A AT )T = AT (AT )T = AT A = C (b) A = 1 2 (A + AT )+ 1 2 (A AT )

A =     23 23     ,B =     14 14     , x =     1 1     then Ax = Bx =     5 5    

however, A = B

A =     10 00     and B =     00 01    

A and B mustbesingular,however, A + B

thenitiseasytoseethatboth

,whichis nonsingular.

1 =((

T )

AB)

)T .It

5

7. A canbereducedtotheidentitymatrixusingthreerowoperations

Theelementarymatricescorrespondingtothethreerowoperationsare

12 Chapter1 • MatricesandSystemsofEquations

ELEMENTARYMATRICES

(a)    01 10   ,typeI (b)Thegivenmatrixisnotanelementarymatrix.Itsinverseisgivenby      1 2 0 0 1 3      (c)        100 010 501        ,typeIII (d)        100 01/50 001        ,typeII

(c)Since C = FB = FEA where F and E areelementarymatrices,itfollowsthat C isrowequivalentto A.

(b) E 1 1 =        100 310 001        , E 1 2 =        100 010 201        , E 1 3 =        100 010 0 11        Theproduct L = E 1 1 E 1 2 E 1 3 islowertriangular. L =        100 310 2 11       

    21 64     →     21 01     →     20 01     →     10 01    

E1 =     10 31     ,E2 =     1 1 01     ,E3 =     1 2 0 01     So E3E2E1A = I andhence A = E 1 1 E 1 3 E 1 3 =     10 31         11 01         20 01     and A 1 = E3E2E1. 8. (b)     10 11         24 05     (d)        100 210 3 21               212 032 002        9. (a)        101 334 223               12 3 11 1 0 23        =        100 010 001        Copyright © 2015 Pearson Education, Inc.

13. (a)If E isanelementarymatrixoftypeIortypeII,then E issymmetric.Thus ET = E is anelementarymatrixofthesametype.If E istheelementarymatrixoftypeIIIformed byadding α timesthe ithrowoftheidentitymatrixtothe jthrow,then ET isthe elementarymatrixoftypeIIIformedfromtheidentitymatrixbyadding α timesthe jth rowtothe ithrow.

(b)Ingeneral,theproductoftwoelementarymatriceswillnotbeanelementarymatrix. Generally,theproductoftwoelementarymatriceswillbeamatrixformedfromthe identitymatrixbytheperformanceoftworowoperations.Forexample,if

Section5 • ElementaryMatrices 13        12 3 11 1 0 2 3               101 334 223        =        100 010 001        10. (e)        1 10 01 1 001        12. (b) XA + B = C X =(C B)A 1 =     8 14 1319     (d) XA + C = X XA XI = C X(A I)= C X = C(A I) 1 =     2 4 36    

E1 =        100 210 000        and E2 =        100 010 201        then E1 and E2 areelementarymatrices,but E1E2 =        100 210 201        isnotanelementarymatrix. 14. If T = UR,then tij = n k=1 uikrkj Since U and R areuppertriangular ui1 = ui2 = = ui,i 1 =0 rj+1,j = rj+2,j = rnj =0 If i>j,then tij = j k=1 uikrkj + n k=j+1 uikrkj = j k=1 0 rkj + n k=j+1 uik0 =0 Copyright © 2015 Pearson Education, Inc.

Therefore T isuppertriangular. If i = j,then

15. Ifweset x =(2, 1 4)T ,then Ax =2a1 +1a2 4a3 = 0

Thus x isanonzerosolutiontothesystem Ax = 0.Butifahomogeneoussystemhasa nonzerosolution,thenitmusthaveinﬁnitelymanysolutions.Inparticular,if c isanyscalar, then cx isalsoasolutiontothesystemsince

A(cx)= cAx = c0 = 0

Since Ax = 0 and x = 0,itfollowsthatthematrix A mustbesingular.(SeeTheorem1.5.2)

16. If a1 =3a2 2a3,then

Therefore x =(1, 3, 2)T isanontrivialsolutionto Ax = 0.ItfollowsfromTheorem1.5.2 that A mustbesingular.

17. If x0 = 0 and Ax0 = Bx0,then Cx0 = 0 anditfollowsfromTheorem1.5.2that C mustbe singular.

18. If B issingular,thenitfollowsfromTheorem1.5.2thatthereexistsanonzerovector x such that Bx = 0.If C = AB,then

Cx = ABx = A0 = 0

Thus,byTheorem1.5.2, C mustalsobesingular.

19. (a)If U isuppertriangularwithnonzerodiagonalentries,thenusingrowoperationII, U can betransformedintoanuppertriangularmatrixwith1’sonthediagonal.Rowoperation IIIcanthenbeusedtoeliminatealloftheentriesabovethediagonal.Thus, U isrow equivalentto I andhenceisnonsingular.

(b)Thesamerowoperationsthatwereusedtoreduce U totheidentitymatrixwilltransform I into U 1.RowoperationIIappliedto I willjustchangethevaluesofthediagonal entries.WhentherowoperationIIIstepsreferredtoinpart(a)areappliedtoadiagonal matrix,theentriesabovethediagonalareﬁlledin.Theresultingmatrix, U 1,willbe uppertriangular.

20. Since A isnonsingularitisrowequivalentto I.Hence,thereexistelementarymatrices

E1,E2,...,Ek suchthat

Itfollowsthat

Ek E1A = I

A 1 = Ek E1 and

Ek ··· E1B = A 1B = C

Thesamerowoperationsthatreduce A to I,willtransform B to C.Therefore,thereduced rowechelonformof(A | B)willbe(I | C).

14 Chapter1 •

MatricesandSystemsofEquations

tjj = tij = i 1 k=1 uikrkj + ujj rjj + n k=j+1 uikrkj = i 1 k=1 0 rkj + ujj rjj + n k=j+1 uik0 = ujj rjj Therefore tjj = ujj rjj j =1,...,n

3a2 +2a3 = 0

Copyright © 2015 Pearson Education, Inc. Linear Algebra with Applications 9th Edition Leon Solutions Manual Full Download: http://testbanktip.com/download/linear-algebra-with-applications-9th-edition-leon-solutions-manual/ Download all pages and all chapters at: TestBankTip.com