Dedication
To all the students who have used the various editions of our book over the years
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Appendix
Appendix
Appendix
StudentManualforAndrilliandHecker- ElementaryLinearAlgebra,5theditionSection1.1
(4)(a)First,we findthevector v havingthegiveninitialandterminalpointbysubtracting: [10
10 11] [ 4
9]= v .Next,thedesiredpointisfoundbyadding 2 3 v (which is 2 3 thelengthof v)tothevectorfortheinitialpoint
(5)Let v representthegivenvector.Thenthedesiredunitvector
u isshorterthan v since k
; neithervectorislongerbecause u = v
(6)Twononzerovectorsareparallelifandonlyifoneisascalarmultipleoftheother.
(a) [12 16] and [9 12] areparallelbecause 3 4 [12 16]=[9 12]. (c) [ 2 3 1] and [6 4 3] arenotparallel.Toshowwhynot,supposetheyare.Thentherewould bea ∈ R suchthat [ 2 3 1]=[6 4 3].Comparing firstcoordinatesshowsthat 2 =6, or = 3.Howevercomparingsecondcoordinatesshowsthat 3 = 4,or = 4 3 instead.But cannothavebothvalues.
(7)(a) 3[ 2 4 5]=[3( 2) 3(4) 3(5)]=[ 6 12 15]
(c) [ 2 4 5]+[ 5 3 6]=[( 2+( 5)) (4+( 3)) (5+6)]=[ 7 1 11] (e) 4[ 5 3 6] 5[ 2 4 5]=[4( 5) 4( 3) 4(6)] [5( 2) 5(4) 5(5)]=[ 20 12 24] [ 10 20 25]= [( 20 ( 10)) ( 12 20) (24 25)]=[ 10 32 1]
(8)(a) x+y =[ 1 5]+[2 4]=[( 1+2) (5 4)]=[1 1], x y =[ 1 5] [2 4]=[( 1 2) (5 ( 4))]= [ 3 9], y x =[2 4] [ 1 5]=[(2 ( 1)) (( 4) 5)]=[3 9] (seeFigure3,nextpage)
(c) x + y =[2 5 3]+[ 1 3 2]=[(2+( 1)) (5+3) (( 3)+( 2))]=[1 8 5], x y =[2 5 3] [ 1 3 2]=[(2 ( 1)) (5 3) (( 3) ( 2))]=[3 2 1], y x =[ 1 3 2] [2 5 3]=[(( 1) 2) (3 5) (( 2) ( 3))]=[ 3 2 1] (seeFigure4, nextpage)
(10)Ineachpart,considerthecenteroftheclocktobetheorigin.
(a)At 12 PM,thetipoftheminutehandisat (0 10).At 12:15 PM,thetipoftheminutehandis at (10 0).To findthedisplacementvector,wesubtractthevectorfortheinitialpointfromthe vectorfortheterminalpoint,yielding [10 0] [0 10]=[10 10]
(b)At 12 PM,thetipoftheminutehandisat (0 10).At 12:40 PM,theminutehandmakesa 210◦ anglewiththepositive -axis.So,asshowninFigure1.10inthetextbook,theminutehandmakes thevector v =[kvk cos kv k sin ]=[10cos(210◦ ) 10sin(210◦ )]= h10 ³ √3 2 ´ 10 ¡ 1 2 ¢i = [ 5√3 5].To findthedisplacementvector,wesubtractthevectorfortheinitialpointfromthe vectorfortheterminalpoint,yielding [ 5√3 5] [0 10]=[ 5√3 15]
Copyright c ° 2016ElsevierLtd.Allrightsreserved.2
StudentManualforAndrilliandHecker-
¤.Finally,
[0 0462 0 2646 0 3015]
(20)AsshowninFigure1.10inthetextbook,foranyvector v ∈ R2 v =[kvk cos
,where
is theangle v makeswiththepositive -axis.Thus,ifwelet =
and
,then
.Now,since theweightisnotmoving,
.Thatis,
.Using the firstcoordinatesgivesus
Also,
(23)If =0,wearedone.Otherwise,
.Usingthesecondcoordinatesgivesus
produces
(bypart(1)ofTheorem1.3) = 0,bytheabove.
)x = 0 (bypart(7)ofTheorem 1.3) =⇒ x = 0.Thuseither =0 or x = 0
(25)(a)False.Thelengthof
.Theformulagivenintheproblemismissingthe squareroot.So,forexample,thelengthof [0 3 4] isactually 5, not 02 +32 +42 =25.
(b)True.Thisiseasilyprovedusingparts(1)and(2)ofTheorem1.3.
(c)True. [2 0 3]=2[1 0 0]+( 3)[0 0 1]
(d)False.Twononzerovectorsareparallelifand onlyifoneisascalarmultipleoftheother.To showwhy [3 5 2] and [6 10 5] arenotparallel,supposetheyare.Thentherewouldbea ∈ R suchthat [3 5 2]=[6 10 5].Comparing firstcoordinatesshowsthat 3 =6,or =2.But thisvalueof mustworkinallcoordinates.However,itdoesnotworkinthethirdcoordinate, since (2)(2) =5
(e)True.Multiplybothsidesof x = 0 by 1 .
(f)False.Parallelvectorscanbeinoppositedirections.Forexample, [1 0] and [ 2 0] areparallel, butarenotinthesamedirection.
(g)False.ThepropertiesofvectorsinTheorem1.3areindependentofthelocationoftheinitial pointsofthevectors.
(h)False. ||[9 8 4]|| = p92 +( 8)2 +( 4)2 = √81+64+16= √161 1
Copyright c ° 2016ElsevierLtd.Allrightsreserved.4
Section1.2
(1)Ineachpart,weusetheformula cos = x y kxkkyk
(a) cos =
=arccos( 27 5√37 ),orabout 152 6◦ ,or 2 66 radians.
(c) cos = x y k
=arccos(0),whichis 90
,or 2 radians.
.Andso,usingacalculatoryields
(4)Whenaforce f movesanobjectbyadisplacement d,theworkperformedis f d.
(b)First,weneedtocompute f .Aunitvectorinthedirectionof
,wherewehaverationalizedthedenominators.
Next,weneedtocompute d.Aunitvectorinthedirectionof
Hence,theworkperformedis
(6)Inallparts,let
(b)
.Andso,
isasumofsquares,eachofwhichmustbenonnegative.Hence,thesumisalsononnegative,and soitssquarerootisdefined.Thus,
(c)Suppose x x =0.Frompart(b),
,foreach
,sincealltermsinthe sumarenonnegative.Hence,
,becauseitisasquare.Henceeach
=0.Therefore, x = 0.
StudentManualforAndrilliandHecker- ElementaryLinearAlgebra,5theditionSection1.2
(8)Bypart(2)ofTheorem1.5, (a b) (a b)= ka bk2 ≥ 0.Hence,byparts(5)and(6)of Theorem1.5, (a a) (b a) (a b)+(b b) ≥ 0.Byparts(1)and(2)ofTheorem1.5,wehave kak2 2(a b)+ kbk2 ≥ 0.Thus, 1 2(a b)+1 ≥ 0 because a and b areunitvectors,andso a b ≤ 1
(16)Positionthecubesothatoneofitscornersisattheoriginandthethreeedgesfromthatcornerare alongthepositive -, -,and -axes.Thenthecornerdiagonallyoppositefromtheoriginisatthe point ().Thevectorrepresentingthisdiagonalis d =[
].
(a)Thelengthofthediagonal = kdk = √
(b)Theangle d makeswithasideofthecubeisequaltotheanglebetween d and [ 0 0],which is arccos ³
0 955 radians.
(20)Ineachpart,if v isthegivenvector,thenbyTheorem1.11, x = projv
),where projv x isparallelto v and (x projv x) isorthogonalto v (a)Inthiscase, x
Andso projv x isclearlyparallelto v since projv x = ¡
¢ v .Also, (x projv x)=[
and (x projv x) areorthogonal,since projv x (
,andthat
and
(23)(a)Notethat projx y =[ 8 5 6 5
isorthogonalto x.Let w = y projx y Thensince y = projx y +(y projx y)
wehave y = 2 5 x + w where w isorthogonal to x
(25)(a)Truebypart(4)ofTheorem1.5.
(b)True.Theorem1.7(theCauchy-SchwarzInequality)statesthat |x y| ≤ kxkkyk.Since x y ≤ |x y |,wehave x y ≤ kxkkyk.Butthen,because kxk 0,wecandividebothsidesby kxk toobtainthedesiredinequality
(c)False.Forexample,if x =[1 0 0] and y =[0 1 0],then kx yk = k[1 1 0]k = p12 +( 1)2 +02 = √2,and kxk kyk = k[1 0 0]k k[0 1 0k = √12 +02 +02 √02 +12 +02 =1 1=0.However, √2 £ 0.
(d)False.Theorem1.9showsthatif 2 ,then x · y 0.(Remember, isdefinedtobe ≤ .)
(e)True.If = ,then e e =0(0)+ +1(0) |{z} thterms + +0(1) |{z} thterms +0(0)+ +0(0)=0+ +0=0
(f)False. proja b = ³ a b kak2 ´ a,andso proja b isparallelto a,sinceitisascalarmultipleof a.Thus,if proja b = b, a and b areparallel,notperpendicular.Foraparticularexample,suppose a =[1 0] and b =[2 0].Then proja b
0] k[1
0]k ´ [1 0]= ³ 1(2)+0(0) √12 +02 ´ [1 0]=2[1 0]=[2 0]= b.However, a =[1 0] and b =[2 0] areparallel(since b =2a)and not perpendicular(since a b =[1 0] [2 0]=1(2)+0(0)=2 =0.
Section1.3
(1)(b)Let =max{|| |
Theproofisasfollows:
y
kxk + k±y k bytheTriangleInequality
byTheorem1.1
and
(2)(b)Theconverseis:Ifanintegerhastheform 3 +1,thenitalsohastheform 6 5,where and areintegers.Foracounterexample,considerthatthenumber 4=3 +1 with =1,butthere isnointeger suchthat 4=6 5,sincethisequationimplies = 3 2 ,whichisnotaninteger.
Copyright c ° 2016ElsevierLtd.Allrightsreserved.7
StudentManualforAndrilliandHecker- ElementaryLinearAlgebra,5theditionSection1.4
(c) 4 ⎡ ⎣ 423 05 1 61 2 ⎤ ⎦ = ⎡ ⎣ 4( 4)4(2)4(3) 4(0)4(5)4( 1) 4(6)4(1)4( 2)
(e)Impossible. C isa 2 × 2 matrix, F isa 3 × 2 matrix,and E isa 3 × 3 matrix.Hence, 3F is 3 × 2 and E is 3 × 3.Thus, C, 3F,and E havedifferentsizes.However,inordertoaddmatrices, theymustbethesamesize.
(g) 2
2( 4)2(2)2(3) 2(0)2(5)2( 1) 2(6)2(1)2( 2)
( 8)+( 9)+( 6)4+9+16+( 15)+0 0+( 3)+( 2)10+0+( 2)( 2)+6+4 12+( 18)+( 3)2+( 21)+1( 4)+6+( 1)
(( 4)+3)(0+1)(6+6) (2+( 3))(5+0)(1+7) (3+5)(( 1)+( 2))(( 2)+( 2))
(l)Impossible.Since C isa 2 × 2 matrix,both C and 2C are 2 × 2.Also,since F isa 3 × 2 matrix, 3F is 3 × 2.Thus, 2C and 3F havedifferentsizes.However,inordertoaddmatrices,they mustbethesamesize.
(n) (B A) = ⎡
(2)Foramatrixtobesquare,itmusthavethesamenumberofrowsascolumns.Amatrix X isdiagonal ifandonlyifitissquareand =0 for = .Amatrix X isuppertriangularifandonlyifitis squareand =0 for .Amatrix X islowertriangularifandonlyifitissquareand =0 for .Amatrix X issymmetricifandonlyifitissquareand = forall .Amatrix X is skew-symmetricifandonlyifitissquareand = forall .Noticethatthisimpliesthatif X isskew-symmetric,then =0 forall .Finally,thetransposeofan × matrix X isthe × matrixwhose( )thentryis .Wenowcheckeachmatrixinturn. A isa 3 × 2 matrix,soisnotsquare.Thereforeitcannotbediagonal,uppertriangular,lower triangular,symmetric,orskew-symmetric.Finally, A = ∙ 106 410 ¸
Copyright c ° 2016ElsevierLtd.Allrightsreserved.10
(5)(d)Thematrix(callit A)mustbeasquarezeromatrix;thatis, O ,forsome .First,since A is diagonal,itmustbesquare(bydefinitionofadiagonalmatrix).Toprovethat =0 forall , weconsidertwocases:First,if = ,then =0 since A isdiagonal.Second,if = ,then = =0 since A isskew-symmetric(
= forall ,implying =0 forall ).Hence, everyentry =0,andso A isazeromatrix.
(10)(a)Part(1):Let B = A and C =(A ) .Then =
=
.Sincethisistrueforall and , wehave C = A;thatis, (A ) = A
(b)Part(2),SubtractionCase:Let D = A B, F =(A B) and G = A B .Then
=
=
=
.Sincethisistrueforall and ,wehave F = G;thatis, (A B) = A B .
(c)Part(3):Let B = (A ), D = A and F =(A) .Then
=
=
.Sincethisis trueforall and ,wehave F = B;thatis, (A) = (A ).
(13)(a)Trace(B) = 11 + 22 =2+( 1)=1,trace(C) = 11 + 22 =( 1)+1=0, trace(E) = 11 + 22 + 33 =0+( 6)+0= 6,trace(F) = 11 + 22 + 33 + 44 =1+0+0+1=2,trace(G) = 11 + 22 + 33 =6+6+6=18,trace(H) = 11 + 22 + 33 + 44 = 0+0+0+0=0,trace(J) = 11 + 22 + 33 + 44 =0+0+1+0=1,trace(K) = 11 + 22 + 33 + 44 =1+1+1+1=4,trace(L) = 11 + 22 + 33 =1+1+1=3,trace(M) = 11 + 22 + 33 = 0+0+0=0,trace(N) = 11 + 22 + 33 =1+1+1=3,trace(P) = 11 + 22 =0+0=0, trace(Q) = 11 + 22 + 33 =( 2)+0+3=1.
(c)Notnecessarilytruewhen 1.Considermatrices L and N inExercise2.Frompart(a)ofthis exercise,trace(L) =3= trace(N).However, L = N. (14)(a)False.Themaindiagonalentriesofa 5 × 6 matrix A are 11 , 22 , 33 , 44 ,and 55 .Thus,there are 5 entriesonthemaindiagonal.(Thereisnoentry 66 becausethereisno 6throw.)
(b)True.Suppose A islowertriangular.Then =0 whenever .If B = A ,then = =0, if ,andso B isuppertriangular.
(c)False.Thesquarezeromatrix O isbothskew-symmetricanddiagonal. (d)True.Thisfollowsfromthedefinitionofaskew-symmetricmatrix.
(e)True. ¡ ¡A + B¢¢ = ³ ¡A + B¢ ´ (bypart(3)ofTheorem1.13) = ³ ³¡A ¢ + B ´´ (bypart(2)ofTheorem1.13) = ¡A + B ¢ (bypart(1)ofTheorem1.13) = A + B (bypart (5)ofTheorem1.12) = B + A (bypart(1)ofTheorem1.12).
Section1.5
(1)(b) BA = ⎡ ⎣ 536 380 204 ⎤ ⎦ ⎡ ⎣ 23 65 1 4 ⎤ ⎦ = ⎡ ⎣ (( 5)( 2)+(3)(6)+(6)(1))(( 5)(3)+(3)(5)+(6)( 4)) ((3)( 2)+(8)(6)+(0)(1))((3)(3)+(8)(5)+(0)( 4)) (( 2)( 2)+(0)(6)+(4)(1))(( 2)(3)+(0)(5)+(4)( 4)) ⎤ ⎦ = ⎡
Copyright c ° 2016ElsevierLtd.Allrightsreserved.13
StudentManualforAndrilliandHecker- ElementaryLinearAlgebra,5theditionSection1.5
Also, PN = ∙ 3 1 47 ¸ ∙ 00 00 ¸ = ∙ ((3)(0)+( 1)(0))((3)(0)+( 1)(0)) ((4)(0)+(7)(0))((4)(0)+(7)(0)) ¸ = ∙ 00 00 ¸.
Hence, NP = PN,andso N and P commute.
(3)(a)(2ndrowof BG)=(2ndrowof B)G = £ 380 ¤ ⎡ ⎣ 510 0 2 1 103
.Theentriesobtainedfrom thisare:
1stcolumnentry =((3)(5)+(8)(0)+(0)(1))=15, 2ndcolumnentry =((3)(1)+(8)( 2)+(0)(0))= 13, 3rdcolumnentry =((3)(0)+(8)( 1)+(0)(3))= 8
Hence,(2ndrowof BG) =[15 13 8]
(c)(1stcolumnof SE) = S(1stcolumnof E)= £ 6 432 ¤
=[(6)(1)+( 4)(1)+(3)(0)+(2)(1)]=[4]
(4)(a)Valid,byTheorem1.16,part(1).
(b)Invalid.Theequationclaimsthatallpairsofmatricesthatcanbemultipliedinbothorders commute.However,parts(a)and(c)ofExercise2illustratetwodifferentpairsofmatricesthat donotcommute(seeabove).
(c)Valid,byTheorem1.16,part(1).
(d)Valid,byTheorem1.16,part(2).
(e)Valid,byTheorem1.18.
(f)Invalid.Foracounterexample,considerthematrices L and M fromExercises1through3.
Usingourcomputationof ML fromExercise2(a),above, L(ML)=
∙ ((10)(62)+(9)(134))((10)(56)+(9)(120)) ((8)(62)+(7)(134))((8)(56)+(7)(120)) ¸ = ∙ 18261640 14341288 ¸ , but L2 M =(LL)M = µ∙ 109 87 ¸
=
((10)(10)+(9)(8))((10)(9)+(9)(7)) ((8)(10)+(7)(8))((8)(9)+(7)(7))
136121 ¸ ∙ 7 1 113 ¸ = ∙ ((172)(7)+(153)(11))((172)( 1)+(153)(3)) ((136)(7)+(121)(11))((136)( 1)+(121)(3)) ¸ = ∙ 2887287 2283227 ¸
(g)Valid,byTheorem1.16,part(3).
(h)Valid,byTheorem1.16,part(2).
(i)Invalid.Foracounterexample,considerthematrices A and K fromExercises1through3.Note that A isa 3 × 2 matrixand K isa 2 × 3 matrix.Therefore, AK isa 3 × 3 matrix.Hence, (AK) isa 3 × 3 matrix.However, A isa 2 × 3 matrixand K isa 3 × 2 matrix.Thus, A K isa 2 × 2 matrix,andsocannotequal (AK) Theequationisalsofalseingeneralforsquare matrices,whereitisnotthesizesofthematrices
Copyright c ° 2016ElsevierLtd.Allrightsreserved.16
thatcausetheproblem.Forexample,let A =
.Then,
However,therearesomerareinstancesforwhichtheequationistrue,suchaswhen A = K,or wheneither A or K equals I .Anotherexampleforwhichtheequationholdsisfor A = G and K = H,where G and H arethematricesusedinExercises1,2,and3.
(j)Valid,byTheorem1.16,part(3),andTheorem1.18.
(5)To findthetotalinsalariespaidbyOutlet1,wemustcomputethetotalamountpaidtoexecutives,the totalamountpaidtosalespersons,andthetotalamountpaidtoothers,andthenaddtheseamounts. ButeachofthesetotalsisfoundbymultiplyingthenumberofthattypeofemployeeworkingatOutlet 1bythesalaryforthattypeofemployee.Hence,thetotalsalarypaidatOutlet1is (3)(30000) | {z } Executives +(7)(22500) | {z } Salespersons +(8)(15000)
Notethatthisisthe(1 1)entryobtainedwhenmultiplyingthetwogivenmatrices.Asimilaranalysis showsthatmultiplyingthetwogivenmatricesgivesallthedesiredsalaryandfringebenefittotals.In particular,if
11 =(3)(30000)+(7)(22500)+(8)(15000)=367500
12 =(3)(7500)+(7)(4500)+(8)(3000)=78000
21 =(2)(30000)+(4)(22500)+(5)(15000)=225000
22 =(2)(7500)+(4)(4500)+(5)(3000)=48000
31 =(6)(30000)+(14)(22500)+(18)(15000)=765000
32 =(6)(7500)+(14)(4500)+(18)(3000)=162000
41 =(3)(30000)+(6)(22500)+(9)(15000)=360000
42 =(3)(7500)+(6)(4500)+(9)(3000)=76500
Hence,wehave:
(7)Tocomputethetonnageofaparticularchemicalappliedtoagiven field,foreachtypeoffertilizerwe mustmultiplythepercentconcentrationofthechemicalinthefertilizerbythenumberoftonsofthat fertilizerappliedtothegiven field.Afterthisisdoneforeachfertilizertype,weaddtheseresultsto
(15)SeveralcrucialstepsinthefollowingproofsrelyonTheorem1.5fortheirvalidity.
(a)ProofofPart(2):The ( ) entryof A(B + C)
= (throwof A) ( thcolumnof (B + C))
= (throwof A) ( thcolumnof B + thcolumnof C)
= (throwof A)·( thcolumnof B)
+(throwof A)·( thcolumnof C)
=(( ) entryof AB) + (( ) entryof AC)
= ( ) entryof (AB + AC)
(b)ProofofPart(3):The ( ) entryof (A + B)C
= (throwof (A + B)) ( thcolumnof C)
= ((throwof A)+(throwof B)) ( thcolumnof C)
= (throwof A) ( thcolumnof C)
+(throwof B) ( thcolumnof C)
=(( ) entryof AC) + (( ) entryof BC)
= ( ) entryof (AC + BC)
(c)Forthe firstequationinPart(4),the ( ) entryof (AB)
= ((throwof A)·( thcolumnof B))
=((throwof A))·( thcolumnof B)
=(throwof A) ( thcolumnof B)
=( ) entryof (A)B
Similarly,the ( ) entryof (AB)
= ((throwof A) ( thcolumnof B))
=(throwof A) (( thcolumnof B))
=(throwof A) ( thcolumnof B)
=( ) entryof A(B)
(20)(a)ProofofPart(1):Weuseinductiononthevariable .BaseStep: A+0 = A = A I =
InductiveStep:Assume A+ = A A forsome ≥ 0.Wemustprove A+(
+1) = A A+1 .But A+(+1) = A(+)+1 = A+ A =(A A )A (bytheinductivehypothesis) = A
(A A)= A
(b)ProofofPart(2):Again,weuseinductiononthevariable .BaseStep: (A )0 = I = A0 = A0 .
.
InductiveStep:Assume (A ) = A forsomeinteger ≥ 0.Wemustprove (A )+1 = A(+1) . But (A )+1 =(A ) A (bydefinition) = A A (bytheinductivehypothesis) = A+ (bypart (1)) = A(+1) .Finally,reversingtherolesof and intheproofaboveshowsthat (A ) = A , whichequals A
(29)(a)Consideranymatrix A oftheform ∙ 10 0 ¸.Then A2 = ∙ 10 0 ¸ ∙ 10 0 ¸ = ∙ ((1)(1)+(0)())((1)(0)+(0)(0)) (()(1)+(0)())(()(0)+(0)(0)) ¸ = A.So,forexample, ∙ 10 10 ¸ isidempotent.
Copyright c ° 2016ElsevierLtd.Allrightsreserved.20
(30)(b)Consider
Settingtheseequalshowsthat 21 =0
(()(0)+(12 )(1))(()(0)+(12 )(0)) ((0)(0)+()(1))((0)(0)+()(0)) ¸
((0)()+(0)(0))((0)(12 )+(0)()) ((1)()+(0)(0))((1)(12 )+(0)())
12 ¸
Setting AD = DA showsthat 12 =0,andso A = I2 . Finally,wemustshowthat I2 actuallydoescommutewithevery 2 × 2 matrix.If M isa 2 × 2 matrix, then (I2 )M = (I2 M)= M.Similarly, M(I2 )= (MI2 )= M.Hence I2 commuteswith M (32)(a)True.Thisisa“boxed”resultgiveninthesection.
(b)True. D (A + B)= DA + DB (bypart(2)ofTheorem1.16) = DB + DA (bypart(1)ofTheorem 1.12)
(c)True.Thisispart(4)ofTheorem1.16.
(d)False.Let D = ∙ 11 0 1 ¸ ,andlet E = ∙ 1 2 01 ¸.Now, DE = ∙ 11 0 1 ¸ ∙ 1 2 01 ¸ = ∙ ((1)(1)+(1)(0))((1)( 2)+(1)(1)) ((0)(1)+( 1)(0))((0)( 2)+( 1)(1)) ¸ = ∙ 1 1 0 1 ¸.Hence, (DE)2 = ∙ 1 1 0 1 ¸ ∙ 1 1 0 1 ¸ = ∙ ((1)(1)+( 1)(0))((1)( 1)+( 1)( 1)) ((0)(1)+( 1)(0))((0)( 1)+( 1)( 1)) ¸ = I2 But, D2 = ∙ 11 0 1 ¸ ∙ 11 0 1 ¸ = ∙ ((1)(1)+(1)(0))((1)(1)+(1)( 1)) ((0)(1)+( 1)(0))((0)(1)+( 1)( 1)) ¸ = I2 .Also, E2 = ∙ 1 2 01 ¸ ∙ 1 2 01 ¸ = ∙ ((1)(1)+( 2)(0))((1)( 2)+( 2)(1)) ((0)(1)+(1)(0))((0)( 2)+(1)(1)) ¸ = ∙ 1 4 01 ¸
Hence, D2 E2 = I2 E2 = E2 = I2 .Thus, (DE)2 = D2 E2 .Theproblemhereisthat D and E do
Copyright c ° 2016ElsevierLtd.Allrightsreserved.21
StudentManualforAndrilliandHecker- ElementaryLinearAlgebra,5theditionChapter1Review
notcommute.For, iftheydid,wewouldhave (DE)2 =(DE)(DE)= D(ED)E = D(DE)E = (DD)(EE)= D2 E2
(e)False.Seetheanswerforpart(i)ofExercise4foracounterexample.WeknowfromTheorem 1.18that (DE) = E D
.Soif D and E
donotcommute,then (DE) = D E .
(f)False.Seetheanswerforpart(b)ofExercise30foracounterexample. DE = O wheneverevery rowof D isorthogonaltoeverycolumnof E,butneithermatrixisforcedtobezerofor DE = O tobetrue.
(g)False.Consider
Chapter1ReviewExercises
(2) kxk =
0 5955 0 7444].Thisisslightlylongerthan x becausewehave divided x byascalarwithabsolutevaluelessthan 1,whichamountstomultiplying x byascalarhavingabsolutevaluegreaterthan 1.
(4)WewilluseNewton’sSecondLawwhichstatesthat f = a,orequivalently, a = 1 f .Fortheforce
6].Similarly,forthe
(8)Work=
(10)First, x = 0,orelse projx y isnotdefined.Also, y = 0,sincethatwouldimply projx y = y.Now, assume x k y.Then,thereisascalar =0 suchthat y = x.Hence, projx y =
kxk
kxk2 ´ x = x = y ,contradictingtheassumptionthat y = projx y
