Solution Manual for Elementary Linear Algebra, 12th Edition Howard Anton by Examexperts

1.1 Introduction to Systems of Linear Equations

1. (a) Thisisalinearequationin1 x ,2 x ,and3 x .

(b) Thisisnotalinearequationin1 x ,2 x ,and3 x becauseoftheterm13 xx

(d) Thisisnotalinearequationin1 x ,2 x ,and3 x becauseoftheterm2 1x

(e) Thisisnotalinearequationin1 x ,2 x ,and3 x becauseoftheterm3/5 1x .

(f) Thisisalinearequationin1 x ,2 x ,and3 x

2. (a) Thisisalinearequationin x and y .

(b) Thisisnotalinearequationin x and y becauseoftheterms1/3 2 x and3 y .

(d) Thisisnotalinearequationin x and y becauseoftheterm 7cos x .

(e) Thisisnotalinearequationin x and y becauseoftheterm xy

(f) Wecanrewritethisequationintheform 7 xy thusitisalinearequationin x and y

3. (a)

(b)

1111221 2112222 axaxb axaxb

1111221331 2112222332 3113223333 axaxaxb axaxaxb axaxaxb (c)

1111221331441 2112222332442 axaxaxaxb axaxaxaxb

4. (a)

11121 21222 aab aab

1112131 2122232 3132333 aaab aaab aaab

111213141 212223242 aaaab aaaab

9. Thevaluesin(a),(d),and(e)satisfyallthreeequations–these3-tuplesaresolutionsofthesystem. The3-tuplesin(b)and(c)arenotsolutionsofthesystem.

10. Thevaluesin(b),(d),and(e)satisfyallthreeequations–these3-tuplesaresolutionsofthesystem. The3-tuplesin(a)and(c)arenotsolutionsofthesystem.

11. (a) Wecaneliminate x fromthesecondequationbyadding2timesthefirstequationtothesecond.Thisyields thesystem

Thesecondequationiscontradictory,sotheoriginalsystemhasnosolutions.Thelinesrepresentedbythe equationsinthatsystemhavenopointsofintersection(thelinesareparallelanddistinct).

(b) Wecaneliminate x fromthesecondequationbyadding2timesthefirstequationtothesecond.Thisyields thesystem

Thesecondequationdoesnotimposeanyrestrictionon x and y thereforewecanomitit.Thelines representedbytheoriginalsystemhaveinfinitelymanypointsofintersection.Solvingthefirstequationfor x weobtain  1 22 xy .Thisallowsustorepresentthesolutionusingparametricequations

1 2, 2 xtyt

wheretheparameter t isanarbitraryrealnumber.

20 28 xy y

Fromthesecondequationweobtain  4 y .Substituting4for y intothefirstequationresultsin  8 x Therefore,theoriginalsystemhastheuniquesolution

8,4xy

Therepresentedbytheequationsinthatsystemhaveonepointofintersection:   8,4

12. Wecaneliminate x fromthesecondequationbyadding2timesthefirstequationtothesecond.Thisyields thesystem

 23 02 xya ba

If 20ba (i.e.,  2 ba )thenthesecondequationimposesnorestrictionon x and y ;consequently,the systemhasinfinitelymanysolutions.

If 20ba (i.e.,  2 ba )thenthesecondequationbecomescontradictorythusthesystemhasnosolutions. Therearenovaluesof a and b forwhichthesystemhasonesolution.

13. (a) Solvingtheequationfor x weobtain 35 77 xy thereforethesolutionsetoftheoriginalequationcanbe describedbytheparametricequations

  35 77, xtyt

wheretheparameter t isanarbitraryrealnumber.

(b) Solvingtheequationfor1 x weobtain 754 123 333 xxx thereforethesolutionsetoftheoriginalequationcan bedescribedbytheparametricequations

 123 754 333,, xrsxrxs

wheretheparameters r and s arearbitraryrealnumbers.

 1153 1234 8484 xxxx thereforethesolutionsetoftheoriginal

equationcanbedescribedbytheparametricequations

 1234 1153 8484,,, xrstxrxsxt

wheretheparameters r , s ,and t arearbitraryrealnumbers.

(d) Solvingtheequationfor v weobtain   8214 3333 vwxyz thereforethesolutionsetoftheoriginalequation canbedescribedbytheparametricequations

 12341234 8214 3333,,,, vttttwtxtytzt

wheretheparameters1 t ,2 t ,3 t ,and4 t arearbitraryrealnumbers.

14. (a) Solvingtheequationfor x weobtain  210 xy thereforethesolutionsetoftheoriginalequationcanbe describedbytheparametricequations

 210, xtyt

wheretheparameter t isanarbitraryrealnumber.

(b) Solvingtheequationfor1 x weobtain  123 3312 xxx thereforethesolutionsetoftheoriginalequationcan bedescribedbytheparametricequations

 123 3312,, xrsxrxs

wheretheparameters r and s arearbitraryrealnumbers.

(c) Solvingtheequationfor1 x weobtain  3 11 1234 5244 xxxx thereforethesolutionsetoftheoriginal equationcanbedescribedbytheparametricequations  12 131 5,,, 244 xrstxryszt

wheretheparameters r , s ,and t arearbitraryrealnumbers.

(d) Solvingtheequationfor v weobtain  57 vwxyz thereforethesolutionsetoftheoriginalequation canbedescribedbytheparametricequations

 12341234 57,,,, vttttwtxtytzt wheretheparameters1 t ,2 t ,3 t ,and4 t arearbitraryrealnumbers.

15. (a) Wecaneliminate x fromthesecondequationbyadding3timesthefirstequationtothesecond.Thisyields thesystem   231 00 xy

Thesecondequationdoesnotimposeanyrestrictionon x and y thereforewecanomitit.Solvingthefirst equationfor x weobtain 13 22 xy .Thisallowsustorepresentthesolutionusingparametricequations   13 22, xtyt

wheretheparameter t isanarbitraryrealnumber.

(b) Wecanseethatthesecondandthethirdequationaremultiplesofthefirst:adding3timesthefirstequation tothesecond,thenaddingthefirstequationtothethirdyieldsthesystem

Thelasttwoequationsdonotimposeanyrestrictionontheunknownsthereforewecanomitthem.Solvingthe firstequationfor1 x weobtain  123 43 xxx .Thisallowsustorepresentthesolutionusingparametric equations

123 43,, xrsxrxs

wheretheparameters r and s arearbitraryrealnumbers.

16. (a) Wecaneliminate1 x fromthefirstequationbyadding2timesthesecondequationtothefirst.Thisyields thesystem

12 34 xx

Thefirstequationdoesnotimposeanyrestrictionon1 x and2 x thereforewecanomitit.Solvingthesecond equationfor1 x weobtain  41 12 33 xx .Thisallowsustorepresentthesolutionusingparametricequations

 12 41 33, xtxt

wheretheparameter t isanarbitraryrealnumber.

(b) Wecanseethatthesecondandthethirdequationaremultiplesofthefirst:adding3timesthefirstequation tothesecond,thenadding2timesthefirstequationtothethirdyieldsthesystem

224 xyz

Thelasttwoequationsdonotimposeanyrestrictionontheunknownsthereforewecanomitthem.Solvingthe firstequationfor x weobtain  1 22 xyz .Thisallowsustorepresentthesolutionusingparametric equations

wheretheparameters r and s arearbitraryrealnumbers.

17. (a) Add2timesthesecondrowtothefirsttoobtain

(b) Addthethirdrowtothefirsttoobtain

(anothersolution:interchangethefirstrowandthethirdrowtoobtain

18. (a) Multiplythefirstrowby12toobtain

(b) Addthethirdrowtothefirsttoobtain

(anothersolution:add2timesthesecondrowtothefirsttoobtain

19. (a) Add4timesthefirstrowtothesecondtoobtain

whichcorrespondstothesystem

If  2 k thenthesecondequationbecomes  018,whichiscontradictorythusthesystembecomes inconsistent.

If  2 k thenwecansolvethesecondequationfor y andproceedtosubstitutethisvalueintothefirstequation andsolvefor x .

Consequently,forallvaluesof  2 k thegivenaugmentedmatrixcorrespondstoaconsistentlinearsystem.

(b) Add4timesthefirstrowtothesecondtoobtain

20. (a)

If  2 k thenthesecondequationbecomes  00,whichdoesnotimposeanyrestrictionon x and y therefore wecanomititandproceedtodeterminethesolutionsetusingthefirstequation.Thereareinfinitelymany solutionsinthisset.

If  2 k thenthesecondequationyields  0 y andthefirstequationbecomes 1 x .

Consequently,forallvaluesof k thegivenaugmentedmatrixcorrespondstoaconsistentlinearsystem.

Add2timesthefirstrowtothesecondtoobtain

If  5 2 k thenthesecondequationbecomes  00,whichdoesnotimposeanyrestrictionon x and y thereforewecanomititandproceedtodeterminethesolutionsetusingthefirstequation.Thereareinfinitely manysolutionsinthisset.

If  5 2 k thenthesecondequationiscontradictorythusthesystembecomesinconsistent.

Consequently,thegivenaugmentedmatrixcorrespondstoaconsistentlinearsystemonlywhen  5 2 k . (b) Addthefirstrowtothesecondtoobtain

whichcorrespondstothesystem

If 4 k thenthesecondequationbecomes  00,whichdoesnotimposeanyrestrictionon x and y thereforewecanomititandproceedtodeterminethesolutionsetusingthefirstequation.Thereareinfinitely manysolutionsinthisset.

If 4 k thenthesecondequationyields  0 x andthefirstequationbecomes 2 y . Consequently,forallvaluesof k thegivenaugmentedmatrixcorrespondstoaconsistentlinearsystem.

21. Substitutingthecoordinatesofthefirstpointintotheequationofthecurveweobtain

2 111 yaxbxc

Repeatingthisfortheothertwopointsandrearrangingthethreeequationsyields

Thisisalinearsystemintheunknowns a , b ,and c .Itsaugmentedmatrixis

23. Solvingthefirstequationfor1 x weobtain  12 xckx thereforethesolutionsetoftheoriginalequationcanbe describedbytheparametricequations

12 , xcktxt

wheretheparameter t isanarbitraryrealnumber.

Substitutingtheseintothesecondequationyields

cktltd

whichcanberewrittenas

cktdlt

Thisequationmustholdtrueforallrealvalues t ,whichrequiresthatthecoefficientsassociatedwiththesamepower of t onbothsidesmustbeequal.Consequently,  cd and  kl

24. (a) Thesystemhasnosolutionsifeither

 atleasttwoofthethreelinesareparallelanddistinctor

 eachpairoflinesintersectsatadifferentpoint(withoutanylinesbeingparallel)

(b) Thesystemhasexactlyonesolutionifeither

 twolinescoincideandthethirdoneintersectsthemor

 allthreelinesintersectatasinglepoint(withoutanylinesbeingparallel)

237 239 42516 xyz xyz xyz

26. WesetupthelinearsystemasdiscussedinExercise21:

Onesolutionisexpected,sinceexactlyoneparabolapassesthroughanythreegivenpoints  11 , xy ,   22 , xy , 

33 , xy if1 x ,2 x ,and3 x aredistinct. 27.

True-False Exercises

(a) True.   0,0,,0isasolution.

(b) False.Onlymultiplicationbya nonzeroconstantisavalidelementaryrowoperation.

(d) True.Accordingtothedefinition, 1122 nn axaxaxb isalinearequationifthe a'sarenotallzero.Letus assume  0 j a .Thevaluesofall x'sexceptfor j x canbesettobearbitraryparameters,andtheequationcanbeused toexpress j x intermsofthoseparameters.

(e) False.E.g.iftheequationsareallhomogeneousthenthesystemmustbeconsistent.(SeeTrue-FalseExercise(a) above.)

(f) False.If  0 c thenthenewsystemhasthesamesolutionsetastheoriginalone.

(g) True.Adding1timesonerowtoanotheramountstothesamethingassubtractingonerowfromanother.

(h) False.Thesecondrowcorrespondstotheequation  01,whichiscontradictory.

1.2 Gaussian Elimination

1. (a) Thismatrixhasproperties1-4.Itisinreducedrowechelonform,thereforeitisalsoinrowechelonform.

(b) Thismatrixhasproperties1-4.Itisinreducedrowechelonform,thereforeitisalsoinrowechelonform.

(d) Thismatrixhasproperties1-4.Itisinreducedrowechelonform,thereforeitisalsoinrowechelonform.

(e) Thismatrixhasproperties1-4.Itisinreducedrowechelonform,thereforeitisalsoinrowechelonform.

(f) Thismatrixhasproperties1-4.Itisinreducedrowechelonform,thereforeitisalsoinrowechelonform.

(g) Thismatrixhasproperties1-3butdoesnothaveproperty4:thesecondcolumncontainsaleading1anda nonzeronumber(7)aboveit.Thematrixisinrowechelonformbutnotreducedrowechelonform.

2. (a) Thismatrixhasproperties1-3butdoesnothaveproperty4:thesecondcolumncontainsaleading1anda nonzeronumber(2)aboveit.Thematrixisinrowechelonformbutnotreducedrowechelonform.

(b) Thismatrixdoesnothaveproperty1sinceitsfirstnonzeronumberinthethirdrow(2)isnota1.Thematrixis notinrowechelonform,thereforeitisnotinreducedrowechelonformeither.

(c) Thismatrixhasproperties1-3butdoesnothaveproperty4:thethirdcolumncontainsaleading1anda nonzeronumber(4)aboveit.Thematrixisinrowechelonformbutnotreducedrowechelonform.

(d) Thismatrixhasproperties1-3butdoesnothaveproperty4:thesecondcolumncontainsaleading1anda nonzeronumber(5)aboveit.Thematrixisinrowechelonformbutnotreducedrowechelonform.

(e) Thismatrixdoesnothaveproperty2sincetherowthatconsistsentirelyofzerosisnotatthebottomofthe matrix.Thematrixisnotinrowechelonform,thereforeitisnotinreducedrowechelonformeither.

(f) Thismatrixdoesnothaveproperty3sincetheleading1inthesecondrowisdirectlybelowtheleading1in thefirst(insteadofbeingfarthertotheright).Thematrixisnotinrowechelonform,thereforeitisnotin reducedrowechelonformeither.

(g) Thismatrixhasproperties1-4.Itisinreducedrowechelonform,thereforeitisalsoinrowechelonform.

(a) Thefirstthreecolumnsarepivotcolumnsandallthreerowsarepivotrows.Thelinearsystem

andsolvedbyback-substitution:

thereforetheoriginallinearsystemhasauniquesolution:  37 x ,  8 y

(b) Thefirstthreecolumnsarepivotcolumnsandallthreerowsarepivotrows.Thelinearsystem

685 349 2 wyz xyz yz Let  zt .Then

thereforetheoriginallinearsystemhasinfinitelymanysolutions:

where t isanarbitraryvalue.

Let  2 xs and  5 xt .Then

 4 3 1 93 593643 3724381172 xt xttt xsttst

thereforetheoriginallinearsystemhasinfinitelymanysolutions:

 12345 1172,,43,93, xstxsxtxtxt where s and t arearbitraryvalues.

(d) Thefirsttwocolumnsarepivotcolumnsandthefirsttworowsarepivotrows.Thesystemisinconsistentsince thethirdrowoftheaugmentedmatrixcorrespondstotheequation

  0001. xyz

4. (a) Thefirstthreecolumnsarepivotcolumnsandallthreerowsarepivotrows.Auniquesolution:  3 x ,  0 y ,  7 z

(b) Thefirstthreecolumnsarepivotcolumnsandallthreerowsarepivotrows.Infinitelymanysolutions:

87 wt , 23 xt ,  5 yt ,  zt where t isanarbitraryvalue.

263 vst ,  ws , 74 xt ,  85 yt ,  zt where s and t arearbitraryvalues.

(d) Columns1and3arepivotcolumns.Thefirsttworowsarepivotrows.Thesystemisinconsistentsincethe thirdrowoftheaugmentedmatrixcorrespondstotheequation

  0001. xyz

1128 1231 37410

 1128 0159 37410

1128 0159 010214

1128 0159 0052104

Theaugmentedmatrixforthesystem.

Thefirstrowwasaddedtothesecondrow.

3timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby1

10timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby152

Thesystemofequationscorrespondingtothisaugmentedmatrixinrowechelonformis

Back-substitutionyields

Thelinearsystemhasauniquesolution:

Theaugmentedmatrixforthesystem.

1110 2521 8141

1110 0741 8141

1110 0741 0741

41 77 1110 01 0741

Thefirstrowwasmultipliedby1 2

2timesthefirstrowwasaddedtothesecondrow.

8timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby17.

41 77 1110 01 0000 7timesthesecondrowwasaddedtothethirdrow.

Thesystemofequationscorrespondingtothisaugmentedmatrixinrowechelonformis

Solvetheequationsfortheleadingvariables

thensubstitutethesecondequationintothefirst

Ifweassign3 x anarbitraryvalue t ,thegeneralsolutionisgivenbytheformulas

11211 21222 12411 30033

11211 03600 12411 30033

11211 03600 01200 30033

11211 03600 01200 03600

Theaugmentedmatrixforthesystem.

2timesthefirstrowwasaddedtothesecondrow.

Thefirstrowwasaddedtothethirdrow.

3timesthefirstrowwasaddedtothefourthrow.



11211

01200 01200 03600

11211

01200 00000 03600

Thesecondrowwasmultipliedby1 3

1timesthesecondrowwasaddedtothethirdrow.



11211

01200 00000 00000

3timesthesecondrowwasaddedtothefourthrow.

Thesystemofequationscorrespondingtothisaugmentedmatrixinrowechelonformis

Solvetheequationsfortheleadingvariables

thensubstitutethesecondequationintothefirst

Ifweassign z and w thearbitraryvalues s and t ,respectively,thegeneralsolutionisgivenbytheformulas

0231 3632 6635

3632 0231 6635

Theaugmentedmatrixforthesystem.

Thefirstandsecondrowswereinterchanged.

2 1213 0231 6635

2 1213 0231 0699

3 31 22 121 01 0699

2 3 31 22 121 01 0006

3 31 22 121 01 0001

Thefirstrowwasmultipliedby1 3

6timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby1 2

6timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby16.

Thesystemofequationscorrespondingtothisaugmentedmatrixinrowechelonform

isclearlyinconsistent.

1128 1231 37410

1128 0159 37410

Theaugmentedmatrixforthesystem.

Thefirstrowwasaddedtothesecondrow.

1128 0159 010214 3timesthefirstrowwasaddedtothethirdrow.

1128 0159 010214

Thesecondrowwasmultipliedby1.

1128 0159 0052104

10timesthesecondrowwasaddedtothethirdrow. 

1128

0159 0012

1128 0101 0012

1104 0101 0012

Thethirdrowwasmultipliedby152

5timesthethirdrowwasaddedtothesecondrow.

2timesthethirdrowwasaddedtothefirstrow. 

1003 0101 0012

1timesthesecondrowwasaddedtothefirstrow.

Thelinearsystemhasauniquesolution:  13x ,  21 x ,  32x

2220 2521 8141

1110 2521 8141

1110 0741 8141

Theaugmentedmatrixforthesystem.

Thefirstrowwasmultipliedby1 2

2timesthefirstrowwasaddedtothesecondrow.

1110 0741 0741 8timesthefirstrowwasaddedtothethirdrow.

41 77 1110 01 0741

Thesecondrowwasmultipliedby1 7

41 77 1110 01 0000

7timesthesecondrowwasaddedtothethirdrow. 

31 77 41 77 10 01 0000

1timesthesecondrowwasaddedtothefirstrow. Infinitelymanysolutions:

11211 21222 12411 30033

11211 03600 12411 30033

11211 03600 01200 30033

Theaugmentedmatrixforthesystem.

2timesthefirstrowwasaddedtothesecondrow.

thefirstrowwasaddedtothethirdrow.



11211 03600 01200 03600

11211 01200 01200 03600

3timesthefirstrowwasaddedtothefourthrow.

Thesecondrowwasmultipliedby13.



11211 01200 00000 03600

1timesthesecondrowwasaddedtothethirdrow.

  

 11211 01200 00000 00000

3timesthesecondrowwasaddedtothefourthrow.

10011 01200 00000 00000

thesecondrowwasaddedtothefirstrow.

Thesystemofequationscorrespondingtothisaugmentedmatrixinrowechelonformis

Solvetheequationsfortheleadingvariables

Ifweassign z and w thearbitraryvalues s and t ,respectively,thegeneralsolutionisgivenbytheformulas

0231 3632 6635

3632 0231 6635

2 1213 0231 6635

1213 0231 0699

3 31 22 121 01 0699

3 31 22 121 01 0006

Theaugmentedmatrixforthesystem.

Thefirstandsecondrowswereinterchanged.

Thefirstrowwasmultipliedby1 3

6timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby12.

6timesthesecondrowwasaddedtothethirdrow.

2 3 31 22 121 01 0001

Thethirdrowwasmultipliedby1 6

2 3 3 2 121 010 0001 1 2timesthethirdrowwasaddedtothesecondrow. 

3 2 1210 010 0001 2 3timesthethirdrowwasaddedtothefirstrow.

3 2 1020 010 0001 2timesthesecondrowwasaddedtothefirstrow.

Thelastrowcorrespondstotheequation

0001 abc thereforethesystemisinconsistent.

(Note:thiswasalreadyevidentafterthefifthelementaryrowoperation.)

13. Sincethenumberofunknowns(4)exceedsthenumberofequations(3),itfollowsfromTheorem1.2.2thatthis systemhasinfinitelymanysolutions.Thoseincludethetrivialsolutionandinfinitelymanynontrivialsolutions.

14. Thesystemdoesnothavenontrivialsolutions.

(Thethirdequationrequires  30x ,whichsubstitutedintothesecondequationyields  20.x Bothofthese substitutedintothefirstequationresultin  10x .)

15. Wepresenttwodifferentsolutions. SolutionIusesGauss-Jordanelimination

2130 1200 0110

13 22 10 1200 0110

Theaugmentedmatrixforthesystem.

Thefirstrowwasmultipliedby12.

13 22 33 22 10 00 0110 1timesthefirstrowwasaddedtothesecondrow.

13 22 10

0110 0110

Thesecondrowwasmultipliedby2 3 

13 22 10 0110 0020 1timesthesecondrowwasaddedtothethirdrow. 

13 22 10 0110 0010

1 2 100 0100 0010



Thethirdrowwasmultipliedby1 2

Thethirdrowwasaddedtothesecondrow and32timesthethirdrowwasaddedtothefirstrow

1000 0100 0010 1 2timesthesecondrowwasaddedtothefirstrow.

Uniquesolution:  10x ,  20x ,  30x

SolutionII.Thistime,weshallchoosetheorderoftheelementaryrowoperationsdifferentlyinordertoavoid introducingfractionsintothecomputation.(Sinceeverymatrixhasauniquereducedrowechelonform,theexact sequenceofelementaryrowoperationsbeinguseddoesnotmatter–seepart1ofthediscussion“SomeFactsAbout EchelonForms”inSection1.2)



2130 1200 0110

1200 2130 0110

Theaugmentedmatrixforthesystem.

Thefirstandsecondrowswereinterchanged (toavoidintroducingfractionsintothefirstrow). 

1200 0330 0110

1200 0110 0110

2timesthefirstrowwasaddedtothesecondrow.

Thesecondrowwasmultipliedby1 3



1200 0110 0020

1timesthesecondrowwasaddedtothethirdrow. 

1200 0110 0010

Thethirdrowwasmultipliedby1 2 

1200 0100 0010

1000 0100 0010

Uniquesolution:

16. Wepresenttwodifferentsolutions.

SolutionIusesGauss-Jordanelimination

2130 1230 1140

13 22 10 1230 1140

13 22 39 22 10 00 1140

13 22 39 22 311 22 10 00 00

13 22 311 22 10 0130 00

Thethirdrowwasaddedtothesecondrow.

2timesthesecondrowwasaddedtothefirstrow.

Theaugmentedmatrixforthesystem.

Thefirstrowwasmultipliedby12.

Thefirstrowwasaddedtothesecondrow.

1timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby23. 

13 22 10 0130 00100 3 2timesthesecondrowwasaddedtothethirdrow.

13 22 10

0130 0010

Thethirdrowwasmultipliedby110

3timesthethirdrowwasaddedtothesecondrow and32timesthethirdrowwasaddedtothefirstrow

Uniquesolution:

Theaugmentedmatrixforthesystem.

1140 1230 2130

Thefirstandthirdrowswereinterchanged (toavoidintroducingfractionsintothefirstrow). 

1140 0310 2130

Thefirstrowwasaddedtothesecondrow. 

1140 0310 03110

2timesthefirstrowwasaddedtothethirdrow. 

1140 0310 00100

 1140 0310 0010

Thesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby110

Uniquesolution:



1140 0300 0010

1timesthethirdrowwasaddedtothesecondrow. 

1100 0300 0010

4timesthethirdrowwasaddedtothefirstrow. 

1100 0100 0010

1000 0100 0010

Thesecondrowwasmultipliedby13.

1timesthesecondrowwasaddedtothefirstrow.

31110 51110Theaugmentedmatrixforthesystem.

111 333 10 51110Thefirstrowwasmultipliedby1 3 

111 333 88 2 333 10 005timesthefirstrowwasaddedtothesecondrow.

 111 333 1 4 10 0110Thesecondrowwasmultipliedby38.

1 4 1 4 1000 0110 1 3timesthesecondrowwasaddedtothefirstrow.

Ifweassign3 x and4 x thearbitraryvalues s and t ,respectively,thegeneralsolutionisgivenbytheformulas

1234 11 ,,,44 xsxstxsxt (Notethatfractionsinthesolutioncouldbeavoidedifweassigned  34 xs instead,whichalongwith  4 xt would yield  1 xs ,  2 xst ,  34 xs ,  4 xt .)



01320 21430 23210 43540

 21430 01320 23210 43540

Theaugmentedmatrixforthesystem.

Thefirstandsecondrowswereinterchanged.



13 22 120 01320 23210 43540

13 22 120 01320 02640 01320

Thefirstrowwasmultipliedby12.

2timesthefirstrowwasaddedtothethirdrow and4timesthefirstrowwasaddedtothefourthrow.



13 22 120 01320 00000 00000

75 22 100 01320 00000 00000

2timesthesecondrowwasaddedtothethirdrowand thesecondrowwasaddedtothefourthrow.

1 2timesthesecondrowwasaddedtothefirstrow.

Ifweassign w and x thearbitraryvalues s and t ,respectively,thegeneralsolutionisgivenbytheformulas  75 ,32,, 22 ustvstwsxt

02240 10130 23110 21320

Theaugmentedmatrixforthesystem.

      10130 02240 23110 21320

      10130 02240 03370 01180

Thefirstandsecondrowswereinterchanged.

2timesthefirstrowwasaddedtothethirdrow and2timesthefirstrowwasaddedtothefourthrow.

10130 01120 03370 01180

 

Thesecondrowwasmultipliedby12. 

 10130 01120 00010 000100

      10130 01120 00010 00000

3timesthesecondrowwasaddedtothethirdand 1timesthesecondrowwasaddedtothefourthrow.

10timesthethirdrowwasaddedtothefourthrow.

 

10100 01100 00010 00000

2timesthethirdrowwasaddedtothesecondand 3timesthethirdrowwasaddedtothefirstrow.

Ifweassign y anarbitraryvalue t thegeneralsolutionisgivenbytheformulas

13010 14200 02210 24110 12110

Theaugmentedmatrixforthesystem.

1timesthefirstrowwasaddedtothesecondrow, 2timesthefirstrowwasaddedtothefourthrow, and1timesthefirstrowwasaddedtothefifthrow. 

2timesthesecondrowwasaddedtothethirdrow, 10timesthesecondrowwasaddedtothefourthrow, and5timesthesecondrowwasaddedtothefifthrow.



Thethirdrowwasmultipliedby1 2

21timesthethirdrowwasaddedtothefourthrow and9timesthethirdrowwasaddedtothefifthrow.

Thefourthrowwasmultipliedby241.

Usingback-substitution,weobtaintheuniquesolutionofthissystem

21349 102711 33158 214410

Theaugmentedmatrixforthesystem. 

102711 21349 33158 214410

Thefirstandsecondrowswereinterchanged (toavoidintroducingfractionsintothefirstrow). 

102711 0171013 0371625 0181012

2timesthefirstrowwasaddedtothesecondrow, 3timesthefirstrowwasaddedtothethirdrow, and2timesthefirstrowwasaddedtothefourth.

Thesecondrowwasmultipliedby1



102711

0171013

00141414 00152025

3timesthesecondrowwasaddedtothethirdrowand 1timesthesecondrowwasaddedtothefourthrow.



102711 0171013 00111 00152025

Thethirdrowwasmultipliedby114



102711 0171013

00111 000510

102711 0171013 00111 00012

15timesthethirdrowwasaddedtothefourthrow.

Thefourthrowwasmultipliedby1 5

10203 01707 00101 00012

10001 01000 00101 00012

Uniquesolution:

001110 112310 112010 221010

112010 112310 001110 221010

112010 000300 001110 003030

Thefourthrowwasaddedtothethirdrow, 10timesthefourthrowwasaddedtothesecond, and7timesthefourthrowwasaddedtothefirst.

7timesthethirdrowwasaddedtothesecondrow, and2timesthethirdrowwasaddedtothefirstrow.

Theaugmentedmatrixforthesystem.

Thefirstandthirdrowswereinterchanged.

Thefirstrowwasaddedtothesecondrow and2timesthefirstrowwasaddedtothelastrow. 

112010 001110 000300 003030

112010 001110 000300 000300

112010 001110 000100 000300

Thesecondandthirdrowswereinterchanged.

3timesthesecondrowwasaddedtothefourthrow.

Thethirdrowwasmultipliedby1 3



112010

001110

000100 000000

3timesthethirdrowwasaddedtothefourthrow.



112010

001010

000100 000000

1timesthethirdrowwasaddedtothesecondrow.



110010

001010

000100 000000

2timesthesecondrowwasaddedtothefirstrow.

Ifweassign2 Z and5 Z thearbitraryvalues s and t ,respectively,thegeneralsolutionisgivenbytheformulas

23. (a) Thesystemisconsistent;ithasauniquesolution(back-substitutioncanbeusedtosolveforallthree unknowns).

(b) Thesystemisconsistent;ithasinfinitelymanysolutions(thethirdunknowncanbeassignedanarbitraryvalue t ,thenback-substitutioncanbeusedtosolveforthefirsttwounknowns).

(d) Thereisinsufficientinformationtodecidewhetherthesystemisconsistentasillustratedbytheseexamples:

For

1 0000 001 thesystemisconsistentwithinfinitelymanysolutions.

1 0010 0011 thesystemisinconsistent(thematrixcanbereducedto

1 0010 0001 ).

24. (a) Thesystemisconsistent;ithasauniquesolution(back-substitutioncanbeusedtosolveforallthree unknowns).

(b) Thesystemisconsistent;ithasauniquesolution(solvethefirstequationforthefirstunknown,thenproceed tosolvethesecondequationforthesecondunknownandsolvethethirdequationlast.)

1000 0001

;thesecond equation  01iscontradictory).

(d) Thereisinsufficientinformationtodecidewhetherthesystemisconsistentasillustratedbytheseexamples:

For

1001

1001 1001 thesystemisconsistentwithinfinitelymanysolutions.

1002

For

1001 1001

2 1234 3152 41142 aa

thesystemisinconsistent(thematrixcanbereducedto

Theaugmentedmatrixforthesystem. 

2 1234 071410 07214 aa

2 1234 071410 00164 aa

10 7 2 1234 012 00164 aa

1002 0001 0000 ).

3timesthefirstrowwasaddedtothesecondrow and4timesthefirstrowwasaddedtothethirdrow.

1timesthesecondrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby17.

Thesystemhasnosolutionswhen 4 a (sincethethirdrowofourlastmatrixwouldthencorrespondtoa contradictoryequation  08).

Thesystemhasinfinitelymanysolutionswhen  4 a (sincethethirdrowofourlastmatrixwouldthencorrespond totheequation  00).

Forallremainingvaluesof a (i.e., 4 a and  4 a )thesystemhasexactlyonesolution.

2 1212 2231 12(3)aa

2 1212 0613 0022 aa

11 62 2 1212 01 0022 aa

Theaugmentedmatrixforthesystem.

2timesthefirstrowwasaddedtothesecondrow and1timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby1 6

Thesystemhasnosolutionswhen  2 a or  2 a (sincethethirdrowofourlastmatrixwouldthencorrespond toacontradictoryequation).

Forallremainingvaluesof a (i.e.,  2 a and  2 a )thesystemhasexactlyonesolution. Thereisnovalueof a forwhichthissystemhasinfinitelymanysolutions.

131 112 023 a b c

Theaugmentedmatrixforthesystem.

131 023 023 a ab c 1timesthefirstrowwasaddedtothesecondrow.



131 023 000 a ab abc

3 222 131 01 000 ab a abc

Thesecondrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby12.

If  0 abc thenthelinearsystemisconsistent.Otherwise(if  0 abc )itisinconsistent.

131 121 371 a b c

131 012 0243 a ab ac

131 012 0002 a ab abc

Theaugmentedmatrixforthesystem.

Thefirstrowwasaddedtothesecondrowand 3timesthefirstrowwasaddedtothethirdrow.

2timesthesecondrowwasaddedtothethirdrow. If

20abc thenthelinearsystemisconsistent.Otherwise(if 20abc )itisinconsistent. 29.

21 36 a b

11 122 36 a b

Theaugmentedmatrixforthesystem.

Thefirstrowwasmultipliedby12.

11 22 93 22 1 0 a ab

11 22 12 39 1 01 a ab

3timesthefirstrowwasaddedtothesecondrow.

Thethirdrowwasmultipliedby29.

21 39 12 39 10 01 ab ab 1 2timesthesecondrowwasaddedtothefirstrow.

Thesystemhasexactlyonesolution:  21 39 xab and 12 39 yab

Theaugmentedmatrixforthesystem.

111 0202 033 a ab c

2timesthefirstrowwasaddedtothesecondrow.

Thesecondrowwasmultipliedby1 2

2 3 2 111 010 0033 b a a abc

3timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1 3

23 2 23

Adding3timesitssecondrowtothefirstresultsin

213 0229 345

1timesthefirstrowwasaddedtothethirdrow.

Thefirstandthirdrowswereinterchanged.

132 0229 051

2timesthefirstrowwasaddedtothethirdrow.

132 0186 0229

132 0186 00143

132 0186 001

130 010 001

100 010 001

3timesthesecondrowwasaddedtothethirdrow.

Thesecondandthirdrowswereinterchanged.

2timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1143

86timesthethirdrowwasaddedtothesecondrow and2timesthethirdrowwasaddedtothefirstrow.

3timesthesecondrowwasaddedtothefirstrow.

1230 2530 1550

Theaugmentedmatrixforthesystem. 

1230

0130 0380

1230 0130 0010

1200 0100 0010

1000 0100 0010

2timesthefirstrowwasaddedtothesecondrow andthefirstrowwasaddedtothethirdrow.

3timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1.

3timesthethirdrowwasaddedtothesecondrowand 3timesthethirdrowwasaddedtothefirstrow.

2timesthesecondrowwasaddedtothefirstrow.

Thissystemhasexactlyonesolution 0,0,0.xyz

Ontheinterval    02,theequation   sin0hasthreesolutions:   0,    ,and    2.

Ontheinterval    02,theequation   cos0hastwosolutions:    2and    3 2

Ontheinterval    02,theequation   tan0hasthreesolutions:   0,    ,and    2

Overall,  32318solutions    ,,canbeobtainedbycombiningthevaluesof  ,  ,and  listedabove:

0,,0,,,0 22 ,etc.

34. Webeginbysubstituting   sin x ,   cos y ,and   tan z sothatthesystembecomes



2133 4222 6319

Theaugmentedmatrixforthesystem. 

2133

0484 0080

2timesthefirstrowwasaddedtothesecondrow and3timesthefirstrowwasaddedtothethirdrow.



2133

0484 0010

2103

0404 0010

Thethirdrowwasmultipliedby18.

8timesthethirdrowwasaddedtothesecondrow and3timesthethirdrowwasaddedtothefirstrow. 

2103

0101 0010

2002 0101 0010

1001 0101 0010

Thesecondrowwasmultipliedby14.

Thesecondrowwasaddedtothefirstrow.

Thefirstrowwasmultipliedby12. Thissystemhasexactlyonesolution

1116

0214 0139

1116

0139 0214

1116

0139 0214

1116 0139 00714

1116 0139 0012

1104 0103 0012

1001 0103 0012

1timesthefirstrowwasaddedtothesecondrow and2timesthefirstrowwasaddedtothethirdrow.

Thesecondandthirdrowswereinterchanged (toavoidintroducingfractionsintothesecondrow).

Thesecondrowwasmultipliedby1

2timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby17.

3timesthethirdrowwasaddedtothesecondrow and1timesthethirdrowwasaddedtothefirstrow.

1timesthesecondrowwasaddedtothefirstrow.



1241

01162 01166

1241

01162 01166

2timesthefirstrowwasaddedtothesecondrow andthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby1.



1241

01162 0018216

11timesthesecondrowwasaddedtothethirdrow.

8 91 1241

01162 001

Usingback-substitution,weobtain

Thethirdrowwasmultipliedby1182.

37. Eachpointonthecurveyieldsanequation,thereforewehaveasystemoffourequations

equationcorrespondingto1,7:7

equationcorrespondingto3,11:279311

equationcorrespondingto4,14:6416414

equationcorrespondingto0,10:10 abcd abcd abcd d

11117 2793111 64164114 000110

11117 0182426200 0486063462 000110

Theaugmentedmatrixforthesystem.

27timesthefirstrowwasaddedtothesecondrow and64timesthefirstrowwasaddedtothethird.

413100 399 11117 01 0486063462 000110

Thesecondrowwasmultipliedby118



413100 399 19214 33 11117 01 004 000110

48timesthesecondrowwasaddedtothethirdrow.



413100 399 19107 126 11117 01 001 000110

410 33 11103 010 00102 000110

11005 01006 00102 000110



Thethirdrowwasmultipliedby14.

19 12timesthefourthrowwasaddedtothethirdrow, 139timesthefourthrowwasaddedtothesecondrow, and1timesthefourthrowwasaddedtothefirst.

4 3timesthethirdrowwasaddedtothesecondrowand 1timesthethirdrowwasaddedtothefirstrow.

10001 01006 00102 000110 1timesthesecondrowwasaddedtothefirstrow.

Thelinearsystemhasauniquesolution:

d .Thesearethecoefficientvaluesrequired forthecurve  32 yaxbxcxd topassthroughthefourgivenpoints.

38. Eachpointonthecurveyieldsanequation,thereforewehaveasystemofthreeequations

equationcorrespondingto2,7:53270 equationcorrespondingto4,5:41450 equationcorrespondingto4,3:25430 abcd abcd abcd

Ifweassign d anarbitraryvalue t ,thegeneralsolutionisgivenbytheformulas

292929

(Forinstance,lettingthefreevariable d havethevalue29yields

39. Sincethehomogeneoussystemhasonlythetrivialsolution,itsaugmentedmatrixmustbepossibletoreduceviaa sequenceofelementaryrowoperationstothereducedrowechelonform

Applyingthe same sequenceofelementaryrowoperationstotheaugmentedmatrixofthenonhomogeneoussystem yieldsthereducedrowechelonform

, s ,and t aresomerealnumbers.Therefore,the nonhomogeneoussystemhasonesolution.

40. (a) 3(thiswillbethenumberofleading1'sifthematrixhasnorowsofzeros)

(b) 5(ifallentriesin B are0)

41. (a) Thereareeightpossiblereducedrowechelonforms:

where r and s canbeanyrealnumbers.

(b) Therearesixteenpossiblereducedrowechelonforms:

where r , s , t ,and u canbeanyrealnumbers.

42. (a) Eitherthethreelinesproperlyintersectattheorigin,ortwoofthemcompletelyoverlapandtheotherone intersectsthemattheorigin.

(b) Allthreelinescompletelyoverlaponeanother.

43. (a) Weconsidertwopossiblecases:(i)  0 a ,and(ii)  0 a

(i)If  0 a thentheassumption  0 adbc impliesthat  0 b and  0 c .Gauss-Jordaneliminationyields

Weassumed  0 a

Therowswereinterchanged.

Thefirstrowwasmultipliedby1 c and thesecondrowwasmultipliedby1 b (Notethat  ,0.)bc

timesthesecondrowwasaddedtothefirstrow.

(ii)If  0 a thenweperformGauss-Jordaneliminationasfollows:

Thefirstrowwasmultipliedby1 a .

Thesecondrowwasmultipliedby a adbc . (Notethatboth a and adbc arenonzero.)

  10 01 b a timesthesecondrowwasaddedtothefirstrow.

Inbothcases(  0 a aswellas  0 a )weestablishedthatthereducedrowechelonformof

providedthat  0 adbc .

(b) Applyingthe same elementaryrowoperationstepsasinpart(a)theaugmentedmatrix

transformedtoamatrixinreducedrowechelonform

where p and q aresomerealnumbers.We concludethatthegivenlinearsystemhasexactlyonesolution:  xp ,  yq

True-False Exercises

(a) True.Amatrixinreducedrowechelonformhasallpropertiesrequiredfortherowechelonform.

(b) False.Forinstance,interchangingtherowsof

(d) True.Inareducedrowechelonform,thenumberofnonzerorowsequalstothenumberofleading1's.Theresult followsfromTheorem1.2.1.

(e) True.Thisisimpliedbythethirdpropertyofarowechelonform(seeSection1.2).

(f) False.Nonzeroentriesarepermittedabovetheleading1'sinarowechelonform.

(g) True.Inareducedrowechelonform,thenumberofnonzerorowsequalstothenumberofleading1's.From Theorem1.2.1weconcludethatthesystemhas  0 nn freevariables,i.e.ithasonlythetrivialsolution.

(h) False.Therowofzerosimposesnorestrictionontheunknownsandcanbeomitted.Whetherthesystemhas infinitelymany,one,ornosolution(s)depends solely onthenonzerorowsofthereducedrowechelonform.

(i) False.Forexample,thefollowingsystemisclearlyinconsistent:

1.3 Matrices and Matrix Operations

1. (a) Undefined(thenumberofcolumnsin B doesnotmatchthenumberofrowsin A ) (b) Defined;  44matrix

(d) Defined;  52matrix

(e) Defined;  45matrix

(f) Defined;  55matrix

(a) Defined;  54matrix

(b) Undefined(thenumberofcolumnsin D doesnotmatchthenumberofrowsin C )

(d) Defined;  24matrix

(e) Defined;  52matrix

(f) Undefined( BT A isa  44matrix,whichcannotbeaddedtoa  42matrix D )

3. (a)

(b)

165123765 110112213 342143737

5350150 5152510 515155

71747272814 73717521735

(e) Undefined(a  23matrix C cannotbesubtractedfroma  22matrix2 B )

(l) Undefined(traceisonlydefinedforsquarematrices)

(b)

113614161134501 502111510121411 214323231243111

652669 18569182

(j) Undefined(a  22matrix B cannotbemultipliedbya  32matrix A ) (k)

152614 tr101111 324323

  

 

(l) Undefined( BC isa  23matrix;traceisonlydefinedforsquarematrices)

(b) Undefined(thenumberofcolumnsof B doesnotmatchthenumberofrowsin A )

181319318530921823194 313163353062323164 121319312530921223194

        

4210875 12321 367863

45435041524151 311 162163201612216121 021 88838081828181

133011321131262124 tr431041124111212121 235021522151232223

354122815312 tr1225222tr14071501328 687646121213

1136133021621123330 25021121225120122212 2144131122421124311

43330433131403231 11121211311211101221 01511031151001251

161133111132141133 560123510122540123 261143211142241143

 

7. (a) firstrowof  A B [firstrowof A ]

 674141

(b) thirdrowof  A B [thirdrowof A ]

064097024197044395

 636757

(d) firstcolumnof  BAB [firstcolumnof A ]

(e) thirdrowof  AA [thirdrowof A

thirdcolumnof  AAA [thirdcolumnof A ]

(b) thirdcolumnof  BBB [thirdcolumnof B ]

212218

(d) firstcolumnof  AAA [firstcolumnof A ]

32733326703 654663564048 049003469024

(e) thirdcolumnof  ABA [thirdcolumnof B ]

(f) firstrowof  BA [firstrowof

secondcolumnof

thirdcolumnof

(b) firstcolumnof

secondcolumnof

thirdcolumnof

secondcolumnof

02000604604 141302 43416490613162 19.

123126815182228 123456 45648152030364964

0420016022182 214011 12521805516

Aftersubtractingfirstequationfromthefourth,addingthesecondtothethird,andback-substituting,weobtainthe

24. Thegivenmatrixequationisequivalenttothelinearsystem

Aftersubtractingfirstequationfromthesecond,addingthethirdtothefourth,andback-substituting,weobtainthe

25. (a) Ifthe i throwvectorof A is   00thenitfollowsfromFormula(9)inSection1.3that i throwvector of

(b) Ifthe j thcolumnvectorof B is

thenitfollowsfromFormula(8)inSection1.3thatthe j thcolumn

Assumingtheentriesof A arerealnumbersthatdonotdependon x , y ,and z ,thisrequiresthatthecoefficients correspondingtothesamevariableonbothsidesofeachequationmustmatch.Therefore,theonlymatrixsatisfying thegivenconditionis

Assumingtheentriesof A arerealnumbersthatdonotdependon x , y ,and z ,itfollowsthatnorealnumbers11 a , 12a ,and13 a existforwhichthefirstequationissatisfiedforall x , y ,and z .Thereforenomatrix A withreal numberentriescansatisfythegivencondition.

(Notethatif A werepermittedtodependon x , y ,and z ,thensolutionsdoexiste.g.,

29. (a)

(b) Foursquarerootscanbefound:

32. (a)

33. Thegivenmatrixproductrepresents

thetotalcostofitemspurchasedinJanuary

thetotalcostofitemspurchasedinFebruary thetotalcostofitemspurchasedinMarch thetotalcostofitemspurchasedinApril

34. (a) The  43matrix  MJ representssalesoverthetwomonthperiod.

(b) The  43matrix MJ representsthedecreaseinsalesofeachitemfromMaytoJune.

(c)

(d)

(e) Theentryinthe  11matrix My x representsthetotalnumberofitemssoldinMay.

True-False Exercises

(a) True.Themaindiagonalisonlydefinedforsquarematrices.

(b) False.An  mn matrixhas m rowvectorsand n columnvectors.

(d) False.The i throwvectorof AB canbecomputedbymultiplyingthe i throwvectorof A by B

(e) True.UsingFormula(14),

(f) False.E.g.,if

(g) False.E.g.,if

(h) True.Themaindiagonalentriesinasquarematrix A arethesameasthosein TA .

(i) True.Since TA isa  46matrix,itfollowsfrom TTBA beinga  26matrixthat TB mustbea  24matrix. Consequently, B isa  42matrix.

(j) True.

(k) True.Theequalityofthematrices AC and BC impliesthat  ijijijij acbc forall i and j .Adding ij c to bothsidesyields  ijijab forall i and j .Consequently,thematrices A and B areequal.

(l) False.E.g.,if

eventhough  AB .

(m) True.If A isa  pq matrixand B isan  rs matrixthen AB beingdefinedrequires  qr and BA beingdefined requires  sp .Forthe  pp matrix AB tobepossibletoaddtothe  qq matrix BA ,wemusthave  pq

(n) True.Ifthe j thcolumnvectorof B is

thenitfollowsfromFormula(8)inSection1.3that the j thcolumnvectorof

(o) False.E.g.,if

1.4 Inverses; Algebraic Properties of Matrices

5. Thedeterminantof

15.

Thedeterminantofthematrixis

Thematrices

and t ,respectively,thegeneralsolutionisgivenbytheformulas

Therefore,

00 10 commuteif

Ifweassign c and d thearbitraryvalues s and t ,respectively,thegeneralsolutionisgivenbytheformulas

51231 123 3524 x ,

345117 18 1351 x ,

301221 12211 6314 x ,

44242412 1105 2421 x

11143 28 1351 x

62406 12 22211 6314 x

33 AIAI

333 AAIIAI

(3)39 AAIIAII

(3)39 AAAI

29 AIpA

UsingthepropertiesinTheorem1.4.1wecanwrite

Theorem1.4.1(e)

32. Wecanlet A beoneofthefollowingeightmatrices:

Notethattheseeightarenottheonlysolutions-e.g., A canbe

(a) Wecanrewritetheequation

If  3 AI thenitfollowsthat  2 AAI therefore A mustbeinvertible(  12).AA 35. Ifthe i throwvectorof A is

 00thenitfollowsfromFormula(9)inSection1.3that i throwvectorof

0000ABB

Consequentlynomatrix B canbefoundtomaketheproduct  ABI thus A doesnothaveaninverse.

Ifthe j thcolumnvectorof A is

0 0 thenitfollowsfromFormula(8)inSection1.3that

the j thcolumnvectorof  BAB

Consequentlynomatrix B canbefoundtomaketheproduct  BAI thus A doesnothaveaninverse.

36. Ifthe i thand j throwvectorsof A areequalthenitfollowsfromFormula(9)inSection1.3that i throwvectorof  ABj throwvectorof AB

Consequentlynomatrix B canbefoundtomaketheproduct  ABI thus A doesnothaveaninverse.

Ifthe i thand j thcolumnvectorsof A areequalthenitfollowsfromFormula(8)inSection1.3that the i thcolumnvectorof  BA the j thcolumnvectorof BA

Consequentlynomatrix B canbefoundtomaketheproduct  BAI thus A doesnothaveaninverse.

37. Letting

,thematrixequation  AXI becomes

Settingthefirstcolumnsonbothsidesequalyieldsthesystem

Subtractingthesecondandthirdequationsfromthefirstleadsto  21 21 x .Therefore

and(after substitutingthisintotheremainingequations)

Thesecondandthethirdcolumnscanbetreatedinasimilarmannertoresultin

Althoughthiscorrespondstoasystemofnineequations,itissufficienttoexaminejustthethreeequations correspondingtothefirstcolumn

39.

toseethatsubtractingthesecondandthirdequationsfromthefirstleadstoacontradiction  01. Weconcludethat A isnotinvertible.

 111111 ABACDCD

 11 111111()BAACCDD

 1111()BAACCDD

 1111 BAACCDD

Theorem1.4.6

Theorem1.4.7(a)

Theorem1.4.1(c)

11 1111 ACACACAD

1111 CAACCAAD

Theorem1.4.6

Theorem1.4.7(a)

Theorem1.4.1(c)

42. Yes,itistrue.Frompart(e)ofTheorem1.4.8,itfollowsthat

.Thisstatementcanbe extendedto n factors(seeSection1.4)sothat

43. (a) Assuming A isinvertible,wecanmultiply(ontheleft)eachsideoftheequationby1 A :

 11 AABAAC

 IBIC

 BC

1.4Inverses;AlgebraicPropertiesofMatrices64

Multiply(ontheleft)eachsideby1 A

Theorem1.4.1(c)

Formula(1)inSection1.4

Property  AIIAA onSection1.4

(b) If A isnotaninvertiblematrixthen  ABAC doesnotgenerallyimply  BC asevidencedbyExample3.

44. Invertibilityof A impliesthat A isasquarematrix,whichisallthatisrequired. ByrepeatedapplicationofTheorem1.4.1(m)and(l),wehave

45. (a)

111AABBAB

111AABABBAB

1 IBAIAB

1 BAAB

1 ABAB

 I

Theorem1.4.1(d)and(e)

Formula(1)inSection1.4

Property  AIIAA inSection1.4

Theorem1.4.1(a)

Formula(1)inSection1.4

(b) Wecanmultiplyeachsideoftheequalityfrompart(a)ontheleftby1 A ,thenontherightby A toobtain

111 ABBABAI

whichshowsthatif A , B ,and  AB areinvertiblethensois  11AB Furthermore, 

1111 ABBABA .

46. (a)

  2 IA

 IAIA

 IIIAAIAA

 2IAAA

Theorem1.4.1(f)and(g)

Property  AIIAA inSection1.4

IAAAA isidempotentso  2 AA

IA

(b)

    22AIAI

2222 AAAIIAII

Theorem1.4.1(f)and(g)  2 422AAAI

Theorem1.4.1(l)and(m); Property  AIIAA inSection1.4  44AAIA isidempotentso  2 AA

47. ApplyingTheorem1.4.1(d)and(g),property   AIIAA ,andtheassumption  k AO wecanwrite

I 48.

ababab AadAadbcIadadbc cdcdcd

210

22 22 000 000 abcabbdadaabbdadbc cadccbdacdcaddadbc

True-False Exercises

(a) False. A and B areinversesofoneanotherifandonlyif   ABBAI

(b) False.

222 ABABABAABBAB doesnotgenerallyequal

22 2 AABB since AB maynot equal BA

22 ABABAABBAB doesnotgenerallyequal22 AB since AB maynotequal BA .

(d) False.   111 ABBA doesnotgenerallyequal11 AB .

(e) False.

 ATTT BBA doesnotgenerallyequal TTAB

(f) True.ThisfollowsfromTheorem1.4.5.

(g) True.ThisfollowsfromTheorem1.4.8.

(h) True.ThisfollowsfromTheorem1.4.9.(Theinverseof TA isthetransposeof1 A .)

(i) False.

 012 pm IaaaaI .

(j) True.

Ifthe i throwvectorof A is    00thenitfollowsfromFormula(9)inSection1.3that i throwvectorof  0000ABB . Consequentlynomatrix B canbefoundtomaketheproduct  ABI thus A doesnothaveaninverse.

Ifthe j thcolumnvectorof A is

thenitfollowsfromFormula(8)inSection1.3that the j thcolumnvectorof  BAB

Consequentlynomatrix B canbefoundtomaketheproduct  BAI thus A doesnothaveaninverse. (k) False.E.g. I and I arebothinvertiblebut

 IIO isnot.

1.5 Elementary Matrices and a Method for Finding A-1

1. (a) Elementarymatrix(correspondstoadding5timesthefirstrowtothesecondrow)

(b) Notanelementarymatrix

(d) Notanelementarymatrix

2. (a) Elementarymatrix(correspondstomultiplyingthesecondrowby3)

(b) Elementarymatrix(correspondstointerchangingthefirstrowandthethirdrow)

(d) Notanelementarymatrix

3. (a) Add3timesthesecondrowtothefirstrow:

(b) Multiplythefirstrowby17:

(d) Interchangethefirstandthirdrows:

4. (a) Add3timesthefirstrowtothesecondrow:

(b) Multiplythethirdrowby1

(d) Add17timesthethirdrowtothefirstrow:

5. (a) Interchangethefirstandsecondrows:

(b) Add3timesthesecondrowtothethirdrow:

6. (a) Multiplythefirstrowby6:

(b) Add4timesthefirstrowtothesecondrow:

7. (a)

(b)

001

010 100 ( B wasobtainedfrom A byinterchangingthefirstrowandthethirdrow)

001

010 100 ( A wasobtainedfrom B byinterchangingthefirstrowandthethirdrow)

100

(c)

010 201 ( C wasobtainedfrom A byadding2timesthefirstrowtothethirdrow)

100

(d)

8. (a)

(b)

010 201 ( A wasobtainedfrom C byadding2timesthefirstrowtothethirdrow)

100 030 001 ( D wasobtainedfrom B bymultiplyingthesecondrowby3)

1 3 100 00 001 ( B wasobtainedfrom D bymultiplyingthesecondrowby13)

100 012 001 ( F wasobtainedfrom B byadding2timesthethirdrowtothesecondrow)

(d)

100 012 001 ( B wasobtainedfrom F byadding2timesthethirdrowtothesecondrow)

9. (a) (MethodI:usingTheorem1.4.5)

Thedeterminantof

(MethodII:usingtheinversionalgorithm)

,isnonzero.Therefore A isinvertibleanditsinverseis

1410 0121 2timesthefirstrowwasaddedtothesecondrow.

   1410 0121Thesecondrowwasmultipliedby1

   1074 0121,4timesthesecondrowwasaddedtothefirstrow.

Theinverseis    74 21

(b) (MethodI:usingTheorem1.4.5)

Thedeterminantof A ,

det28440 A .Therefore A isnotinvertible.

(MethodII:usingtheinversionalgorithm)

   2410 4801Theidentitymatrixwasadjoinedtothegivenmatrix. 

2410 0021

2timesthefirstrowwasaddedtothesecondrow.

Arowofzeroswasobtainedontheleftside,therefore A isnotinvertible.

10. (a) (MethodI:usingTheorem1.4.5)

Thedeterminantof A ,

det116531 A ,isnonzero.Therefore A isinvertibleanditsinverse is

11 1 165165 3131 A

(MethodII:usingtheinversionalgorithm)



  1510 31601Theidentitymatrixwasadjoinedtothegivenmatrix.

   1510 0131 3timesthefirstrowwasaddedtothesecondrow.

   1510 0131Thesecondrowwasmultipliedby1.

   10165 0131,5timesthesecondrowwasaddedtothefirstrow.

Theinverseis

  165 31

(b) (MethodI:usingTheorem1.4.5)

Thedeterminantof A ,

det62430 A .Therefore A isnotinvertible.

(MethodII:usingtheinversionalgorithm)

   6410 3201Theidentitymatrixwasadjoinedtothegivenmatrix.

0012 3201

2timesthesecondrowwasaddedtothefirstrow.

Arowofzeroswasobtainedontheleftside,thereforethematrixisnotinvertible. 11. (a)

123100 253010 108001

123100 013210 025101

123100 013210 001521

1201463 0101353 001521

10040169 0101353 001521

Theidentitymatrixwasadjoinedtothegivenmatrix.

2timesthefirstrowwasaddedtothesecondrowand 1timesthefirstrowwasaddedtothethirdrow.

2timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1.

3timesthethirdrowwasaddedtothesecondrowand 3timesthethirdrowwasaddedtothefirstrow.

2timesthesecondrowwasaddedtothefirstrow. Theinverseis

40169 1353 521

Theidentitymatrixwasadjoinedtothegivenmatrix.

Thefirstrowwasmultipliedby1

2timesthefirstrowwasaddedtothesecondrowand 4timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasaddedtothethirdrow.

Arowofzeroswasobtainedontheleftside,thereforethematrixisnotinvertible.

Theidentitymatrixwasadjoinedtothegivenmatrix.

Eachrowwasmultipliedby5

1timesthefirstrowwasaddedtothesecondand 1timesthefirstrowwasaddedtothethirdrow.

Thesecondandthirdrowswereinterchanged.

Thesecondrowwasmultipliedby15and thethirdrowwasmultipliedby2 5

1 2timesthethirdrowwasaddedtothesecondrowand 2timesthethirdrowwasaddedtothefirstrow.

1timesthesecondrowwasaddedtothefirstrow.

Theidentitymatrixwasadjoinedtothegivenmatrix.

Eachrowwasmultipliedby5

2timesthefirstrowwasaddedtothesecondand 1timesthefirstrowwasaddedtothethirdrow.

1timesthesecondrowwasaddedtothethirdrow.

Arowofzeroswasobtainedontheleftside,thereforethematrixisnotinvertible.

Theidentitymatrixwasadjoinedtothegivenmatrix.

1timesthefirstrowwasaddedtothethirdrow.

011101

002111

1timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1 2

001

1timesthethirdrowwasaddedtothesecondand 1timesthethirdrowwasaddedtothefirstrow Theinverseis

Theidentitymatrixwasadjoinedtothegivenmatrix.

Eachofthefirsttworowswasmultipliedby12 .

4timesthefirstrowwasaddedtothesecondrow.

1 2 222 1326 13000 0100 001001

Thesecondrowwasmultipliedby113. 

232 2626 222 1326 1000 0100 001001

3timesthesecondrowwasaddedtothefirstrow. Theinverseis

232 2626 222 1326 0 0 001 .

266100

010110 001011

260166

010110 001011

7 2 10003 010110 001011

Theinverseis

10001000 13000100 13500010 13570001

10001000 03001100 03501010 03571001

1timesthesecondrowwasaddedtothethirdrow.

6timesthethirdrowwasaddedtothefirstrow

6timesthesecondrowwasaddedtothefirstrow

Thefirstrowwasmultipliedby12.

Theidentitymatrixwasadjoinedtothegivenmatrix.

1timesthefirstrowwasaddedtoeachoftheremaining rows. 

10001000 03001100 00500110 00570101

10001000 03001100 00500110 00070011

1timesthesecondrowwasaddedtothethirdrowand tothefourthrow.

1timesthethirdrowwasaddedtothefourthrow

Theinverseis

121200100 24001000 00200010 01450001

121200100 082401200 00200010 01450001

121200100 01450001 00200010 082401200

121200100 01450001 00200010 008401208

Thesecondrowwasmultipliedby13, thethirdrowwasmultipliedby15,and thefourthrowwasmultipliedby1 7

Theidentitymatrixwasadjoinedtothegiven matrix.

Thefirstandsecondrowswereinterchanged.

2timesthefirstrowwasaddedtothesecond.

Thesecondandfourthrowswereinterchanged.

Thesecondrowwasmultipliedby1.

8timesthesecondrowwasaddedtothefourth.

Thethirdrowwasmultipliedby1 2

8timesthethirdrowwasadded tothefourthrow.

Thefourthrowwasmultipliedby140

5timesthefourthrowwasadded tothesecondrow. 

113 842 1 2 1111 4020105 12000160 01000

4timesthethirdrowwasadded tothesecondrowand 12timesthethirdrowwasadded tothefirstrow.

2timesthesecondrowwasadded tothefirstrow.



10010100

00201000 01300010 21530001

10010100

 

00201000 01300010 01550201

10010100 01300010

00201000 01550201

Thefirstandsecondrowswereinterchanged.

2timesthefirstrowwasaddedtothefourthrow andtothefourthrow.

Thesecondandthirdrowswereinterchanged.



10010100 01300010

00201000 01550201

Thesecondrowwasmultipliedby1.



10010100 01300010

00201000 00850211

1timesthesecondrowwasadded tothefourthrow. 

10010100

01300010

00201000 00054211

4timesthethirdrowwasadded tothefourthrow. 

1 2 4211 5555 10010100 01300010 0010000 0001

Thethirdrowwasmultipliedby12and thefourthrowwasmultipliedby1 5 

3 411 5555 3 2 1 2 4211 5555 1000 0100010 0010000 0001

1timesthefourthrowwasaddedtothefirstrow and 3timesthethirdrowwasaddedtothesecond.

Theinverseis

Theidentitymatrixwasadjoinedtothegivenmatrix.

Thefirstrowwasmultipliedby11/, k thesecondrowwasmultipliedby21/, k thethirdrowwasmultipliedby31/, k and thefourthrowwasmultipliedby41/. k

Theinverseis

11 11 100000 01000100 001000 00010001 kk kk

Theidentitymatrixwasadjoinedtothegivenmatrix.

Firstrowandthirdrowwerebothmultipliedby1/ k .

1 k timesthefourthrowwasadded tothethirdrowand 1 k timesthesecondrowwasadded tothefirstrow. Theinverseis

11 11 00 0100 00 0001 kk kk

1 2 3 4 0001000

4 3 2 1 0000001 0000010 0000100 0001000 k k k k

Theinverseis

0001000 1000100 0100010 0010001 k k k k

1 11 11 11 1000000 100000 010000 001000 k kk kk kk

2 1 11 11 11 1000000 010000 010000 001000 k kk kk

Theidentitymatrixwasadjoinedtothegivenmatrix.

Thefirstandfourthrowswereinterchanged; thesecondandthirdrowswereinterchanged.

Thefirstrowwasmultipliedby41/, k thesecondrowwasmultipliedby31/, k thethirdrowwasmultipliedby21/, k and thefourthrowwasmultipliedby11/. k

Theidentitymatrixwasadjoinedtothegivenmatrix.

Eachrowwasmultipliedby1/ k .

1 k timesthefirstrowwasadded tothesecondrow.

2 32 1 11 111 11 1000000 010000 00100 001000 k kk kkk kk 1 k timesthesecondrowwasadded tothethirdrow.

1 k timesthethirdrowwasadded tothefourthrow.

Theinverseis

21. Itfollowsfromparts(a)and(c)ofTheorem1.5.3thatasquarematrixisinvertibleifandonlyifitsreducedrow echelonformisidentity.

Thefirstandthirdrowswereinterchanged.

c cc

1timesthefirstrowwasaddedtothesecondrowand c timesthefirstrowwasaddedtothethirdrow.

210cccc or 10 c ,i.e.if  0 c or  1 c

thelastmatrixcontainsatleastonerowofzeros,therefore itcannotbereducedto I byelementaryrowoperations.

Otherwise(if  0 c and  1 c ),multiplyingthesecondrowby  1 1 c andmultiplyingthethirdrowby2 1 cc would resultinarowechelonformwith1’sonthemaindiagonal.Subsequentelementaryrowoperationswouldthenlead totheidentitymatrix.

Weconcludethatforanyvalueof c otherthan0and1thematrixisinvertible.

22. Itfollowsfromparts(a)and(c)ofTheorem1.5.3thatasquarematrixisinvertibleifandonlyifitsreducedrow echelonformisidentity.

11 10 01 c c c

11 01 10 c c c

Thefirstandsecondrowswereinterchanged.

Thesecondandthirdrowswereinterchanged.

2 11 01 01 c c cc c timesthefirstrowwasaddedtothethirdrow.

3 11 01 002 c c cc 21 c timesthesecondrowwasaddedtothethird.

thelastmatrixcontainsarowofzeros,thereforeitcannot bereducedto I byelementaryrowoperations.

Otherwise(if  320cc ),multiplyingthelastrowby31 2 cc wouldresultinarowechelonformwith1’sonthemain diagonal.Subsequentelementaryrowoperationswouldthenleadtotheidentitymatrix.

Weconcludethatforanyvalueof c otherthan0,2and2thematrixisinvertible.

23. Weperformasequenceofelementaryrowoperationstoreducethegivenmatrixtotheidentitymatrix.Aswedoso, wekeeptrackofeachcorrespondingelementarymatrix:

15 222timesthesecondrowwasaddedtothefirst.

15 082timesthefirstrowwasaddedtothesecond.

15 01Thesecondrowwasmultipliedby18.

10 01 5timesthesecondrowwasaddedtothefirst.

Notethatthisanswerisnotuniquesinceadifferentsequenceofelementaryrowoperations(andthecorresponding elementarymatrices)couldbeusedinstead.

24. Weperformasequenceofelementaryrowoperationstoreducethegivenmatrixtotheidentitymatrix.Aswedoso, wekeeptrackofeachcorrespondingelementarymatrix:

Notethatthisanswerisnotuniquesinceadifferentsequenceofelementaryrowoperations(andthecorresponding elementarymatrices)couldbeusedinstead.

25. Weperformasequenceofelementaryrowoperationstoreducethegivenmatrixtotheidentitymatrix.Aswedoso, wekeeptrackofeachcorrespondingelementarymatrix:

Notethatthisanswerisnotuniquesinceadifferentsequenceofelementaryrowoperations(andthecorresponding elementarymatrices)couldbeusedinstead.

26. Weperformasequenceofelementaryrowoperationstoreducethegivenmatrixtotheidentitymatrix.Aswedoso, wekeeptrackofeachcorrespondingelementarymatrix:

011 001

110 010 001

1timesthefirstrowwasaddedtothesecondrow.

Thesecondandthirdrowswereinterchanged

1timesthethirdrowwasaddedtothesecond.

1timesthesecondrowwasaddedtothefirstrow.

correspondingelementarymatrix:

123 141 219 A

123 022 219

105 022 219

105 022 114 B

1timesthefirstrowwasaddedtothesecondrow.

1timesthesecondrowwasaddedtothefirstrow.

1timesthefirstrowwasaddedtothethirdrow.

Since  321 EEEAB ,theequality  CAB issatisfiedbythematrix

Notethatthisanswerisnotuniquesinceadifferentsequenceofelementaryrowoperations(andthecorresponding elementarymatrices)couldbeusedinstead.

28. Letusperformasequenceofelementaryrowoperationstoproduce B from A .Aswedoso,wekeeptrackofeach correspondingelementarymatrix:

2timesthefirstrowwasaddedtothethirdrow.

694 510 121 B

4timesthethirdrowwasaddedtothefirstrow.

29.

Since  321 EEEAB ,theequality  CAB issatisfiedbythematrix

Notethatadifferentsequenceofelementaryrowoperations(andthecorrespondingelementarymatrices)couldbe usedinstead.(However,sinceboth A and B inthisexerciseareinvertible, C isuniquelydeterminedbythe formula  1 CBA .)

010 A abc cannotresultfrominterchangingtworowsof3 I (sincethatwouldcreateanonzeroentryabovethe maindiagonal).

A canresultfrommultiplyingthethirdrowof3 I byanonzeronumber c (inthiscase, 0,0abc ).

Theotherpossibilitiesarethat A canbeobtainedbyadding a timesthefirstrowtothethird  (0,1) bc orby adding b timesthesecondrowtothethird     0,1ac

Inallthreecases,atleastoneentryinthethirdrowmustbezero.

30. Considerthreecases:

 If  0 a then A hasarowofzeros(firstrow).

 If  0 a and  0 h then A hasarowofzeros(fifthrow).

 If  0 a and  0 h thenadding d a timesthefirstrowtothethird,andadding e h timesthefifthrowtothethird resultsinthethirdrowbecomingarowofzeros.

Inallthreecases,thereducedrowechelonformof A isnot5 I .ByTheorem1.5.3, A isnotinvertible.

True-False Exercises

(a) False.Anelementarymatrixresultsfromperforminga single elementaryrowoperationonanidentitymatrix;a productoftwoelementarymatriceswouldcorrespondtoasequenceoftwosuchoperationsinstead,whichgenerally isnotequivalenttoasingleelementaryoperation.

(b) True.ThisfollowsfromTheorem1.5.2.

(c) True.If A and B arerowequivalentthenthereexistelementarymatrices  1,, p EE suchthat   1 p BEEA . Likewise,if B and C arerowequivalentthenthereexistelementarymatrices  ** 1,, q EE suchthat   ** 1 q CEEB . Combiningthetwoequalitiesyields   ** 11qp CEEEEA therefore A and C arerowequivalent.

(d) True.Ahomogeneoussystem  0 Ax haseitheronesolution(thetrivialsolution)orinfinitelymanysolutions.If A isnotinvertible,thenbyTheorem1.5.3thesystemcannothavejustonesolution.Consequently,itmusthave infinitelymanysolutions.

(e) True.Ifthematrix A isnotinvertiblethenbyTheorem1.5.3itsreducedrowechelonformisnot n I .However,the matrixresultingfrominterchangingtworowsof A (anelementaryrowoperation)musthavethesamereducedrow echelonformas A does,sobyTheorem1.5.3thatmatrixisnotinvertibleeither.

(f) True.Addingamultipleofthefirstrowofamatrixtoitssecondrowisanelementaryrowoperation.Denotingby E bethecorrespondingelementarymatrixwecanwrite    111 EAAE sotheresultingmatrix EA isinvertibleif A is.

(g) False.Forinstance,

1.6 More on Linear Systems and Invertible Matrices

1. Thegivensystemcanbewritteninmatrixformas  Axb ,where

Webeginbyinvertingthecoefficientmatrix A

5timesthefirstrowwasaddedtothesecondrow.

1timesthesecondrowwasaddedtothefirstrow.

Webeginbyinvertingthecoefficientmatrix A

4310 2501Theidentitymatrixwasadjoinedtothecoefficientmatrix.

2501 4310Thefirstandsecondrowswereinterchanged.

2501 0712 2timesthefirstrowwasaddedtothesecondrow.

51 22 12 77 10 01

Thefirstrowwasmultipliedby12and thesecondrowwasmultipliedby1 7

53 1414 12 77 10 01 5 2timesthesecondrowwasaddedtothefirstrow. Since

53 11414 12 77 A ,Theorem1.6.2statesthatthesystemhasexactlyonesolution  1 A xb :

53 11414 12 277 33 93 x x ,i.e.,  123xx .

3. Thegivensystemcanbewritteninmatrixformas  Axb ,where

4 1 3 b .Webeginbyinvertingthecoefficientmatrix A

131100 221010 231001

131100 041210 031201

131100 041210 010011

131100 010011 041210

131100 010011 001234

Theidentitymatrixwasadjoinedtothecoefficientmatrix.

2timesthefirstrowwasaddedtothesecondand 2timesthefirstrowwasaddedtothethirdrow.

1timesthesecondrowwasaddedtothethirdrow.

Thesecondandthirdrowswereinterchanged.

4timesthesecondrowwasaddedtothethirdrow.

010011 001234

130134 010011 001234

010011 001234

Thethirdrowwasmultipliedby1

1timesthethirdrowwasaddedtothefirstrow.

3timesthesecondrowwasaddedtothefirstrow.

532100 332010 011001Theidentitymatrixwasadjoinedtothecoefficientmatrix.

200110 332010 011001 1timesthesecondrowwasaddedtothefirstrow.

11 22 1000 332010 011001

11 22 35 22 1000 0320 011001

Thefirstrowwasmultipliedby1 2

3timesthefirstrowwasaddedtothesecondrow.

111100 005110 055401

111100 055401 005110

Thesecondandthirdrowswereinterchanged.

3timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1

1timesthethirdrowwasaddedtothesecondrow.

Theidentitymatrixwasadjoinedtothecoefficientmatrix.

1timesthefirstrowwasaddedtothesecondrowand 4timesthefirstrowwasaddedtothethirdrow.

Thesecondandthirdrowswereinterchanged.

0110 0010

41 55 11 55 111100

41 55 311 555 11 55 1100 010 0010

11 55 311 555 11 55 1000 010 0010

Webeginbyinvertingthecoefficientmatrix

01231000 11440100 13790010 12460001

11440100 01231000 13790010 12460001

11440100 01231000 02350110 01020101

Thesecondrowwasmultipliedby15and thethirdrowwasmultipliedby15.

1timesthethirdrowwasaddedtothesecondrow andtothefirstrow.

1timesthesecondrowwasaddedtothefirstrow.

Theidentitymatrixwasadjoinedtothecoefficient matrix.

Thefirstandsecondrowswereinterchanged.

1timesthefirstrowwasaddedtothethirdrowand thefirstrowwasaddedtothefourthrow.



11440100

01231000 02350110 01020101

Thesecondrowwasmultipliedby1



11440100 01231000 00112110 00211101

2timesthesecondrowwasaddedtothethirdrow andthesecondrowwasaddedtothefourth.

Thethirdrowwasmultipliedby1



11440100 01231000 00112110 00013121

2timesthethirdrowwasaddedtothefourth.



11440100 01231000 00112110 00013121

114012384 01208363 00101011 00013121

Thefourthrowwasmultipliedby1.

1timesthelastrowwasaddedtothethirdrow, 3timesthelastrowwasaddedtothesecondrow and4timesthelastrowwasaddedtothefirst.

 

11008340 01006341 00101011 00013121

10002001 01006341 00101011 00013121

2timesthethirdrowwasaddedtothesecondrow and 4timesthethirdrowwasaddedtothefirstrow.

1timesthesecondrowwasaddedtothefirst.

1 2 b b b .Webeginbyinvertingthecoefficientmatrix A

3510 1201Theidentitymatrixwasadjoinedtothecoefficientmatrix.

1201 3510Thefirstandsecondrowswereinterchanged.

1201 0113

3timesthefirstrowwasaddedtothesecondrow.

1201 0113Thesecondrowwasmultipliedby1.

1025 0113

2timesthesecondrowwasaddedtothefirstrow. Since

125 13 A ,Theorem1.6.2statesthatthesystemhasexactlyonesolution

123100 255010 358001

123100 011210 011301

Theidentitymatrixwasadjoinedtothecoefficientmatrix.

2timesthefirstrowwasaddedtothesecondrowand 3timesthefirstrowwasaddedtothethirdrow. 

123100

011210 002511

Thesecondrowwasaddedtothethirdrow. 

123100 011210 001

Thethirdrowwasmultipliedby12.

511 222



1333 222 111 222 511 222 120 010 001

Thethirdrowwasaddedtothesecondrowand 3timesthethirdrowwasaddedtothefirstrow.

155 1 222 111 222 511 222 100 010 001



2timesthesecondrowwasaddedtothefirstrow. Since

155 1 222 1111 222 511 222 A ,Theorem1.6.2statesthatthesystemhasexactlyonesolution  1 A xb :

constantsontherighthandsidesofthesystems (i)and(ii)–refertoExample2.

3timesthefirstrowwasaddedtothesecondrow.

Weconcludethatthesolutionsofthetwosystemsare:

Weaugmentedthecoefficientmatrixwithtwocolumnsof constantsontherighthandsidesofthesystems (i)and(ii)–refertoExample2.

Thefirstrowwasmultipliedby1.

1timesthefirstrowwasaddedtothesecondrowand 6timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby113.

28timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby13 2

1 13timesthethirdrowwasaddedtothesecondrow andthethirdrowwasaddedtothefirstrow.

4timesthesecondrowwasaddedtothefirstrow.

Weconcludethatthesolutionsofthetwosystemsare:

470415 121631Weaugmentedthecoefficientmatrixwithfourcolumns ofconstantsontherighthandsidesofthesystems(i), (ii),(iii),and(iv)–refertoExample2.

121631 470415Thefirstandsecondrowswereinterchanged.

121631 015428139 4timesthefirstrowwasaddedtothesecondrow.

428133 1515155 121631 01Thesecondrowwasmultipliedby115

734191 1515155 428133 1515155 10 012timesthesecondrowwasaddedtothefirstrow.

Weconcludethatthesolutionsofthefoursystemsare:

135101 120011 254110

135101 015112 016312

135101 015112 001220

Weaugmentedthecoefficientmatrixwiththreecolumns ofconstantsontherighthandsidesofthesystems (i),(ii)and(iii)–refertoExample2.

Thefirstrowwasaddedtothesecondrowand 2timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1

1309101

0109112 001220

5timesthethirdrowwasaddedtothefirstrow andtothesecondrow.

3timesthesecondrowwasaddedtothefirstrow.

Weconcludethatthesolutionsofthethreesystemsare:

(i)

118,x  29x ,  32x (ii)

123x ,  211x ,  32x (iii)

15,x  22x ,  30x

Thesystemisconsistentforallvaluesof1 b and2 b .

2timesthefirstrowwasaddedtothesecondrow.

Thesecondrowwasmultipliedby1 7

Thesystemisconsistentifandonlyif

Theaugmentedmatrixforthesystem.

Thefirstrowwasmultipliedby1

3timesthefirstrowwasaddedtothesecondrow.

4timesthefirstrowwasaddedtothesecondrow and3timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby1

Theaugmentedmatrixforthesystem.

4timesthefirstrowwasaddedtothesecondrow

Thesecondandthirdrowswereinterchanged.

Thesecondrowwasmultipliedby1.

3timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby12.

Theaugmentedmatrixforthesystem.

2timesthefirstrowwasaddedtothesecondrow, 3timesthefirstrowwasaddedtothethirdrow,and 4timesthefirstrowwasaddedtothefourthrow.

Thesecondrowwasmultipliedby1.

Thesecondrowwasaddedtothethirdrowand 1timesthesecondrowwasaddedtothefourthrow.

Thesystemisconsistentforallvaluesof1 b ,2 b ,3 b ,and4 b thatsatisfytheequations

Theseequationsformalinearsysteminthevariables1 b ,2 b ,3 b ,and4 b whoseaugmentedmatrix

11100 21010 hasthereducedrowechelonform

134 bbb and  234 2 bbb

10110 01210.Thereforethesystemisconsistentif

18. (a) Theequation  xx A canberewrittenas  xxAI ,whichyields  xx0AI and

 x0 AI .

Thisisamatrixformofahomogeneouslinearsystem-tosolveit,wereduceitsaugmentedmatrixtoarow echelonform.

Theaugmentedmatrixforthehomogeneoussystem

 x0 AI .

2timesthefirstrowwasaddedtothesecondrow and3timesthefirstrowwasaddedtothethirdrow.

1120 0160 0260

1120 0160 0060

1120 0160 0010

Thesecondrowwasmultipliedby1.

2timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1 6

Usingback-substitution,weobtaintheuniquesolution:   1230 xxx (b) Aswasdoneinpart(a),theequation  x4x A canberewrittenas    4x0AI .Wesolvethelattersystem byGauss-Jordanelimination

2120 2220 3130

2220 2120 3130

Theaugmentedmatrixforthehomogeneoussystem

  4x0AI .

Thefirstandsecondrowswereinterchanged.

Thefirstrowwasmultipliedby12.

2timesthefirstrowwasaddedtothesecondrowand 3timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby1

4timesthesecondrowwasaddedtothethirdrowand thesecondrowwasaddedtothefirstrow.

Ifweassign3 x anarbitraryvalue t ,thegeneralsolutionisgivenbytheformulas

021001

Theidentitymatrixwasadjoinedtothematrix.

2timesthefirstrowwasaddedtothesecondrow.

2timesthethirdrowwasaddedtothesecondrow.

021001

2timesthesecondrowwasaddedtothethirdrow.

111100 010212 001425

110525 010212 001425

100313 010212 001425

1141379 X .Letusfind

201100 011010 114001

114001 011010 201100

114001 011010 027102

Thethirdrowwasmultipliedby1

1timesthethirdrowwasaddedtothefirstrow.

Thesecondrowwasaddedtothefirstrow.

Theidentitymatrixwasadjoinedtothematrix.

Thefirstandthirdrowswereinterchanged.

2timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasmultipliedby1

114001

011010 009122

2timesthesecondrowwasaddedtothethirdrow.

122 999 114001

011010 001

8 41 999 7 12 999 122 999 110 010 001

511 999 7 12 999 122 999 100 010 001

Thethirdrowwasmultipliedby19.

1timesthethirdrowwasaddedtothesecondrowand 4timesthethirdrowwasaddedtothefirstrow.

1timesthesecondrowwasaddedtothefirstrow. Using

1511 999 7 12 999 122 999 201 011 114 weobtain

5252523 11 999999 74040 1244 999999 122233237 999999 43213 67894 13792 X

True-False Exercises

(a) True.ByTheorem1.6.1,ifasystemoflinearequationhasmorethanonesolutionthenitmusthaveinfinitelymany.

(b) True.If A isasquarematrixsuchthat  Axb hasauniquesolutionthenthereducedrowechelonformof A must be I .Consequently,  Axc musthaveauniquesolutionaswell.

(d) True.Since A and B arerowequivalentmatrices,itmustbepossibletoperformasequenceofelementaryrow operationson A resultingin B .Let E betheproductofthecorrespondingelementarymatrices,i.e.,  EAB .Note that E mustbeaninvertiblematrixthus  1 AEB .

Anysolutionof  0 Ax isalsoasolutionof  0 Bx since  00 BEAE xx . Likewise,anysolutionof  0 Bx isalsoasolutionof  0 Ax since   1100AEBE xx

(e) True.If   1 SAS xb then   1 SSASASS xxb .Consequently,  S yx isasolutionof  ASyb .

(f)

(g)

True.  4 A xx isequivalentto  4 n AIxx ,whichcanberewrittenas    40 n AI x .ByTheorem1.6.4,this homogeneoussystemhasauniquesolution(thetrivialsolution)ifandonlyifitscoefficientmatrix4 n AI is invertible.

True.If A B wereinvertible,thenbyTheorem1.6.5both A and B wouldbeinvertible.

1.7 Diagonal, Triangular, and Symmetric Matrices

1. (a) Thematrixisuppertriangular.Itisinvertible(itsdiagonalentriesarebothnonzero).

(b) Thematrixislowertriangular.Itisnotinvertible(itsdiagonalentriesarezero).

(d) Thematrixisuppertriangular.Itisnotinvertible(itsdiagonalentriesincludeazero).

2. (a) Thematrixislowertriangular.Itisinvertible(itsdiagonalentriesarebothnonzero).

(b) Thematrixisuppertriangular.Itisnotinvertible(itsdiagonalentriesarezero).

(d) Thematrixislowertriangular.Itisnotinvertible(itsdiagonalentriesincludeazero).

(a)

21 13

03 30

19. Frompart(c)ofTheorem1.7.1,atriangularmatrixisinvertibleifandonlyifitsdiagonalentriesareallnonzero. Sincethisuppertriangularmatrixhasa0onitsdiagonal,itisnotinvertible.

20. Frompart(c)ofTheorem1.7.1,atriangularmatrixisinvertibleifandonlyifitsdiagonalentriesareallnonzero. Sincethisuppertriangularmatrixhasallthreediagonalentriesnonzero,itisinvertible.

21. Frompart(c)ofTheorem1.7.1,atriangularmatrixisinvertibleifandonlyifitsdiagonalentriesareallnonzero. Sincethislowertriangularmatrixhasallfourdiagonalentriesnonzero,itisinvertible.

22. Frompart(c)ofTheorem1.7.1,atriangularmatrixisinvertibleifandonlyifitsdiagonalentriesareallnonzero. Sincethislowertriangularmatrixhasa0onitsdiagonal,itisnotinvertible.

AB .Thediagonalentriesof A B are:3,5,6.

015 0016

050

AB .Thediagonalentriesof A B are:24,0,42.

25. Thematrixissymmetricifandonlyif 53 a .Inorderfor A tobesymmetric,wemusthave  8 a

26. Thematrixissymmetricifandonlyifthefollowingequationsmustbesatisfied

WesolvethissystembyGauss-Jordanelimination

1223 2110 1012

1012 2110 1223

1012 0114 0215

1012 0114 00113

10011 0109 00113

Theaugmentedmatrixforthesystem.

Thefirstandthirdrowswereinterchanged.

2timesthefirstrowwasaddedtothesecondrow and1timesthefirstrowwasaddedtothethird.

2timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1

Thethirdrowwasaddedtothesecondrow and1timesthethirdrowwasaddedtothefirst.

Inorderfor A tobesymmetric,wemusthave  11 a ,  9 b ,and  13 c .

27. Frompart(c)ofTheorem1.7.1,atriangularmatrixisinvertibleifandonlyifitsdiagonalentriesareallnonzero. Therefore,thegivenuppertriangularmatrixisinvertibleforanyrealnumber

28. Frompart(c)ofTheorem1.7.1,atriangularmatrixisinvertibleifandonlyifitsdiagonalentriesareallnonzero. Therefore,thegivenlowertriangularmatrixisinvertibleforanyrealnumber x suchthat 

29. ByTheorem1.7.1,1 A isalsoanuppertriangularorlowertriangularinvertiblematrix.Itsdiagonalentriesmustall benonzero-theyarereciprocalsofthecorrespondingdiagonalentriesofthematrix A

30. ByTheorem1.4.8(e),   ATTT BBA .Thereforewehave:

BTTTT BBBBB ,and

32. Forexample

since A issymmetric.

A (therearesevenotherpossibleanswers,e.g.,

122050182250102153 021030081230001133 020040080240000143 AB

 

21217 0210 0012 .Sincethisisanuppertriangularmatrix,wehaveverifiedTheorem1.7.1(b).

34. (a) Theorem1.4.8(e)statesthat   ATTT BBA (ifthemultiplicationcanbeperformed).Therefore,

whichshowsthat2 A issymmetric. (b)

 2 222 Th.Th.and 1.4.81.4.8are (b-d)(e)symmetric 23232323 TTTTTTT AI AAIAAIAAIAAI whichshowsthat  2 23AAI issymmetric.

35. (a)

issymmetric,thereforeweverifiedTheorem1.7.4.

217010 374001

Theidentitymatrixwasadjoinedtothematrix A

2timesthefirstrowwasaddedtothesecondrowand 3timesthefirstrowwasaddedtothethirdrow.

123100 015301 031210

Thesecondandthirdrowswereinterchanged.

Thesecondrowwasmultipliedby1

3timesthesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby114.

1113 141414

1939 141414 1351 141414 1113 141414

5timesthethirdrowwasaddedtothesecondrowand 3timesthethirdrowwasaddedtothefirstrow.

2timesthesecondrowwasaddedtothefirstrow.

Thisisazeromatrixwheneverthevalueof a , b ,and c iseither4or1.Weconcludethatthefollowingareall  33diagonalmatricesthatsatisfytheequation:

400400400100 040,040,010,040, 004001004004

400100100100 010,040,010,010 001001004001

37. (a)

2222 jiij ajiija forall i and j therefore A issymmetric. (b)

22 ji aji doesnotgenerallyequal  22 ij aij for

(c)

2222 jiij ajiija forall i and j therefore A issymmetric. (d)

2322 ji aji doesnotgenerallyequal

2322 ij aij for  ij therefore A isnotsymmetric(unless 

Thefirstandthethirdequationsyield

Substitutingtheseintothesecondequationleadsto

Weconcludethattheonlyuppertriangularmatrix A

40. (a)

Step1.Solve

Step2.Solve

(b)

Step1.Solve

Step2.Solve

Theaugmentedmatrix

44.

23102 35503 58605 00010 hasthereducedrowechelonform

101001 01700 00010 00000 .

Ifweassign c thearbitraryvalue t ,thegeneralsolutionisgivenbytheformulas

whichdoesnotgenerally equal AB .(Theproductofskew-symmetricmatricesthatcommuteissymmetric.)

1 2 AT A issymmetricsince

1 2 AT A isskew-symmetricsince

thereforetheresultfollowsfromtheidentity

11 22ATTAAAA

45. (a)

1 T A

1 TA

1 A

 1 A

(b) TTA

 A

 TA

Theorem1.4.9(d)

Theassumption: A isskew-symmetric

Theorem1.4.7(c)

Theorem1.4.8(a)

Theassumption: A isskew-symmetric    AT B

TTAB

AB

Theorem1.4.8(b)

Theassumption: A and B areskew-symmetric

Theorem1.4.1(h)

AT B

 TTAB

AB

AB

Theorem1.4.8(c)

Theassumption: A and B areskew-symmetric

Theorem1.4.1(i)

kAT  TkA

Theorem1.4.8(d)

Theassumption: A isskew-symmetric

Theorem1.4.1(l) 47.

True-False Exercises

(a) True.Everydiagonalmatrixissymmetric:itstransposeequalstotheoriginalmatrix.

(b) False.Thetransposeofanuppertriangularmatrixisa lower triangularmatrix.

(d) True.Mirrorimagesofentriesacrossthemaindiagonalmustbeequal-seethemarginnotenexttoExample4.

(e) True.Allentriesbelowthemaindiagonalmustbezero.

(f) False.ByTheorem1.7.1(d),theinverseofaninvertiblelowertriangularmatrixisalowertriangularmatrix.

(g) False.Adiagonalmatrixisinvertibleifandonlyifalloritsdiagonalentriesarenonzero(positiveornegative).

(h) True.Theentriesabovethemaindiagonalarezero.

(i) True.If A isuppertriangularthen TA islowertriangular.However,if A isalsosymmetricthenitfollowsthat  T AA mustbebothuppertriangularandlowertriangular.Thisrequires A tobeadiagonalmatrix.

(j) False.Forinstance,neither

(k) False.Forinstance,neither

(l) False.Forinstance,

00 10 A isnotsymmetriceventhough



200 00 A is.

(m) True.ByTheorem1.4.8(d),   TTkAkA .Since kA issymmetric,wealsohave    T kAkA .Fornonzero k the equalityoftherighthandsides  T kAkA implies  T AA .

1.8 Matrix Transformations

1. (a)    A TAxx mapsanyvector x in2 R intoavector  A wx in3 R .

Thedomainof AT is2 R ;thecodomainis3 R .

(b)

(c)

(d)

2. (a)

 A TAxx mapsanyvector x in3 R intoavector  A wx in2 R .

Thedomainof AT is3 R ;thecodomainis2 R

  A TAxx mapsanyvector x in3 R intoavector  A wx in3 R

Thedomainof AT is3 R ;thecodomainis3 R

  A TAxx mapsanyvector x in6 R intoavector  A wx in  1 RR

Thedomainof AT is6 R ;thecodomainis R

  A TAxx mapsanyvector x in5 R intoavector  A wx in4 R

Thedomainof AT is5 R ;thecodomainis4 R .

(b)    A TAxx mapsanyvector x in4 R intoavector  A wx in5 R .

Thedomainof AT is4 R ;thecodomainis5 R .

Thedomainof AT is4 R ;thecodomainis4 R

(d)    A TAxx mapsanyvector x in  1 RR intoavector  A wx in3 R

Thedomainof AT is R ;thecodomainis3 R

3. (a) Thetransformationmapsanyvector x in2 R intoavector w in2 R Itsdomainis2 R ;thecodomainis2 R

(b) Thetransformationmapsanyvector x in2 R intoavector w in3 R . Itsdomainis2 R ;thecodomainis3 R .

4. (a) Thetransformationmapsanyvector x in3 R intoavector w in3 R . Itsdomainis3 R ;thecodomainis3 R .

(b) Thetransformationmapsanyvector x in3 R intoavector w in2 R Itsdomainis3 R ;thecodomainis2 R

5. (a) Thetransformationmapsanyvector x in3 R intoavectorin2 R Itsdomainis3 R ;thecodomainis2 R

(b) Thetransformationmapsanyvector x in2 R intoavectorin3 R Itsdomainis2 R ;thecodomainis3 R

6. (a) Thetransformationmapsanyvector x in2 R intoavectorin2 R . Itsdomainis2 R ;thecodomainis2 R

(b) Thetransformationmapsanyvector x in3 R intoavectorin3 R . Itsdomainis3 R ;thecodomainis3 R

7. (a) Thetransformationmapsanyvector x in2 R intoavectorin2 R . Itsdomainis2 R ;thecodomainis2 R .

(b) Thetransformationmapsanyvector x in3 R intoavectorin2 R . Itsdomainis3 R ;thecodomainis2 R .

8. (a) Thetransformationmapsanyvector x in4 R intoavectorin2 R . Itsdomainis4 R ;thecodomainis2 R .

(b) Thetransformationmapsanyvector x in3 R intoavectorin3 R . Itsdomainis3 R ;thecodomainis3 R .

9. Thetransformationmapsanyvector x in2 R intoavectorin3 R .Itsdomainis2 R ;thecodomainis3 R

10. Thetransformationmapsanyvector x in3 R intoavectorin4 R .Itsdomainis3 R ;thecodomainis4 R

11. (a) Thegivenequationscanbeexpressedinmatrixformas

(b) Thegivenequationscanbeexpressedinmatrixformas

12. (a) Thegivenequationscanbeexpressedinmatrixformas

(b) Thegivenequationscanbeexpressedinmatrixformas

15. Thegivenequationscanbeexpressedinmatrixformas

Bymatrixmultiplication,

24112142w

33122143w

21. (a) If

(b) If

22. (a) If

1212121212 ,2,2, TkTkukukukukukukuuuukT uu

123 ,, uuu u and

123 ,, vvv v then

112233 ,, TTuvuvuv uv

11331122 ,, uvuvuvuv

13121312 ,,,, uuuuvvvv

TTuv

12313121312 ,,,,,, TkTkukukukukukukukuuuukT uu .

23. (a) Thehomogeneitypropertyfailstoholdsince  222 (,)((),)(,) Tkxkykxkykxky doesnotgenerallyequal

22 ,,, kTxykxykxky .(Itcanbeshownthattheadditivitypropertyfailstoholdaswell.)

(b) Thehomogeneitypropertyfailstoholdsince 

 2 ,,,,,, Tkxkykzkxkykxkzkxkykxz doesnotgenerally equal

,,,,,, kTxyzkxyxzkxkykxz .(Itcanbeshownthattheadditivitypropertyfailstoholdaswell.)

24. (a) Thehomogeneitypropertyfailstoholdsince

 ,,1Tkxkykxky doesnotgenerallyequal

,,1, kTxykxykxkyk .(Itcanbeshownthattheadditivitypropertyfailstoholdaswell.)

(b) Thehomogeneitypropertyfailstoholdsince 

 123123 ,,,, Tkxkxkxkxkxkx doesnotgenerallyequal

123123123 ,,,,,, kTxxxkxxxkxkxkx .(Itcanbeshownthattheadditivitypropertyfailstoholdas well.)

25. Thehomogeneitypropertyfailstoholdsincefor  0 b ,

  fkxmkxb doesnotgenerallyequal

kfxkmxbkmxkb .(Itcanbeshownthattheadditivitypropertyfailstoholdaswell.)

Ontheotherhand,bothpropertiesholdfor  0 b :

 fxymxymxmyfxfy and

fkxmkxkmxkfx

Consequently, f isnotamatrixtransformationon R unless  0 b

26. BothpropertiesofTheorem1.8.2holdfor

 ,0,0Txy :

 ,,,0,00,00,0,, TxyxyTxxyyTxyTxy

 ,,0,00,0, TkxyTkxkykkTxy

Ontheotherhand,neitherpropertyholdsingeneralfor

 ,1,1Txy ,e.g.,

,,x,1,1TxyxyTxyy doesnotequal

.ByFormula(13),thestandardmatrixfor

38. (a)

39. ByFormula(13),thestandardmatrixfor

(a)

(b) Since A T isamatrixtransformation,

43. Reflectionaboutthe x y -plane:

sinsin).Thegeometriceffectofmultiplying TA by x istorotatethevectorthroughtheangle  (i.e., torotatethroughtheangle  clockwise).

Theidentitymatrixwasadjoinedtotheoriginalmatrix.

2timesthefirstrowwasaddedtothethirdrow.

Thesecondrowwasaddedtothethirdrow.

Thethirdrowwasmultipliedby1 7

Thethirdrowwasaddedtothesecondrow.

3timesthethirdrowwasaddedtothefirstrow.

1timesthesecondrowwasaddedtothefirstrow.

Likewise,

Therefore,thestandardmatrixforTis

47. Theterminalpointofthevectorisfirstrotatedabouttheoriginthroughtheangle ,thenitis translatedbythevector0 x .No,thisisnotamatrixtransformation,forinstanceitfailstheadditivity property:

49. Since

22 cossincos2and

effectofmultiplying A by x istorotatethevectorthroughtheangle  2.

True-False Exercises

(a) False.Thedomainof A T is3 R .

(b) False.Thecodomainof A T is m R .

(d) False.(RefertoTheorem1.8.3.)

(e) True.Thecolumnsof A are   0 Ti e .

(f) False.Thegivenequalitymustholdforeverymatrixtransformationsinceitfollowsfromthehomogeneityproperty.

(g) False.Thehomogeneitypropertyfailstoholdsince

  Tkkxxb doesnotgenerallyequal

1. (a) FromTables1and3inSection1.8,

Forthesetransformations,

(b) FromTable1inSection1.8,

FromTable3inSection1.8,

Forthesetransformations,

(b) FromTables5and1inSection1.8,

Forthesetransformations,

3. FromTables2and4inSection1.8,

4. FromTable4inSection1.8,

7. (a) Wearelookingforthestandardmatrixof   21TTT where1 T isarotationof  90and2 T isareflectionabout theline  yx .FromTables5and1inSection1.8,

(b) Wearelookingforthestandardmatrixof   21TTT where1 T isanorthogonalprojectionontothe y -axisand 2T isarotationof  45abouttheorigin.FromTables3and5inSection1.8,

(c) Wearelookingforthestandardmatrixof   21TTT where1 T isareflectionaboutthe x -axisand2 T isa rotationof  60abouttheorigin.FromTables1and5inSection1.8,

(a) Wearelookingforthestandardmatrixof

321 TTTT where1 T isarotationof  60,2

(b) Wearelookingforthestandardmatrixof   321 TTTT where1 T isanorthogonalprojectionontothex-axis,

2T isarotationof  45,and3 T isareflectionaboutthe y -axis.FromTables3,5,and1inSection1.8,

(c)

Wearelookingforthestandardmatrixof   321 TTTT where1 T isarotationof  15,2 T isarotationof

105,and3 T isarotationof  60.Theneteffectofthethreerotationsisasinglerotationof

1510560180.FromTable5inSection1.8,

9. (a) Wearelookingforthestandardmatrixof   21TTT where1 T isareflectionaboutthe yz -planeand2 T isan orthogonalprojectionontothe x z -plane.FromTables2and4inSection1.8,

(b) Wearelookingforthestandardmatrixof   21TTT where1 T isareflectionaboutthe x y -planeand2 T isan orthogonalprojectionontothe x y -plane.FromTables2and4inSection1.8,

21TTT where1 T isanorthogonalprojectiononthe x y -planeand 2T isareflectionaboutthe yz -plane.FromTables4and2inSection1.8,

10. (a) Wearelookingforthestandardmatrixof   321 TTTT where1 T isareflectionaboutthe x y -plane,2 T isan orthogonalprojectionontothe x z -plane,and3 T isthetransformationsuchthat

FromTables2and4insection1.8,

Invectorform,

(b) Wearelookingforthestandardmatrixof

where1 T isareflectionaboutthe x y -plane,2 T isa reflectionaboutthe x z -plane,and3 T isanorthogonalprojectiononthe yz -plane.FromTables2and4in Section1.8,

where1 T isanorthogonalprojectionontothe yzplane,2 T isthetransformationsuchthat

,and3 T isareflectionaboutthe x y -plane.

FromTables4and2insection1.8,

12. (a) Invectorform,

Likewise,

isnotdefinedbecausetheoutputsfrom2 T arevectorsin4 R buttheinputsfor1 T arevectorsin2 R

theinverse:

4timesthesecondrowwassubtractedfromthefirstrow.

Sinceweobtainedarowofzerosontheleftside,theoperatorisnotone-to-one.

Theidentitymatrixwasadjoinedtothecoefficientmatrix.

2timesthefirstrowwasaddedtothesecondrowandthe firstrowwasaddedtothethirdrow.

Thesecondrowwassubtractedfromthethirdrow.

Sinceweobtainedarowofzerosontheleftside,theoperatorisnotone-to-one.

(a)

1121 2122 2323 551 wxxx wxxx ;thestandardmatrixis

23 51.UsingTheorem1.5.3(c),weattemptto findtheinverse:

2310 5101Theidentitymatrixwasadjoinedtothecoefficientmatrix.

17013 5101

3timesthesecondrowwasaddedtothefirstrow.

13 101717 5101Thefirstrowwasmultipliedby117

13 1717 52 1717 10 01

5timesthefirstrowwassubtractedfromthesecondrow.

Sincethereducedrowechelonformoftheoperator’sstandardmatrixistheidentity,theoperatorisinvertible.

UsingTheorem1.5.3(c),weattempttofindtheinverse:

123100 253010 108001

123100 013210 025101

Theidentitymatrixwasadjoinedtothematrix A

2timesthefirstrowwasaddedtothesecondrowand thefirstrowwassubtractedfromthethirdrow.

2timesthesecondrowwasaddedtothethirdrow.

001521

123100 013210 001521

Thesecondrowwasmultipliedby1

001521

3timesthethirdrowaddedtothesecondrowand3 timesthethirdrowwassubtractedfromthefirstrow.

2timesthesecondrowwassubtractedfromthefirst row.

Sincethereducedrowechelonformoftheoperator’sstandardmatrixistheidentity,theoperatorisinvertible.

Theorem1.4.5thattheoperatorisinvertible;

Theorem1.4.5thattheoperatorisnotinvertible. 20. (a)

sincethereducedrowechelonformofthematrix

,itfollows fromTheorem1.5.3(c)thattheoperator T isinvertible.Therefore,thestandardmatrixof1 T is

Addingrow1torow2followedbyaddingrow2torow3inthereducedrowechelonformofthematrix

025110 000101 ,itfollowsfromTheorem1.5.3(c)thattheoperator T isnotinvertible.

19. (a)

21. (a) FromTable1inSection1.8,thestandardmatrixis

10 01 ;since   10 10 01,thematrixoperatoris invertible.Theinverseisalsoareflectionaboutthe x -axis.

(b) FromTable5inSection1.8,thestandardmatrixis

13 22 31 22 cos60sin60 sin60cos60 .Since   13 22 31 22 10, thematrixoperatorisinvertible.Theinverseisarotationof  60(equivalentto  300)abouttheorigin.

;since  10 0 00,thematrixoperatorisnot invertible.

22. (a) FromTable1inSection1.8,thestandardmatrixis

10 10 ,the matrixoperatorisinvertible.Theinverseisalsoareflectionabouttheline  yx

(b) FromTable3inSection1.8,thestandardmatrixis

00 01 ;since  00 0 01,thematrixoperatorisnot invertible.

01,thematrixoperatorisinvertible.Theinverseisalsoa reflectionabouttheorigin.

23. (a) Since  12 10 11,itfollowsfromTheorem1.4.5thattheoperator AT isinvertible;

(b) Since  11 0 11,itfollowsfromTheorem1.4.5thattheoperator AT isnotinvertible.

24. (a) Sincethereducedrowechelonformofthematrix

fromTheorem1.5.3thattheoperator AT isnotinvertible.

(b) Sincethereducedrowechelonformofthematrix

fromTheorem1.5.3thattheoperator AT isinvertible.

AT x

25. (a) Invectorform,

01x x,y. 10y A Ty x Thegeometriceffectofapplying thistransformationto x istoreflect x about  yx andthentoreflecttheresultaboutthe origin.

(b) Forinstance,if

26. (a) Since

(thestandardmatrixofthereflectionabout  yx )and

(thestandardmatrixofthereflectionabouttheorigin)then

22 cossincos2and

2sincossin2,wehave

cos2sin2 sin2cos2 A .Thegeometriceffectofapplyingthistransformationto x istorotatethevector throughtheangle  2.

(b) Forinstance,if

True-False Exercises

cossin sincos B (thestandardmatrixoftherotationthroughanangle  )then  

(a) False.Forinstance,Example2showstwomatrixoperatorson2 R whosecompositionisnotcommutative.

(b) True.ThisisstatedasTheorem1.9.1.

(d) False.Forinstance,compositionofanyreflectionoperatorwithitselfistheidentityoperator,whichisnota reflection.

(e) True.Thereflectionofavector

(f) False.ThisfollowsfromExample6.

abouttheline  yx is

soasecondreflectionyields

(g) True.Thereflectionabouttheoriginisgivenbythetransformation    T xx sothat T isitsowninverse. 1.10 Applications of

1. Therearefournodes,whichwedenoteby A , B , C ,and D (seethefigureontheleft).

Wedeterminetheunknownflowrates1 x ,2 x ,and3 x assumingthecounterclockwisedirection(ifanyofthese quantitiesarefoundtobenegativethentheflowdirectionalongthecorrespondingbranchwillbereversed).

Thissystemcanberearrangedasfollows

Byinspection,thissystemhasauniquesolution

.Thisyieldstheflowratesand directionsshowninthefigureontheright.

2. (a) Therearefivenodes–eachofthemcorrespondstoanequation.

Thissystemcanberearrangedasfollows

(b) Theaugmentedmatrixofthelinearsystemobtainedinpart(a)hasthereducedrowechelonform

100101150 010101175

00110150 000011200 0000000 .Ifweassign4 x and6 x thearbitraryvalues s and t ,respectively,thegeneral

solutionisgivenbytheformulas

Thedirectionsoftheflowagreewiththearroworientationsinthediagram.

3. (a) Therearefournodes–eachofthemcorrespondstoanequation.

NetworknodeFlowInFlowOut topleft300400 topright(A)750250 bottomleft100400 bottomright(B)200300 xx

Thissystemcanberearrangedasfollows

(b) Theaugmentedmatrixofthelinearsystemobtainedinpart(a)

hasthereducedrow echelonform

0011500 00000 .Ifweassign4 x thearbitraryvalue s ,thegeneralsolutionisgivenby

theformulas

(c) Inorderforall i x valuestoremainpositive,wemusthave  500 s .Therefore,tokeepthetrafficflowingon allroads,theflowfrom A to B mustexceed500vehiclesperhour.

4. (a) Therearesixintersections–eachofthemcorrespondstoanequation.

IntersectionFlowInFlowOut topleft500300 topmiddle200 topright100600 bottomleft400350 bottommiddle600 bottomright450400

Werewritethesystemasfollows

(b) Theaugmentedmatrixofthelinearsystemobtainedinpart(a)hasthereducedrowechelonform

100001050 0100001450 0010010750 0001011600 000010150 00000000 .Ifweassign6 x and7 x thearbitraryvalues s and t ,respectively,the generalsolutionisgivenbytheformulas

150 xs , 2450 xt , 3750 xs ,

4600 xst , 550 xt ,  6 xs ,  7 xt subject totherestrictionthatallsevenvaluesmustbenonnegative.Obviously,weneedboth 60sx and   70 tx , whichinturnimply  10x and  20x .Additionallyimposingthethreeinequalities  37500 xs ,

 46000 xst ,and  5500 xt resultsinthesetofallowable s and t valuesdepictedinthegrey regiononthegraph.

(c) Setting  10x inthegeneralsolutionobtainedinpart(b)wouldresultinthenegativevalue   650sx whichisnotallowed(thetrafficwouldflowinawrongwayalongthestreetmarkedas6 x .)

5. FromKirchhoff'scurrentlawateachnode,wehave   1230. III Kirchhoff'svoltagelawyields

VoltageRisesVoltageDrops

LeftLoop(clockwise)226

RightLoop(clockwise)248 II II

12 23

(Anequationcorrespondingtotheouterloopisacombinationofthesetwoequations.)

Thelinearsystemcanberewrittenas

Itsaugmentedmatrixhasthereducedrowechelonform

Since2 I isnegative,thiscurrentisoppositetothedirectionshowninthediagram.

6. FromKirchhoff'scurrentlawateachnode,wehave  1230. III Kirchhoff'svoltagelawyields

LeftInsideLoop(clockwise)461 RightInsideLoop(clockwise)224

(Anequationcorrespondingtotheouterloopisacombinationofthesetwoequations.)

Thelinearsystemcanberewrittenas

Since1 I isnegative,thiscurrentisoppositetothedirectionshowninthediagram.

7. FromKirchhoff'scurrentlaw,wehave

Kirchhoff'svoltagelawyields

VoltageRisesVoltageDrops

LeftLoop(clockwise)102020

MiddleLoop(clockwise)20=20

RightLoop(clockwise)201020

(Equationscorrespondingtotheotherloopsarecombinationsofthesethreeequations.)

Thelinearsystemcanberewrittenas

Itsaugmentedmatrixhasthereducedrowechelonform

8. FromKirchhoff'scurrentlawateachnode,wehave  1230. III Kirchhoff'svoltagelawyields

Thecorrespondinglinearsystemcanberewrittenas

Itsaugmentedmatrixhasthereducedrowechelonform

9. Wearelookingforpositiveintegers123 ,, xxx ,and4 x suchthat

Thelinearsystem

Thegeneralsolutionis

10. Wearelookingforpositiveintegers12,,xx and3 x suchthat

Thenumberofatomsofcarbon,hydrogen,andoxygenonbothsidesmustequal:

11. Wearelookingforpositiveintegers123

Thenumberofatomsofcarbon,hydrogen,ox

Thelinearsystem

theunknownsoccurwhen

16. Wearelookingforapolynomialoftheform

Itsaugmentedmatrixhasthereducedrowechelonform

(a) Wearelookingforapolynomialoftheform

Itsaugmentedmatrixhasthereducedrowechelonform

Thegeneralsolutionofthelinearsystemis

isarbitrary. Consequently,thefamilyofallsecond-degreepolynomialsthatpassthrough 0,1and 1,2canbe representedby

112 pxtxtx where t isanarbitraryrealnumber.

(b)

True-False Exercises

(a) False.Ingeneral,networksmayormaynotsatisfythepropertyofflowconservationateachnode(althoughtheones discussedinthissectiondo).

(b) False.Whenacurrentpassesthrougharesistor,thereisadropintheelectricalpotentialinacircuit.

(d) False.Achemicalequationissaidtobebalancedif foreachtypeofatominthereaction,thesamenumberofatoms appearsoneachsideoftheequation.

(e) False.ByTheorem1.10.1,thisistrueifthepointshavedistinct x -coordinates.

1.11 Leontief Input-Output Models

(b) TheLeontiefmatrixis

TheLeontiefequation

.Itsreducedrowechelonformis

Tomeettheconsumerdemand,theeconomymustproduce$300,000worthoffoodand$400,000worthof housing.

(b) TheLeontiefmatrixis

theoutsidedemandvectoris

TheLeontiefequation

0.900.600.401930

0.300.800.303860

0.400.100.805790

Itsreducedrowechelonformis

10031,500 01026,500 00126,300

Theproductionvectorthatwillmeetthegivendemandis

TheLeontiefmatrixis

TheLeontiefequation

Itsreducedrowechelonformis

Theproductionvectorthatwillmeetthegivendemandis

reducedrowechelonformis

arbitrarynonnegative t )tomeetthedemand.

vectorcannotbefoundtomeetthedemand.

(b) Mathematically,thelinearsystemrepresentedby

Clearly,if  20d thesystemhasinfinitelymanysolutions:  11 2 xd ;  2 xt where t isanarbitrary nonnegativenumber.

If  20d thesystemisinconsistent.(NotethattheLeontiefmatrixisnotinvertible.)

Aneconomicexplanationoftheresultinpart(a)isthat

2 0 1 c thereforethesecondsectorconsumesallof itsownoutput,makingitimpossibletomeetanyoutsidedemandforitsproducts.

Iftheopensectordemands k dollarsworthfromeachproduct-producingsector,i.e.theoutsidedemandvectoris

d k k k .TheLeontiefequation

xd IC leadstothelinearsystemwiththeaugmentedmatrix

.Itsreducedrowechelonformis

Weconcludethatthefirstsectormustproducethegreatestdollarvaluetomeetthespecifiedopensectordemand.

9. Fromtheassumption  211211 1 ccc ,itfollowsthatthedeterminantof

1 detdet1

inverseis

isnonzero.Consequently,theLeontiefmatrixisinvertible;its

.Sincetheconsumptionmatrix C hasnonnegativeentriesand

 112112 10 ccc ,weconcludethatallentriesof  1 IC arenonnegativeaswell.Thiseconomyisproductive(see thediscussionaboveTheorem1.10.1)-theequation  C xxd hasauniquesolution

1 IC xd forevery demandvector d

True-False Exercises

(a) False.Sectorsthatdo not produceoutputsarecalledopensectors.

(b) True.

(c) False.The i throwvectorofaconsumptionmatrixcontainsthemonetaryvaluesrequiredofthe i thsectorbythe othersectorsforeachofthemtoproduceonemonetaryunitofoutput.

(d) True.ThisfollowsfromTheorem1.11.1.

(e) True. Chapter 1 Supplementary

   31041 20331Theoriginalaugmentedmatrix.



  11312 20331 1timesthesecondrowwasaddedtothefirstrow.

   11312 029152timesthefirstrowwasaddedtothesecondrow.

   95 1 222 11312 01Thesecondrowwasmultipliedby1 2

Thismatrixisinrowechelonform.Itcorrespondstothesystemofequations

Solvetheequationsfortheleadingvariables

thensubstitutethesecondequationintothefirst

Ifweassign3 x and4 x thearbitraryvalues s and t ,respectively,thegeneralsolutionisgivenbytheformulas

Theoriginalaugmentedmatrix.

2timesthefirstrowwasaddedtothesecondrowand3 timesthefirstrowwasaddedtothethirdrow.

Thismatrixisbothinrowechelonformandinreducedrowechelonform.Itcorrespondstothesystemofequations

Ifweassign2 x anarbitraryvalue t ,thegeneralsolutionisgivenbytheformulas

Theoriginalaugmentedmatrix.

Thefirstrowwasmultipliedby1 2

2 123 08511 0113

2 123 0113 08511

4timesthefirstrowwasaddedtothesecondrow.

Thesecondandthirdrowswereinterchanged.

sincethethirdequationiscontradictory.(Wecouldhaveperformedadditionalelementaryrowoperationstoobtaina

Theaugmentedmatrixcorrespondingtothesystem.

6. Webreakupthesolutionintothreecases: CaseI:   cos0and   sin0

Thesecondrowwasmultipliedby3 5

sin coscos 1 sincos x y

sin coscos 1sin coscos 1 0 x yx

sin coscos 1 01cossin x yx

10cossin 01cossin xy yx

Theaugmentedmatrixcorrespondingtothesystem.

Thefirstrowwasmultipliedby  1 cos

sintimesthefirstrowwasaddedtothesecond (

22 sincos1 coscoscos).

Thesecondrowwasmultipliedby  cos

 sin cos timesthesecondrowwasaddedtothefirstrow (

22 sincos coscoscos cos) xxxx

Thesystemhasexactlyonesolution: 

cossinxxy and      sincosyxy .

CaseII:   cos0whichimplies   2 sin1.Theoriginalsystembecomes

sin xy ,

sin yx .Multiplying bothsidesoftheeachequationby  sinyields   



sin,sinxyyx

CaseIII:   sin0,whichimplies   2 cos1.Theoriginalsystembecomes 

cos xx ,

cos yy .Multiplying bothsidesofeachequationby  cosyields    cos xx ,

cos. yy

NoticethatthesolutionfoundincaseI

cossinxxy and

actuallyappliestoallthreecases.

 1119 151044Theoriginalaugmentedmatrix.

1119 04935 1timesthefirstrowwasaddedtothesecondrow.

935 44 1119 01Thesecondrowwasmultipliedby1 4

51 44 935 44 10 011timesthesecondrowwasaddedtothefirstrow.

Ifweassign z anarbitraryvalue t ,thegeneralsolutionisgivenbytheformulas

15359 ,,4444 xtytzt

Thepositivityofthethreevariablesrequiresthat   15 440 t ,  359 440 t ,and  0 t .Thefirstinequalitycanbe rewrittenas  1 4 t ,whilethesecondinequalityisequivalentto  35 9 t .Allthreeunknownsarepositivewhenever

 3509 t .Therearethreeintegervaluesof  tz inthisinterval:1,2,and3.Ofthose,only  3 zt yieldsinteger valuesfortheremainingvariables:  4 x ,  2 y .

8. Let,, xy and z denotethenumberofpennies,nickels,anddimes,respectively.Sincethereare13coins,wemust have   13. xyz

Ontheotherhand,thetotalvalueofthecoinsis83centssothat   51083.xyz

Theresultingsystemofequationshastheaugmentedmatrix

 59 42 935 42 10 01

11113 151083 whosereducedrowechelonformis

Ifweassign z anarbitraryvalue t ,thegeneralsolutionisgivenbytheformulas

95359 ,,2424 xtytzt

However,allthreeunknownsmustbenonnegativeintegers. Thenonnegativityof x requirestheinequality 95 240 t ,i.e.,  18 5 t

Likewisefor y , 359 240 t yields  70 9 t

When  1870 59 t ,allthreevariablesarenonnegative.Ofthefourinteger  tz valuesinsidethisinterval(4,5,6, and7),only  6 tz yieldsintegervaluesfor x and y .

Weconcludethattheboxhastocontain3pennies,4nickels,and6dimes.

02 44 02 ab aa ab

02 042 02 ab ab ab

02 042 0022 ab ab bb

Theaugmentedmatrixforthesystem.

1timesthefirstrowwasaddedtothesecondrow.

1timesthesecondrowwasaddedtothethirdrow.

(a) thesystemhasauniquesolutionif  0 a and  2 b (multiplyingtherowsby1 a , 1 a ,and12 b ,respectively, yieldsarowechelonformoftheaugmentedmatrix

(b) thesystemhasaone-parametersolutionif  0 a and  2 b (multiplyingthefirsttworowsby 1 a yieldsa reducedrowechelonformoftheaugmentedmatrix

(thereducedrowechelonformoftheaugmentedmatrixis

(d) thesystemhasnosolutionif  0 a and  2 b

(thereducedrowechelonformoftheaugmentedmatrixis

0011 0000 0000 ).

0010 0001 0000 ).

Fromquadraticformulawehave

Thesystemhasnosolutionswhen

(sincethethirdrowofourlastmatrixwouldthencorrespondto acontradictoryequation).

Thesystemhasinfinitelymanysolutionswhen

Novaluesof a resultinasystemwithexactlyonesolution. 11. Fortheproduct AKB tobedefined, K mustbea

Thematrixequation  AKBC canberewrittenasasystemofninelinearequations

whichhasauniquesolution  0 a ,

2 b ,

1 c ,

1 d .(Aneasywaytosolvethissystemistofirstsplititintotwo smallersystems.Thesystem  288 ac ,  466 ac ,  244 ac involves a and c only,whereasthe remainingsixequationsinvolvejust b and d .)Weconcludethat

12. Substitutingthevalues  1, x 1 y ,and  2 z intotheoriginalsystemyieldsasystemofthreeequationsinthe unknowns,, ab and c :

323 2121 3123 ab bc ac thatcanberewrittenas

3 21 20 ab bc ac Theaugmentedmatrixofthissystemhasthereducedrowechelonform

1002 0101 0011 .Weconcludethatforthe originalsystemtohave  1 x , 1 y ,and  2 z asitssolution,wemustlet  2, a 1 b ,and  1 c . (Notethatitcanalsobeshownthatthesystemwith  2,

,and  2 z asits only solution.Onewaytodothatwouldbetoverifythatthereducedrowechelonformofthecoefficientmatrixofthe originalsystemwiththesespecificvaluesof,ab and c istheidentitymatrix.) 13. (a) X mustbea  23matrix.Letting

thereforethegivenmatrixequationcanberewrittenasasystemoflinearequations:

Theaugmentedmatrixofthissystemhasthereducedrowechelonform

sothesystemhasauniquesolution

(Analternativetodealingwiththislargesystemistosplititintotwosmallersystemsinstead:thefirstthree equationsinvolve a , b ,and c only,whereastheremainingthreeequationsinvolvejust d , e ,and f . Sincethecoefficientmatrixforbothsystemsisthesame,wecanfollowtheprocedureofExample2in Section1.6;the

reducedrowechelonformofthematrix

Yetanotherwayofsolvingthisproblemwouldbetodeterminetheinverse

usingthemethodintroducedinSection1.5,thenmultiplybothsidesofthe givenmatrixequationontherightbythisinversetodetermine X :

Theaugmentedmatrixofthissy

(Analternativetodealingwiththislargesystemistosplititintotwosmallersystemsinstead:thefirstthree equationsinvolve a and b only,whereastheremainingthreeequationsinvolvejust c and d .Sincethe coefficientmatrixforbothsystemsisthesame,wecanfollowtheprocedureofExample2inSection1.6;

thereducedrowechelonformofthematrix

Theaugmentedmatrixofthissystemhasthereducedrowechelonform

15. Wearelookingforapolynomialoftheform

Itsaugmentedmatrixhasthereducedrowechelonform

0 429 40 abc abc ab

Thereducedrowechelonformoftheaugmentedmatrixofthissystemis

1001 0104 0015 .Therefore,thevalues

 1 a , 4 b ,and 5 c resultinapolynomialthatsatisfiestheconditionsspecified.

17. Whenmultiplyingthematrix n J byitself,eachentryintheproductequals n .Therefore,  nnn JJnJ

   1 1 nn n IJIJ

211 11nnnnnn IIJJIJJ

 I

 11 11nnnnnn JJJJ

 I  11 11nnnn JJJJ nn

 I

Theorem1.4.1(f)and(g)

Property   AIIAA inSection1.4

Theorem1.4.1(m)

 1 11 n nnnnn JJJ  nnn JJnJ

 1 11 1 n nnn IJ

 11 111 nn nnnn IJ

 I

Theorem1.4.1(j)and(k)