Linear algebra and its applications 4th edition lay solutions manual by phillip.dejulio853

Linear Algebra And Its Applications 4th Edition Lay Solutions Manual Visit to Download in Full: https://testbankdeal.com/download/linear-algebra-and-its-ap plications-4th-edition-lay-solutions-manual/

3.1 SOLUTIONS

Notes:Someexercisesinthissectionprovidepracticeincomputingdeterminants,whileothersallowthe studenttodiscoverthepropertiesofdeterminantswhichwillbestudiedinthenextsection.Determinants aredevelopedthroughthecofactorexpansion,whichisgiveninTheorem1.Exercises33–36inthis sectionprovidethefirststepintheinductiveproofofTheorem3inthenextsection.

1.Expandingalongthefirstrow: 304 322223 2323043(13)4(10)1 510105 051 =−+=−+= Expandingalongthesecondcolumn: 122232 304 223434 232(1)0(1)3(1)53(3)5(2)1 010122 051 +++ =−⋅+−⋅+−⋅=−−−= 2.Expandingalongthefirstrow: 051 304043 4300515(4)1(22)2 412124 241 −=−+=−+= Expandingalongthesecondcolumn: 122232 051 400101 430(1)5(1)(3)(1)45(4)3(2)4(4)2 212140 241 +++ −=−⋅+−⋅−+−⋅=−−−−−= 3.Expandingalongthefirstrow: 243 123231 3122(4)32(9)4(5)(3)(11)5 411114 141 =−−+=−+−+=− Linear Algebra And Its Applications 4th Edition Lay Solutions Manual Visit TestBankDeal.com to get complete for all chapters

168CHAPTER3•Determinants

Expandingalongthesecondcolumn: 122232 243 322323 312(1)(4)(1)1(1)44(5)1(5)4(5)5 111132 141 +++ =−⋅−+−⋅+−⋅=−+−−−=− 4.Expandingalongthefirstrow: 135 112121 2111351(2)3(1)5(5)20 423234 342 =−+=−−+= Expandingalongthesecondcolumn: 122232 135 211515 211(1)3(1)1(1)43(1)1(13)4(9)20 323221 342 +++ =−⋅+−⋅+−⋅=−+−−−= 5.Expandingalongthefirstrow: 234 054540 40523(4)2(5)3(1)4(4)23 165651 516 =−+−=−−−−=− 6.Expandingalongthefirstrow: 524 350503 0355(2)45(1)2(10)4(6)1 472724 247 −=−−+=++−= 7.Expandingalongthefirstrow: 430 526265 6524304(1)3(0)4 739397 973 =−+=−= 8.Expandingalongthefirstrow: 816 034340 4038168(6)1(11)6(8)11 253532 325 =−+=−+−=− 9.Firstexpandalongthethirdrow,thenexpandalongthefirstrowoftheremainingmatrix:

3.1•Solutions169 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 3113 6005 005 172572 (1)27252(1)510(1)10200031 318 8318 ++ =−⋅−=⋅−⋅== 10.Firstexpandalongthesecondrow,thenexpandalongeitherthethirdroworthesecondcolumnof theremainingmatrix. 23 1252 122 0030 2675(1)3265 504 5044 + =−⋅− 31332212 (3)(1)5(1)4(3)(5(2)4(2))6 6526 ++ ⎛⎞ =−−⋅+−⋅=−+−=− ⎜⎟ ⎝⎠ or 23 1252 122 0030 2675(1)3265 504 5044 + =−⋅− 12222512 (3)(1)(2)(1)(6)5454 ++ ⎛⎞ =−−⋅−+−⋅−⎜⎟ ⎝⎠ () (3)2(17)6(6)6 =−−−−=− 11.Therearemanywaystodothisdeterminantefficiently.Onestrategyistoalwaysexpandalongthe firstcolumnofeachmatrix: 1111 3584 237 023715 (1)30153(1)(2) 001502 002 0002 ++ =−⋅ =⋅−⋅− =3(–2)(2)=–12 12.Therearemanywaystodothisdeterminantefficiently.Onestrategyistoalwaysexpandalongthe firstrowofeachmatrix: 1111 4000 100 710030 2630(1)46304(1)(1)43 843 5843 ++ =−⋅=⋅−⋅− =4(–1)(–9)=36

170CHAPTER3•Determinants Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 13.Firstexpandalongeitherthesecondroworthesecondcolumn.Usingthesecondrow, 23 40735 4035 00200 7348 73648(1)25023 50523 0012 00912 + =−⋅ Nowexpandalongthesecondcolumntofind: 2322 4035 435 7348 (1)22(1)3523 5023 012 0012 ++ ⎛⎞ ⎜⎟ −⋅=−−⋅−⎜⎟ ⎜⎟ ⎝⎠ Nowexpandalongeitherthefirstcolumnorthirdrow.Thefirstcolumnisusedbelow. 22 435 2(1)3523 012 + ⎛⎞ ⎜⎟ −−⋅−⎜⎟ ⎜⎟ ⎝⎠ 11212335 6(1)4(1)5(6)(4(1)5(1))6 1212 ++ ⎛⎞ =−−⋅+−⋅=−−= ⎜⎟ ⎝⎠ 14.Firstexpandalongeitherthefourthroworthefifthcolumn.Usingthefifthcolumn, 35 63240 6324 90410 9041 85671(1)13000 30000 4232 42320 + =−⋅ Nowexpandalongthethirdrowtofind: 3531 6324 324 9041 (1)11(1)3041 3000 232 4232 ++ ⎛⎞ ⎜⎟ −⋅=−⋅−⎜⎟ ⎜⎟ ⎝⎠ Nowexpandalongeitherthefirstcolumnorsecondrow.Thefirstcolumnisusedbelow. 31 324 1(1)3041 232 + ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ 11314124 3(1)3(1)2(3)(3(11)2(18))9 3241 ++ ⎛⎞ =−⋅+−⋅=−+= ⎜⎟ ⎝⎠ 15. 304 232 051 = (3)(3)(–1)+(0)(2)(0)+(4)(2)(5)–(0)(3)(4)–(5)(2)(3)–(–1)(2)(0) =–9+0+40–0–30–0=1

k,andthedeterminantismultipliedby k.

21 34 18202, 56 =−=− 34 53643(64)(53)42 kk kk =+−+=− ++

Therowoperationreplacesrow2withktimesrow1plusrow2,andthedeterminantisunchanged.

22., ab adbc cd =− ()() akcbkd akcdcbkdadkcdbckcdadbc cd ++ =+−+=+−−=−

Therowoperationreplacesrow1withktimesrow2plusrow1,andthedeterminantisunchanged.

23.

111 3841(4)1(2)1(7)5, 232 −−=−+−=− 384(4)(2)(7)5 232

kkk kkkk −−=−+−=−

Therowoperationscalesrow1by k,andthedeterminantismultipliedby k.

abc abcabc =−+=−+

24.322(2)(6)(3)263, 656

3.1•Solutions171 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 16. 051 430 241 −= (0)(–3)(1)+(5)(0)(2)+(1)(4)(4)–(2)(–3)(1)–(4)(0)(0)–(1)(4)(5) =0+0+16–(–6)–0–20=2 17. 243 312 141 = (2)(1)(–1)+(–4)(2)(1)+(3)(3)(4)–(1)(1)(3)–(4)(2)(2)–(–1)(3)(–4) =–2+(–8)+36–3–16–12=–5 18. 135 211 342 = (1)(1)(2)+(3)(1)(3)+(5)(2)(4)–(3)(1)(5)–(4)(1)(1)–(2)(2)(3) =2+9+40–15–4–12=20 19., ab adbc cd =− () cd cbdaadbc ab =−=−− Therowoperationswapsrows1and2ofthematrix,andthesignofthedeterminantisreversed. 20., ab adbc cd =− ()()() ab akdkcbkadkbckadbc kckd =−=−=− Therowoperationscalesrow2by

3(65)2(66)2(56)263

Therowoperationswapsrows1and2ofthematrix,andthesignofthedeterminantisreversed.

25.Sincethematrixistriangular,byTheorem2thedeterminantistheproductofthediagonalentries: 100 010(1)(1)(1)1 01 k ==

26.Sincethematrixistriangular,byTheorem2thedeterminantistheproductofthediagonalentries: 100 010(1)(1)(1)1 01 k ==

27.Sincethematrixistriangular,byTheorem2thedeterminantistheproductofthediagonalentries:

28.Sincethematrixistriangular,byTheorem2thedeterminantistheproductofthediagonalentries:

Ineachofthesecases,thematrixistriangularanditsdeterminantistheproductofitsdiagonal entries,whichis1.Thusthedeterminantofa3 × 3elementaryrowreplacementmatrixis1.

172CHAPTER3•Determinants

322

656

=−−−+−=−+−

abcbcacababc

00 010()(1)(1) 001 k kk==

100 00(1)()(1) 001 kkk == 29.Acofactorexpansionalongrow1gives 010 10 10011 01 001 =−=− 30.Acofactorexpansionalongrow1gives 001 01 01011 10 100 ==− 31.A3 × 3elementaryrowreplacementmatrixlookslikeoneofthesixmatrices 1001001001001010 10,010,010,01k,010,010 0010101001001001 kk k kk ⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤ ⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥

⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦

32.A3 × 3elementaryscalingmatrixwith k onthediagonallookslikeoneofthethreematrices

00100100

⎡⎤⎡⎤⎡⎤

010,00,010

00100100

k k k

Ineachofthesecases,thematrixistriangularanditsdeterminantistheproductofitsdiagonal entries,whichisk.Thusthedeterminantofa3 × 3elementaryscalingmatrixwith k onthediagonal is k.

33. 01 10, E ⎡⎤ = ⎢⎥ ⎣⎦ , ab A cd ⎡⎤ = ⎢⎥ ⎣⎦ cd EA ab ⎡⎤ = ⎢⎥ ⎣⎦

det E =–1,det A = ad – bc, det EA = cb – da =–1(ad – bc)=(det E)(det A)

34. 10 0, E k ⎡⎤ = ⎢⎥ ⎣⎦ , ab A cd ⎡⎤ = ⎢⎥ ⎣⎦ ab EA kckd ⎡⎤ = ⎢⎥ ⎣⎦

det E = k,det A = ad – bc, det EA = a(kd)–(kc)b = k(ad – bc)=(det E)(det A)

35. 1 01, k E ⎡⎤ = ⎢⎥ ⎣⎦ , ab A cd ⎡⎤ = ⎢⎥ ⎣⎦ akcbkd EA cd ++ ⎡⎤ = ⎢⎥ ⎣⎦

det E =1,det A = ad – bc, det EA =(a + kc)d – c(b + kd)= ad + kcd – bc – kcd =1(ad – bc)=(det E)(det A)

36 10 1, E k ⎡⎤ = ⎢⎥ ⎣⎦ , ab A cd ⎡⎤ = ⎢⎥ ⎣⎦ ab EA kackbd ⎡⎤ = ⎢⎥ ++ ⎣⎦

det E =1,det A = ad – bc, det EA = a(kb + d)–(ka + c)b = kab + ad – kab – bc =1(ad – bc)=(det E)(det A)

37. 31 42, A ⎡⎤

155 5, 2010 A

det A =2,det5A =50 ≠ 5det A

, kakb kA kckd ⎡⎤ = ⎢⎥ ⎣⎦ det A = ad – bc, 22 det()()()()()det kAkakdkbkckadbckA =−=−=

39a

.True.Seetheparagraphprecedingthedefinitionofthedeterminant.

b.False.Seethedefinitionofcofactor,whichprecedesTheorem1.

40. a.False.SeeTheorem1.

b.False.SeeTheorem2.

3.1•Solutions173

⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦

= ⎢⎥ ⎣⎦

⎡⎤ = ⎢⎥ ⎣⎦

., ab A cd ⎡⎤

⎢⎥ ⎣⎦

41.Theareaoftheparallelogramdeterminedby

0 is6,sincethebaseof theparallelogramhaslength3andtheheightoftheparallelogramis2.Bythesamereasoning,the areaoftheparallelogramdeterminedby

Alsonotethat [] 31 detdet6, 02

matrixwhosecolumnsarethosevectorswhichdefinethesidesoftheparallelogramadjacentto 0 is equaltotheareaoftheparallelogram

42.Theareaoftheparallelogramdeterminedby

theparallelogramhaslengthcandtheheightoftheparallelogramisb.

thematrixwhosecolumnsarethosevectorswhichdefinethesidesoftheparallelogramadjacentto 0 eitherisequaltotheareaoftheparallelogramorisequaltothenegativeoftheareaofthe parallelogram. 43[M] Answerswillvary.Theconclusionshouldbethatdet(A + B) ≠ det A +det B 44[M] Answerswillvary.Theconclusionshouldbethatdet(AB)=(det A)(det B).

174CHAPTER3•Determinants

3 0, ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ u 1 2, ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ vu +

,and

3 0, ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ u 2, x ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ xu + x,and 0 isalso6.

⎡⎤

⎢⎥ ⎣⎦

and [] 3 detdet6. 02 x ⎡⎤ == ⎢⎥ ⎣⎦ ux Thedeterminantofthe

a b ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ u ,0 c ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ v , u + v,and 0

cb,sincethebaseof

Alsonotethat

detdet 0 ac cb b ⎡⎤ ==− ⎢⎥ ⎣⎦ uv ,and [] detdet 0 ca cb b ⎡⎤ == ⎢⎥ ⎣⎦ vu Thedeterminantof

[]

X V UU X2 X2 X1 X1 2 1 2 1 1122 4 X2 a b c X1 U V

45. [M] Answerswillvary.For4 × 4matrices,theconclusionsshouldbethatdetdet, T AA = det(–A)=det A,det(2A)=16det A,and4det(10)10detAA = .For5 × 5matrices,theconclusions shouldbethatdetdet, T AA = det(–A)=–det A,det(2A)=32det A,and5det(10)10det. AA = For6 × 6matrices,theconclusionsshouldbethatdetdet T AA = ,det(–A)=det A,det(2A)=64det A,and 6 det(10)10det. AA =

46[M] Answerswillvary.Theconclusionshouldbethat1det1/det. AA =

3.2 SOLUTIONS

Notes:Thissectionpresentsthemainpropertiesofthedeterminant,includingtheeffectsofrow operationsonthedeterminantofamatrix.ThesepropertiesarefirststudiedbyexamplesinExercises1–20.Thepropertiesaretreatedinamoretheoreticalmannerinlaterexercises.Anefficientmethodfor computingthedeterminantusingrowreductionandselectivecofactorexpansionispresentedinthis sectionandusedinExercises11–14.Theorems4and6areusedextensivelyinChapter5.Thelinearity propertyofthedeterminantstudiedinthetextisoptional,butisusedinmoreadvancedcourses.

1.Rows1and2areinterchanged,sothedeterminantchangessign(Theorem3b.).

2.Theconstant2maybefactoredoutoftheRow1(Theorem3c.).

3.Therowreplacementoperationdoesnotchangethedeterminant(Theorem3a.).

4.Therowreplacementoperationdoesnotchangethedeterminant(Theorem3a.).

3.2•Solutions175

5 156156156 1440120123 279033003 −−=−=−= 6 153153153153 3330181260326032(6)(3)18 2137031031001 −=−=−=−=−=− 7 1302130213021302 2574017801780178 0 35210425003027003027 112304250030270000 ==== 8 133413341334 012501250125 0 254301250000 3752024100000 ===

11.Firstusearowreplacementtocreatezerosinthesecondcolumn,andthenexpanddownthesecond column:

Nowusearowreplacementtocreatezerosinthefirstcolumn,andthenexpanddownthefirst

12.Firstusearowreplacementtocreatezerosinthefourthcolumn,andthenexpanddownthefourth column:

Nowusearowreplacementtocreatezerosinthefirstcolumn,andthenexpanddownthefirst

13.Firstusearowreplacementtocreatezerosinthefourthcolumn,andthenexpanddownthefourth column:

176CHAPTER3•Determinants

9. 1130113011301130 0154015401540154 1285015500010035(3)3 3123027300350001 === −= = 10 131021310213102 024160241602416 262390003500035 373870208100477 3552704821300001 === 13102 02416 00477(24)24 00035 00001 −=−−=

25312531 313 30133013 5649 60496049 021 410410021 ==−−−

313313 23 56495023(5)(3)(5)(3)(8)120 21 021021 −−−=−−=−=−−=

column:

12301230 123 34303430 3343 54663020 302 42434243 ==

column: 123123 1012 33433010123(1)3(1)(38)114 611 3020611 == −= =

3.2•Solutions177 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 25412541 032 47620320 1624 62406240 677 67706770 ==−−− Nowusearowreplacementtocreatezerosinthefirstcolumn,andthenexpanddownthefirst column: 032032 32 16241624(1)(6)(1)(6)(1)6 53 677053 −−−=−−−=−−=−−= 14.Firstusearowreplacementtocreatezerosinthethirdcolumn,andthenexpanddownthethird column: 32143214 133 13031303 1900 34289000 344 34043404 ==− Nowexpandalongthesecondrow: 133 33 19001((9))(1)(9)(0)0 44 344 −=− −== 15 55(7 )35 555 abcabc defdef ghighi === 16 33333(7)21 abcabc defdef ghighi === 17 7 abcabc ghidef defghi =−=− 18.(7)7 ghiabcabc abcghidef defdefghi ⎛⎞ ⎜⎟ =−=−−=−−= ⎜⎟ ⎜⎟ ⎝⎠ 19 22222222(7)14 abcabcabc daebfcdefdef ghighighi +++====

27a.True.SeeTheorem3.

b.True.SeetheparagraphfollowingExample2.

c.True.SeetheparagraphfollowingTheorem4.

d.False.SeethewarningfollowingExample5.

28a.True.SeeTheorem3.

b.False.SeetheparagraphsfollowingExample2.

c.False.SeeExample3.

d.False.SeeTheorem5.

29.ByTheorem6,555det(det)(2)32 BB==−=−

178CHAPTER3•Determinants

20.7 adbecfabc defdef ghighi +++ == 21.Since 230 13410 121 =−≠ ,thematrixisinvertible. 22.Since 501 1320 053 −−= ,thematrixisnotinvertible. 23.Since 2008 1750 0 3860 0754 = ,thematrixisnotinvertible. 24.Since 473 605110 726 −=≠ ,thecolumnsofthematrixformalinearlyindependentset. 25.Since 787 45010 675 −= −≠ ,thecolumnsofthematrixformalinearlyindependentset. 26.Since 3220 5610 0 6030 4703 = ,thecolumnsofthematrixformalinearlydependentset.

30.Supposethetworowsofasquarematrix A areequal.Byswappingthesetworows,thematrix A is notchangedsoitsdeterminantshouldnotchange.Butsinceswappingrowschangesthesignofthe determinant,det A =–det A.Thisisonlypossibleifdet A =0.Thesamemaybeproventruefor columnsbyapplyingtheaboveresultto TA andusingTheorem5.

31.ByTheorem6,1(det)(det)det1 AAI== ,so 1 det1/det. AA =

32.Byfactoringan r outofeachofthe n rows,det()det. n rArA =

33.ByTheorem6,det AB =(det A)(det B)=(det B)(det A)=det BA.

34.ByTheorem6andExercise31, 111 det()(det)(det)(det)(det)(det)(det) PAPPAPPPA == 1 (det)(det)1det det PAA P

det A =

35.ByTheorem6andTheorem5,2 det(det)(det)(det). TT UUUUU == Since, T UUI = detdet1 T UUI== ,so 2 (det)1. U = Thusdet U = ±1.

36.ByTheorem644 det(det) AA = .Since4det0 A = ,then4(det)0 A = .Thusdet A =0,and A isnot invertiblebyTheorem4.

37.OnemaycomputeusingTheorem2thatdet A =3anddet B =8,while 60 174 AB ⎡⎤ = ⎢⎥ ⎣⎦ .Thus det AB =24=3 × 8=(det A)(det B).

38.Onemaycomputethatdet A =0anddet B =–2,while 60 20 AB ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ .Thus det AB =0=0 × –2=(det A)(det B).

39. a.ByTheorem6,det AB =(det A)(det B)=4 × –3=–12.

b.ByExercise32,3det55det1254500 AA==×= .

c.ByTheorem5,detdet3 T BB==− .

d.ByExercise31,1det1/det1/4 AA== .

e.ByTheorem6,333det(det)464 AA===

40. a.ByTheorem6,det AB =(det A)(det B)=–1 × 2=–2.

b.ByTheorem6,555det(det)232 BB===

c.ByExercise32,4det22det16116 AA==×−=− .

d.ByTheorems5and6,det(det)(det)(det)(det)111 TT AAAAAA === × =

3.2•Solutions179

⎛⎞ == ⎜⎟ ⎝⎠

e.ByTheorem6andExercise31, 11

det(det)(det)(det)(1/det)(det)(det)det1 BABBABBABA === =

41.det A =(a + e)d – c(b + f)= ad + ed – bc – cf =(ad – bc)+(ed – cf)=det B +det C

42. 1 det()(1)(1)1detdet 1 ab ABadcbadadcbAadB cd + +==++−=+++−=+++ + ,so det(A + B)=det A +det B ifandonlyif a + d =0.

43.Computedet A byusingacofactorexpansiondownthethirdcolumn:

111322233333 det()det()det()det AuvAuvAuvA =+−+++

113223333113223333 detdetdetdetdetdet uAuAuAvAvAvA =−++−+ detdetBC=+

44.ByTheorem5,detdet(). AEAET = Since()TTT AEEA = ,detdet(). TTAEEA = Now T E isitself anelementarymatrix,sobytheproofofTheorem3,det()(det)(det). TTTT EAEA = Thusitistrue thatdet(det)(det), TT AEEA = andbyapplyingTheorem5,det AE =(det E)(det A).

45. [M] Answerswillvary,butwillshowthatdet T AA alwaysequals0whiledet TAA shouldseldom bezero.Toseewhy T AA -shouldnotbeinvertible(andthusdet0 T AA = ),let A beamatrixwith morecolumnsthanrows.Thenthecolumnsof A mustbelinearlydependent,sotheequation Ax = 0 musthaveanon-trivialsolution x.Thus()(), TTT AAAAA=== xx00 andtheequation() T AA = x0 hasa non-trivialsolution.Since T AA isasquarematrix,theInvertibleMatrixTheoremnowsaysthat T AA isnotinvertible.Noticethatthesameargumentwillnotworkingeneralfor, TAA since TA hasmorerowsthancolumns,soitscolumnsarenotautomaticallylinearlydependent.

46. [M] Onemaycomputeforthismatrixthatdet A =–4008andcond A ≈ 16.3.Notethatthisisthe 2 conditionnumber,whichisusedinSection2.3.Sincedet A ≠ 0,itisinvertibleand

837181207297

175057430654

4008171195871095

2118781639

Thedeterminantisverysensitivetoscaling,as4det1010det40,080,000 AA==− and det0.1 A = 4 (0.1)det0.4008. A =−

Theconditionnumberisnotchangedatallbyscaling: cond(10A)=cond(0.1A)=cond A ≈ 16.3.When4 AI = ,det A=1andcond A =1.Asbeforethe determinantissensitivetoscaling:4det1010det10,000 AA== and4det0.1(0.1)det0.0001. AA==

Yettheconditionnumberisnotchangedbyscaling:cond(10A)=cond(0.1A)=cond A =1.

180CHAPTER3•Determinants

A ⎡⎤ ⎢⎥ ⎢⎥ =− ⎢⎥ ⎢⎥ ⎣⎦

3.3 SOLUTIONS

Notes:Thissectionfeaturesseveralindependenttopicsfromwhichtochoose.Thegeometric interpretationofthedeterminant(Theorem10)providesthekeytochangesofvariablesinmultiple integrals.StudentsofeconomicsandengineeringarelikelytoneedCramer’sRuleinlatercourses. Exercises1–10concernCramer’sRule,exercises11–18dealwiththeadjugate,andexercises19–32 coverthegeometricinterpretationofthedeterminant.Inparticular,Exercise25examinesstudents’ understandingoflinearindependenceandrequiresacarefulexplanation,whichisdiscussedinthe Study Guide.The StudyGuide alsocontainsaheuristicproofofTheorem9for2 × 2matrices.

1212 3753 (),(),det6,det()5,det()1, 1421 AAAAA

6146 (),(),det3,det()5,det()2, 7257 AAAAA

12 det()det()52 ,. det3det3 AA xx AA ====−

3.Thesystemisequivalentto Ax = b,where 32

1212 7237 (),(),det8,det()32,det()20, 5655 AAAAA

=====

12 12 det()det()32205 4, det8det82 AA xx AA ====== bb

4.Thesystemisequivalentto Ax = b,where 53 31 A ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ and 9 5 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ b .Wecompute

=== −==

1212 9359 (),(),det4,det()6,det()2, 5135 AAAAA

12 12 det()det()6321 ,. det42det42 AA xx AA ===−=== bb

3.3•Solutions181

1.Thesystemisequivalentto Ax = b,where 57 24 A ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ and 3 1 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ b .Wecompute

⎡⎤⎡⎤ ===== ⎢⎥⎢⎥ ⎣⎦⎣⎦ bbbb 12 12 det()det()51 ,. det6det6 AA xx AA ====− bb

41 52 A ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ and 6 7 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ b .Wecompute

⎡⎤⎡⎤ ===== ⎢⎥⎢⎥ ⎣⎦⎣⎦ bbbb 12

2.Thesystemisequivalentto Ax = b,where

1212

⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ and 7 5 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ b .Wecompute

⎡⎤⎡⎤

⎢⎥⎢⎥ ⎣⎦⎣⎦ bbbb

⎡⎤⎡⎤

⎢⎥⎢⎥ ⎣⎦⎣⎦

bbbb

123

710270217 ()801,()381,()308, 312032013 AAA

123 det4,det()6,det()16,det()14, AAAA ====− bbb

123 123 det()det()det()6316147 ,4,. det42det4det42 AAA xxx

AAA =========− bbb

411241214 ()202,()122,()102, 213323312 AAA

det4,det()16,det()52,det()4, AAAA ==−==−

123 det()det()det()16524 4,13,1.

bbbb

182CHAPTER3•Determinants

210 301 012 A ⎡ ⎤ ⎢ ⎥ =− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ and 7 8 3 ⎡ ⎤ ⎢ ⎥ =− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ b .Wecompute

5.Thesystemisequivalentto Ax = b,where ⎡⎤⎡⎤⎡⎤ ⎢⎥⎢⎥⎢⎥ =− =−−=−− ⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦ bbb

211 102 313 A ⎡ ⎤ ⎢ ⎥ =− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ and 4 2 2 ⎡ ⎤ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ b .Wecompute

⎡⎤⎡⎤⎡⎤ ⎢⎥⎢⎥⎢⎥ == −= ⎢⎥⎢⎥⎢⎥ ⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦ bbb 123

bbb 123

det4det4det4 AAA xxx AAA ===−======− bbb 7

64 92 s A s ⎡ ⎤

⎢ ⎥ ⎣ ⎦

6.Thesystemisequivalentto Ax = b,where ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ b .Wecompute

123

.Thesystemisequivalentto Ax = b,where

and

⎡⎤⎡⎤

⎣⎦⎣⎦

1212 5465 (),(),det()108,det()1245. 2292 s AAAsAs s

=== += ⎢⎥⎢⎥

++−−−− ====== bb

Since22det123612(3)0 Ass=−=−≠ for3 s ≠± ,thesystemwillhaveauniquesolutionwhen 3 s ≠± .Forsuchasystem,thesolutionwillbe 12 122222 det()det()108541245415 det,. 12(3)6(3)det12(3)4(3)

ssss xx AAssss

⎡ ⎤

⎢ ⎥ ⎣ ⎦

⎡ ⎤ = ⎢ ⎥ ⎣ ⎦

.Wecompute

8.Thesystemisequivalentto Ax = b,where

95 s A s

and

⎡⎤⎡⎤

⎢⎥⎢⎥ ⎣⎦⎣⎦

1212 3533 (),(),det()1510,det()627. 2592 s AAAsAs s

=== +=

Since22det154515(3)0 Ass=+=+≠ forallvaluesof s,thesystemwillhaveauniquesolutionfor allvaluesof s.Forsuchasystem,thesolutionwillbe 12 122222 det()det()15103262729 det,. 15(3)3(3)det15(3)5(3) AAssss xx AAssss ++ ====== ++ ++ bb

9.Thesystemisequivalentto Ax = b,where

1212 121 (),(),det()2,det()43. 4634 ss AAAsAs s ⎡⎤⎡⎤ ==== + ⎢⎥⎢⎥

Since2det666(1)0 Assss=+=+= for s =0,–1,thesystemwillhaveauniquesolutionwhen s ≠ 0, –1.Forsuchasystem,thesolutionwillbe

12 12 det()det()2143

,. det6(1)3(1)det6(1) AAss xx AsssAss + ===== ++ + bb

10.Thesystemisequivalentto Ax = b,where 21 36

1212 1121 (),(),det()62,det(). 2632 s

Since2det1233(41)0 Assss =−=−= for s =0,1/4,thesystemwillhaveauniquesolutionwhen s ≠ 0,1/4.Forsuchasystem,thesolutionwillbe

3.3•Solutions183

2 36 ss

s ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ and 1 4 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ b .Wecompute

⎣⎦⎣⎦

bbbb

ss ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ and 1 2 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ b .Wecompute

s A

⎡⎤⎡⎤ === −= ⎢⎥⎢⎥ ⎣⎦⎣⎦ bbbb

AAAsAs ss

det()det()621 ,. det3(41)det3(41)3(41) AA

xx AssAsss ===== bb 11.Sincedet A =3andthecofactorsofthegivenmatrixare 11 00 0, 11 C == 12 30 3, 11 C =−=− 13 30 3, 11 C == 21 21 1, 11 C =−= 22 01 1, 11 C ==− 23 02 2, 11 C =−= 31 21 0, 00 C == 32 01 3, 30 C =−=− 33 02 6, 30 C == 010 adj313 326 A ⎡⎤ ⎢⎥ =−−−⎢⎥ ⎢⎥ ⎣⎦ and1 01/30 1 detadj11/31. 12/32 AA A ⎡ ⎤ ⎢ ⎥ ==−−− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 12.Sincedet A =5andthecofactorsofthegivenmatrixare 11 21 1, 10 C ==− 12 21 0, 00 C =−= 13 22 2, 01 C == 21 13 3, 10 C =−= 22 13 0, 00 C == 23 11 1, 01 C =−=− 31 13 7, 21 C == 32 13 5, 21 C =−= 33 11 4, 22 C ==− 137 adj005 214 A ⎡⎤ ⎢⎥ = ⎢⎥ ⎢⎥ ⎣⎦ and1 1/53/57/5 1 detadj001. 2/51/54/5 AA A ⎡ ⎤ ⎢ ⎥ == ⎢ ⎥ ⎢ ⎥ ⎣ ⎦

184CHAPTER3•Determinants Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 13.Sincedet A =6andthecofactorsofthegivenmatrixare 11 01 1, 11 C ==− 12 11 1, 21 C =−= 13 10 1, 21 C == 21 54 1, 11 C =−=− 22 34 5, 21 C ==− 23 35 7, 21 C =−= 31 54 5, 01 C == 32 34 1, 11 C =−= 33 35 5, 10 C ==− 115 adj151 175 A ⎡⎤ ⎢⎥ =−⎢⎥ ⎢⎥ ⎣⎦ and1 1/61/65/6 1 detadj1/65/61/6. 1/67/65/6 AA A ⎡ ⎤ ⎢ ⎥ ==− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 14.Sincedet A =–1andthecofactorsofthegivenmatrixare 11 21 5, 34 C == 12 01 2, 24 C =−= 13 02 4, 23 C ==− 21 67 3, 34 C =−=− 22 37 2, 24 C ==− 23 36 3, 23 C =−= 31 67 8, 21 C ==− 32 37 3, 01 C =−=− 33 36 6, 02 C == 538 adj223 436 A ⎡⎤ ⎢⎥ =−−⎢⎥ ⎢⎥ ⎣⎦ and1 538 1 detadj223. 436 AA A ⎡ ⎤ ⎢ ⎥ ==− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 15.Sincedet A =6andthecofactorsofthegivenmatrixare 11 10 2, 32 C == 12 10 2, 22 C =−= 13 11 1, 23 C ==− 21 00 0, 32 C =−= 22 30 6, 22 C == 23 30 9, 23 C =−=− 31 00 0, 10 C == 32 30 0, 10 C =−= 33 30 3, 11 C == 200 adj260 193 A ⎡⎤ ⎢⎥ = ⎢⎥ ⎢⎥ ⎣⎦ and1 1/300 1 detadj1/310. 1/63/21/2 AA A ⎡ ⎤ ⎢ ⎥ == ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 16.Sincedet A =–9andthecofactorsofthegivenmatrixare 11 31 9, 03 C ==− 12 01 0, 03 C =−= 13 03 0, 00 C == 21 24 6, 03 C =−=− 22 14 3, 03 C == 23 12 0, 00 C =−=

18.Eachcofactorof A isanintegersinceitisasumofproductsofentriesin A.Henceallentriesinadj A willbeintegers.Sincedet A =1,theinverseformulainTheorem8showsthatalltheentriesin1 A willbeintegers.

19.Theparallelogramisdeterminedbythecolumnsof 56 24 A

20.Theparallelogramisdeterminedbythecolumnsof 14 35 A

21.Firsttranslateonevertextotheorigin.Forexample,subtract(–1,0)fromeachvertextogetanew parallelogramwithvertices(0,0),(1,5),(2,–4),and(3,1).Thisparallelogramhasthesameareaas theoriginal,andisdeterminedbythecolumnsof 12

22.Firsttranslateonevertextotheorigin.Forexample,subtract(0,–2)fromeachvertextogetanew parallelogramwithvertices(0,0),(6,1),(–3,3),and(3,4).Thisparallelogramhasthesameareaas theoriginal,andisdeterminedbythecolumnsof

23.Theparallelepipedisdeterminedbythecolumnsof

3.3•Solutions185 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 31 24 14, 31 C == 32 14 1, 01 C =−=− 33 12 3, 03 C ==− 9614 adj031 003 A ⎡⎤ ⎢⎥ =−⎢⎥ ⎢⎥ ⎣⎦ and1 12/314/9 1 detadj01/31/9. 001/3 AA A ⎡ ⎤ ⎢ ⎥ ==− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 17.Let ab A cd ⎡⎤ = ⎢⎥ ⎣⎦ .Thenthecofactorsof A are11, Cdd == 12, Ccc =−=− 21 Cbb =−=− ,and 22 Caa == .Thusadj db A ca ⎡⎤ = ⎢⎥ ⎣⎦ .Sincedet A = ad – bc,Theorem8givesthat 111 detadj db AA Aadbcca ⎡⎤ == ⎢⎥ ⎣⎦ .ThisresultisidenticaltothatofTheorem4inSection2.2.

⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ,sotheareaoftheparallelogramis |det A|=|8|=8.

⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ,sotheareaoftheparallelogram is|det A|=|–7|=7.

54 A ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ,sotheareaoftheparallelogramis |det A|=|–14|=14.

63 13 A ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ,sotheareaoftheparallelogramis |det A|=|21|=21.

117 021 240 A ⎡ ⎤ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ,sothevolumeofthe parallelepipedis|det A|=|22|=22.

24.Theparallelepipedisdeterminedbythecolumnsof

parallelepipedis|det A|=|–15|=15.

25.TheInvertibleMatrixTheoremsaysthata3 × 3matrix A isnotinvertibleifandonlyifitscolumns arelinearlydependent.Thiswillhappenifandonlyifoneofthecolumnsisalinearcombinationof theothers;thatis,ifoneofthevectorsisintheplanespannedbytheothertwovectors.Thisis equivalenttotheconditionthattheparallelepipeddeterminedbythethreevectorshaszerovolume, whichisinturnequivalenttotheconditionthatdet A =0.

26.Bydefinition, p + S isthesetofallvectorsoftheform p + v,where v isin S.Applying T toatypical vectorin p + S,wehave T(p + v)= T(p)+ T(v).Thisvectorisinthesetdenotedby T(p)+ T(S).This provesthat T mapstheset p + S intotheset T(p)+ T(S).Conversely,anyvectorin T(p)+ T(S)has theform T(p)+ T(v)forsome v in S.Thisvectormaybewrittenas T(p + v).Thisshowsthatevery vectorin T(p)+ T(S)istheimageunder T ofsomepoint p + v in p + S.

27.Sincetheparallelogram S

S}=6 ⋅ 4=24.Alternatively,onemaycomputethevectorsthatdeterminethe image,namely,thecolumnsof

Thedeterminantofthismatrixis–24,sotheareaoftheimageis24.

A|{areaof S}=5 ⋅ 4=20.Alternatively,onemaycomputethevectorsthatdeterminetheimage, namely,thecolumnsof

7240142 117131

Thedeterminantofthismatrixis20,sotheareaoftheimageis20.

29.Theareaofthetrianglewillbeonehalfoftheareaoftheparallelogramdeterminedby1 v and2 v ByTheorem9,theareaofthetrianglewillbe(1/2)|det A|,where [] 12. A = vv

30.Translate R toanewtriangleofequalareabysubtracting33(,) xy fromeachvertex.Thenewtriangle hasvertices(0,0),1313 (,) xxyy ,and2323 (,). xxyy ByExercise29,theareaofthetriangle willbe

186CHAPTER3•Determinants

121 452 021 A ⎡ ⎤ ⎢ ⎥ =− ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ,sothevolumeofthe

isdeterminedbythecolumnsof 22 35 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ ,theareaof S is 22 det|4|4. 35 ⎡⎤ =−= ⎢⎥ ⎣⎦ Thematrix A has 62 det6 32 A == .ByTheorem10,theareaof T(S) is|det A

[] 12 62221822 32351216 A ⎡⎤⎡⎤⎡⎤ == ⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦ bb

|{areaof

40 71 ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ ,theareaof S

40 det|4|4 71 ⎡⎤ == ⎢⎥ ⎣⎦ .Thematrix

has 72 det5 11 A == .ByTheorem10,theareaof T

[]

28.Sincetheparallelogram S isdeterminedbythecolumnsof

(

)is |det

A ⎡⎤⎡⎤⎡⎤ == ⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦ bb

uxa = ,22/uxb = ,and33/uxc = ,and u liesinside S (or

1231uuu++≤ )ifandonlyif x liesinside T(S)(or 222 123 2221xxx abc ++≤ ).

b.BythegeneralizationofTheorem10, {volumeofellipsoid}{volumeof()} TS =

32a.Alineartransformation T thatmaps S onto S ′ willmap1 e to1, v 2 e to2, v and3 e to3; v thatis, 11() T = ev ,22 () T = ev ,and33 (). T = ev Thestandardmatrixforthistransformationwillbe [] [] 123123 ()()().ATTT ==eeevvv

3.3•Solutions187 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 1323 1323 1 det. 2 xxxx yyyy ⎡⎤ ⎢⎥ ⎣⎦ Nowconsiderusingrowoperationsandacofactorexpansiontocomputethedeterminantinthe formula: 111313 1313 222323 2323 3333 10 det1det0det 11 xyxxyy xxyy xyxxyy xxyy xyxy ⎡⎤⎡⎤ ⎡ ⎤ ⎢⎥⎢⎥ =−−= ⎢ ⎥ ⎢⎥⎢⎥ ⎣ ⎦ ⎢⎥⎢⎥ ⎣⎦⎣⎦ ByTheorem5, 13131323 23231323 detdetxxyyxxxx xxyyyyyy ⎡⎤⎡⎤ = ⎢⎥⎢⎥ ⎣⎦⎣⎦ Sotheaboveobservationallowsustostatethattheareaofthetrianglewillbe 11 1323 22 1323 33 1 11 detdet1 22 1 xy xxxx xy yyyy xy ⎡ ⎤ ⎡⎤ ⎢ ⎥ = ⎢⎥ ⎢ ⎥ ⎣⎦ ⎢ ⎥ ⎣ ⎦ 31. a.Toshowthat T(S)isboundedbytheellipsoidwithequation 222 123 2221xxx abc ++= ,let 1 2 3 u u u ⎡⎤ ⎢⎥ = ⎢⎥ ⎢⎥ ⎣⎦ u and let 1 2 3 x xA x ⎡⎤ ⎢⎥ == ⎢⎥ ⎢⎥ ⎣⎦ xu .Then11/

222

|det|{volumeof}44

ASabc

=⋅==

33 abc

ππ

b.Theareaofthebaseof S is(1/2)(1)(1)=1/2,sothevolumeof S is(1/3)(1/2)(1)=1/6.Byparta.

T(S)= S ′ ,sothegeneralizationofTheorem10givesthatthevolumeof S ′ is|det A|{volumeof S}=(1/6)|det A|.

33. [M] Answerswillvary.InMATLAB,entriesin B –inv(A)areapproximately1510orsmaller.

34[M] Answerswillvary,aswillthecommandswhichproducethesecondentryof x.Forexample,the MATLABcommandis x2 = det([A(:,1) b A(:,3:4)])/det(A) whiletheMathematica commandis x2 = Det[{Transpose[A][[1]],b,Transpose[A][[3]], Transpose[A][[4]]}]/Det[A].

35. [M] MATLABStudentVersion4.0uses57,771flopsforinv A and14,269,045flopsfortheinverse formula.The inv(A) commandrequiresonlyabout0.4%oftheoperationsfortheinverseformula.

Chapter 3 SUPPLEMENTARY EXERCISES

1. a.True.Thecolumnsof A arelinearlydependent.

b.True.SeeExercise30inSection3.2.

c.False.SeeTheorem3(c);inthiscase3det55detAA = .

d.False.Consider

e.False.ByTheorem6,33 det2 A = .

f.False.SeeTheorem3(b).

g.True.SeeTheorem3(c).

h.True.SeeTheorem3(a).

i.False.SeeTheorem5.

j.False.SeeTheorem3(c);thisstatementisfalsefor n × n invertiblematriceswith n aneven integer.

k.True.SeeTheorems6and5;2 det(det) T AAA = .

l.False.Thecoefficientmatrixmustbeinvertible.

m.False.Theareaofthe triangle is5.

n.True.SeeTheorem6;33 det(det) AA = .

o.False.SeeExercise31inSection3.2.

p.True.SeeTheorem6.

188CHAPTER3•Determinants

20 01 A ⎡⎤ = ⎢⎥ ⎣⎦ , 10 03 B ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ ,and 30 04 AB ⎡ ⎤ += ⎢ ⎥ ⎣ ⎦ .

2. 121314121314 1516173330 181920666 == 3. 111 10()()0110 10011 abcabcabc bacbaabbaca cabcaac +++ +=−−=−−−= +−−

Thisisanequationoftheform ax + by + c =0,andsincethepoints11(,) xy and22(,) xy aredistinct, atleastoneof a and b isnotzero.Thustheequationistheequationofaline.Thepoints11(,) xy and 22 (,) xy areontheline,becausewhenthecoordinatesofoneofthepointsaresubstitutedfor x and y, tworowsofthematrixareequalandsothedeterminantiszero.

Chapter3•SupplementaryExercises189 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 4.1110 111 abcabcabc axbxcxxxxxy aybycyyyy +++=== +++ 5 91999 9992 90992405 4050 40050(1)(1)(2)939 9390 90390607 6070 60070 =− =−− 45 (1)(2)(3)(1)(2)(3)(2)12 67 =−−=−−−=− 6 48885 4885 01000485 688745 68887(1)(1)(2)687(1)(2)(3)(1)(2)(3)(2)12 083067 08830030 0200 08200 ===−=−−= 7.Expandalongthefirstrowtoobtain 1111 11 2222 22 1 11 110. 11 1 xy xyyx xyxy xyyx xy =−+=

.Expandalongthefirstrowtoobtain 1111 1111 1 11 111()()(1)0. 1001 01 xy xyyx xyxymxyxmy mm m =−+=−−+= Thisequationmay berewrittenas110, mxymxy−−+= or11().yymxx −=− 9. 222 222 222 111 det100()() 100()() aaaaaa Tbbbababababa cccacacacaca ==−−=−−+ −−+ 2211 ()()01()()01()()() 0100 aaaa bacababacababacacb cacb =−−+=−−+=−−− +−

10.Expandingalongthefirstrowwillshowthat23 ()det0123. ftVcctctct ==+++ ByExercise9,

since1 x ,2 x ,and3 x aredistinct.Thus f (t)isacubicpolynomial.Thepoints1(,0) x ,2(,0) x ,and 3 (,0) x areonthegraphof f,sincewhenanyof1 x ,2 x or3 x aresubstitutedfor t,thematrixhastwo equalrowsandthusitsdeterminant(whichis f (t))iszero.Thus()0 ifx = for i =1,2,3.

11.Totellifaquadrilateraldeterminedbyfourpointsisaparallelogram,firsttranslateoneofthe verticestotheorigin.Ifwelabeltheverticesofthisnewquadrilateralas 0,1 v ,2 v ,and3 v ,then theywillbetheverticesofaparallelogramifoneof1 v ,2 v ,or3 v isthesumoftheothertwo.In thisexample,subtract(1,4)fromeachvertextogetanewparallelogramwithvertices 0 =(0,0), 1(2,1) =− v ,2(2,5) = v ,and3(4,4) = v .Since231 =+ vvv ,thequadrilateralisaparallelogramas stated.Thetranslatedparallelogramhasthesameareaastheoriginal,andisdeterminedbythe columnsof [] 13 24 14 A ⎡⎤ == ⎢⎥

12.A2 × 2matrix A isinvertibleifandonlyiftheparallelogramdeterminedbythecolumnsof A has nonzeroarea.

13.ByTheorem8,1 1 (adj)detAAAAI A ⋅== .BytheInvertibleMatrixTheorem,adj A isinvertible and11 (adj)detAA A = .

14. a.Considerthematrix k k

AO A OI ⎡⎤ = ⎢⎥

,where1 ≤ k ≤ n and O isanappropriatelysizedzeromatrix.

Wewillshowthatdetdet k AA = forall1 ≤ k ≤ n bymathematicalinduction.

Firstlet k =1.Expandalongthelastrowtoobtain

(1)(1) 1 detdet(1)1detdet. 1 nn AO AAA O +++ ⎡⎤ ==−⋅⋅= ⎢⎥ ⎣⎦

Nowlet1< k ≤ n andassumethat1detdet. k AA = Expandalongthelastrowof kA toobtain

AO AAAA OI +++ ⎡⎤ ==−⋅⋅== ⎢⎥ ⎣⎦ Thuswehaveproventhe result,andthedeterminantofthematrixinquestionisdet A.

()() 11 detdet(1)1detdetdet. nknk kkk k

b.Considerthematrix k k k

IO A CD ⎡⎤ = ⎢⎥ ⎣⎦ ,where1 ≤ k ≤ n, kC isan n × k matrixand O isan appropriatelysizedzeromatrix.Wewillshowthatdetdet k AD = forall1 ≤ k ≤ n by mathematicalinduction.

190CHAPTER3•Determinants

2 11 2 322213132 2 33 1 1()()()0 1 xx cxxxxxxxx xx =−=−−−≠

⎣⎦

,sotheareaoftheparallelogramis|det A|=|–12|=12.

⎣⎦

15. a.Computetherightsideoftheequation:

Setthisequaltotheleftsideoftheequation:

Since XA = C and A isinvertible,1 . XCA = Since XB + Y = D,

16. a.Doingthegivenoperationsdoesnotchangethedeterminantof A sincethegivenoperationsare allrowreplacementoperations.Theresultingmatrixis

Chapter3•SupplementaryExercises191 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. Firstlet k =1.Expandalongthefirstrowtoobtain 11 1 1 1 detdet(1)1detdet. O ADD CD + ⎡⎤ == ⋅= ⎢⎥ ⎣⎦ Nowlet1< k ≤ n andassumethat1detdet. k AD = Expandalongthefirstrowof kA toobtain 11 detdet11(1)1detdetdet. k kkk k IO AAAD CD + ⎡⎤ == −⋅ ⋅== ⎢⎥ ⎣⎦ Thuswehaveproventheresult,and thedeterminantofthematrixinquestionisdet D c.Bycombiningpartsa.andb.,wehaveshownthat detdetdet(det)(det). AOAOIO AD CDOICD ⎛⎞⎛⎞ ⎡⎤⎡⎤⎡⎤ == ⎜⎟⎜⎟ ⎢⎥⎢⎥⎢⎥ ⎣⎦⎣⎦⎣⎦ ⎝⎠⎝⎠ FromthisresultandTheorem5,wehave detdetdet(det)(det) TT TT TT ABABAO AD ODODBD ⎡⎤ ⎡⎤⎡⎤ === ⎢⎥ ⎢⎥⎢⎥ ⎣⎦⎣⎦ ⎢⎥ ⎣⎦ (det)(det). AD =

IOABAB XIOYXAXBY ⎡⎤⎡⎤⎡⎤ = ⎢⎥⎢⎥⎢⎥ + ⎣⎦⎣⎦⎣⎦

sothat ABAB XACXBYD CDXAXBY ⎡⎤⎡⎤ == + = ⎢⎥⎢⎥ + ⎣⎦⎣⎦

11 detdetdet ABIOAB CDCAIODCAB ⎡⎤⎡⎤ ⎡⎤ = ⎢⎥⎢⎥ ⎢⎥ ⎣⎦ ⎣⎦⎣⎦ 1

=−

.Fromparta., 11 det(det)(det())det[()] AB ADCABADCAB CD ⎡⎤ =−=− ⎢⎥ ⎣⎦ det[11

ADACABADCAAB=−=− det[] ADCB=−

00 00 000 abab abab ab bbba −−+… ⎡⎤ ⎢⎥ −−+… ⎢⎥ ⎢⎥ −… ⎢⎥ ⎢⎥ ⎢⎥ … ⎣⎦

1 YDXBDCAB =−=− .Thus byExercise14(c),

(det)(det()) ADCAB

]det[]

b.Sincecolumnreplacementoperationsareequivalenttorowoperationson TA anddetdet T AA = , thegivenoperationsdonotchangethedeterminantofthematrix.Theresultingmatrixis

c.Sincetheprecedingmatrixisatriangularmatrixwiththesamedeterminantas A, 1 det()((1)). n Aabanb =−+−

17.Firstconsiderthecase n =2.Inthiscase

=+=−+−=−=−+=−+−

Nowassumethattheformulaholdsforall(k –1) × (k –1)matrices,andlet A, B,and C be k × k matrices.Byacofactorexpansionalongthefirstcolumn,

()()((2))()((2)) kk

sincethematrixintheaboveformulaisa(k –1) × (k –1)matrix.Wecanperformaseriesofrow operationson C to“zeroout”belowthefirstpivot,andproducethefollowingmatrixwhose determinantisdet C:

Sincethisisatriangularmatrix,wehavefoundthat1 det() Cbabk =− .Thus

111 detdetdet()((2))()()((1)), kkk ABCabakbbababakb =+=−+−+−=−+− whichiswhatwastobeshown.Thustheformulahasbeenprovenbymathematicalinduction.

18[M] Sincethefirstmatrixhas a =3, b =8,and n =4,itsdeterminantis

413 (38)(3(41)8)(5)(324)(125)(27)3375. −+−=−+=−=−

Sincethesecondmatrixhas a =8, b = 3,and n =5,itsdeterminantis514(83)(8(51)3)(5)(812)(625)(20)12,500. −+−=+==

192CHAPTER3•Determinants

000 000 000 23(1) ab ab ab bbbanb −… ⎡⎤ ⎢⎥ −… ⎢⎥ ⎢⎥ −… ⎢⎥ ⎢⎥ ⎢⎥ …+− ⎣⎦

det2 0(),det, abbbb BaabCabb aba ==−==−

so detdetdet()22221()()()((21)) ABCaababbababababab

,and theformulaholdsfor n =2.

det()21

abb bab

bba

Babababakbabakb

… =−=−−+−=−+−

00 bbb

… ⎡⎤ ⎢⎥ −… ⎢⎥ ⎢⎥ ⎢⎥ …− ⎢⎥ ⎣⎦

ab ab

Toshowthis,considerusingrowreplacementoperationsto“zeroout”belowthefirstpivot.The resultingmatrixis

Nowuserowreplacementoperationsto“zeroout”belowthesecondpivot,andsoon.Thefinal matrixwhichresultsfromthisprocessis

whichisanuppertriangularmatrixwithdeterminant1.

Chapter3•SupplementaryExercises193

19. [M] Wefindthat 11111 1111 11112222 1222 1221,1,1. 12333 1233 12312344 1234 12345 ===

1111 1222

Ourconjecturethenisthat

12331. 123 n = …

⎡⎤ ⎢⎥ … ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ …− ⎣⎦

1111 0111 0122 0121 n …

0001 ⎡⎤ ⎢⎥ … ⎢⎥ ⎢⎥ … ⎢⎥ ⎢⎥ ⎢⎥ ⎣⎦

1111 0111 0011,

[M] Wefindthat 11111 1111 11113333 1333 1336,18,54. 13666 1366 13613699 1369 136912 === Ourconjecturethenisthat

20.

Toshowthis,considerusingrowreplacementoperationsto“zeroout”belowthefirstpivot.The resultingmatrixis

Nowuserowreplacementoperationsto“zeroout”belowthesecondpivot.Thematrixwhichresults fromthisprocessis

Thismatrixhasthesamedeterminantastheoriginalmatrix,andisrecognizableasablockmatrixof theform

194CHAPTER3•Determinants

2 1111 1333 136623. 1363(1) n n … =⋅ … …−

1111 0222 0255. 0253(1)1 n … ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ …−− ⎣⎦

1111111 0222222 0033333 0036666 0036999 00369123(2) n ⎡⎤ ⎢⎥ ⎢⎥ ⎢⎥ … ⎢⎥ … ⎢⎥ ⎢⎥ … ⎢⎥ ⎢⎥ ⎢⎥ …− ⎣⎦

, AB OD ⎡⎤ ⎢⎥ ⎣⎦ where 3333311111 3666612222 11 and3. 3699912333 02 369123(2)12342 AD nn …… ⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ …… ⎢ ⎥⎢ ⎥ ⎡⎤ ⎢ ⎥⎢ ⎥ == = ⎢⎥ ⎢ ⎥⎢ ⎥ ⎣⎦ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ …− ⎣ ⎦⎣ ⎦ … AsinExercise14(c),thedeterminantofthematrix AB OD ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ is(det A)(det D)=2det D. Since D isan(n –2) × (n –2)matrix,

Chapter3•SupplementaryExercises195 Copyright©2012PearsonEducation,Inc.PublishingasAddison-Wesley. 222 11111 12222 det33(1)3 12333 12342 Dnnn n … === … …− byExercise19.Thusthedeterminantofthematrix AB OD ⎡ ⎤ ⎢ ⎥ ⎣ ⎦ is2 2det23. Dn =⋅

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