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Test Bank For Elementary Linear Algebra - 12th - 2017 All Chapters - 9781305658004.pdf

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Test Bank For Elementary Linear Algebra - 12th - 2017 All Chapters -

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ELEMENTARY LIN g




EAR ALGEBRA g




K. R. MATTHEWS
g g




DEPARTMENTg OFg MATHEMATICS




UNIVERSITYgOFgQUEENSLAND




CorrectedgVersion,g27thgAprilg2013gComme



,Chapter 1 g




LINEAR EQUATIONS g




1.1 Introduction to linear equations
g g g




Aglineargequationgingngunknownsgx1,gx2,g·g·g·g,gxng isgangequationgofgthegform

a1x1g+ga2x2g+g·g·g·g+ganxng =g b,

whereg a1,g a2,g.g.g.g,gan,g bg areg giveng realg numbers.
Forg example,g withg xg andg yg insteadg ofg x1g andg x2,g theg linearg equationg
2xg+g3yg=g6gdescribesgtheglinegpassinggthroughgthegpointsg(3,g0)gandg(0,g2).g
Similarly,g withg x,g yg andg zg insteadg ofg x1,g x2g andg x3,g theg linearg equa-
gtiong 2xg+g3yg +g4zg =g 12g describesg theg planeg passingg throughg theg points

(6,g 0,g 0),g (0,g 4,g 0),g (0,g 0,g 3).
Agsystemgofgmglineargequationsgingngunknownsgx1,gx2,g·g·g·g,gxngisgagfamilyg
ofglineargequations


a11x1g+ga12x2g +g·g·g·g+ga1nxn = b1
a21x1g+ga22x2g +g·g·g·g+ga2nxn = b2
..
am1x1g+gam2x2g+g·g·g·g+gamnxn = bm.

Weg wishg tog determineg ifg suchg ag systemg hasg ag solution,g thatg isg tog findg
outgifgtheregexistgnumbersgx1,gx2,g·g·g·g,gxngwhichgsatisfygeachgofgthegequationsgsi
multaneously.g Wegsaygthatgthegsystemgisgconsistentgifgitghasgagsolution.gOther
wisegthegsystemgisgcalledginconsistent.

1

,2 CHAPTERg 1.g g LINEARg EQUATIONS

Noteg thatg theg aboveg systemg cang beg writteng conciselyg as

Σ
n
aijxjg =gbi, ig=g1,g 2,g·g·g·g,gm.
j=1

Theg matrix
a11 a12 ·g·g·
a1ng a21 a22
·g·g· a2n
. .
am1 am2 ·g·g·g g amn
isgcalledg theg coefficientg matrixg ofg theg system,g whileg theg matrix

a11 a12 ·g·g· a1n
b1g a21a22 ·g·g·
a2n b2
. . .
am1 am2 ·g·g·g g amn bm

isgcalledgthegaugmentedg matrixgofgthegsystem.
Geometrically,gsolvinggagsystemgofglineargequationsgingtwog(orgthree)gunk
nownsgisgequivalentgtogdetermininggwhethergorgnotgagfamilygofglinesg(orgplane
s)ghasgagcommongpointgofgintersection.

EXAMPLEg 1.1.1g Solveg theg equation

2xg+g3yg=g6.

Solution.g g Theg equationg 2xg+g3yg =g 6g isg equivalentg tog 2xg =g 6g—g3yg or
xg=g3g—g 32y,gwheregygisgarbitrary.g Sogtheregareginfinitelygmanygsolutions.

EXAMPLEg 1.1.2g Solveg theg system

xg+gyg+gz =g g 1
xg—gyg+gz =g g 0.

Solution.g Wegsubtractgthegsecondgequationgfromgthegfirst,gtoggetg2yg=g1g
andgyg=g1g.g Thengxg=gyg—gzg=g1g—gz,gwheregzg isgarbitrary.g Againgtheregare
2 2
infinitelyg manyg solutions.

EXAMPLEg 1.1.3g Findgagpolynomialgofgthegformgyg=ga0+a1x+a2x2+a3x3
whichgpassesgthroughgthegpointsg(—3,g—2),g(—1,g2),g(1,g5),g(2,g1).

, 1.1. INTRODUCTIONg TOg LINEARg EQUATIONS 3

Solution.g Whengxghasgthegvaluesg—3,g—
1,g1,g2,gthengygtakesgcorrespondinggvaluesg—
2,g 2,g 5,g 1gandgweggetgfourgequationsgingthegunknownsga0,g a1,g a2,g a3:

a0g—g3a1g+g9a2g—g27a3 = —2
a0g—ga1g+ga2g—ga3 = 2
a0g+ga1g+ga2g+ga3 = 5
a0g+g2a1g+g4a2g+g8a3 = 1,

withguniquegsolutionga0g=g93/20,ga1g=g221/120,ga2g=g—23/20,ga3g=g—41/120.
Sog theg requiredg polynomialg is
93g 221g 23g g41g 3
y = +g xg—g x2g—g x.
20 120 20 120
Ing[26,gpagesg33–
35]gtheregaregexamplesgofgsystemsgofglineargequationsgwhichgarisegfromgsimpl
egelectricalgnetworksgusinggKirchhoff’sglawsgforgelec-gtricalgcircuits.
Solvinggagsystemgconsistinggofgagsingleglineargequationgisgeasy.g Howevergifg
wegaregdealinggwithgtwogorgmoregequations,gitgisgdesirablegtoghavegagsystematicg
methodgofgdetermininggifgthegsystemgisgconsistentgandgtogfindgallgsolutions.
Insteadgofgrestrictinggourselvesgtoglineargequationsgwithgrationalgorgrealg
coefficients,gourgtheoryggoesgovergtogthegmoreggeneralgcasegwheregthegcoef-
gficientsgbelonggtogangarbitrarygfield.g AgfieldgFgisgag setgFg whichg possessesgo

perationsgofgadditiongandgmultiplicationgwhichgsatisfygthegfamiliargrulesgofgr
ationalgarithmetic.g Theregaregtengbasicgpropertiesgthatgagfieldgmustghave:

THEg FIELDg AXIOMS.

1. (ag+gb)g+gcg=gag+g(bg+gc)g forg allg a,g b,g cg ing Fg;
2. (ab)cg=ga(bc)gforgallga,gb,gcgingFg;
3. ag+gbg=gbg+gag forg allg a,g bg ing Fg;
4. abg=gbag forg allg a,g bg ingFg;
5. theregexistsgangelementg0gingFg suchg thatg0g+gag=gagforgallgagingFg;
6. theregexistsgangelementg1gingFg suchgthatg1ag=gagforgallgagingFg;
7. togeverygagingFg,gtheregcorrespondsgangadditiveginverseg—agingFg,gsatis-
gfying

ag+g(—a)g=g0;

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