C H A P T E R
6 Direct Methods forSolving Linear Systems
6.1 Introduction
Systems ofequations areused torepresent physical problems thatinvolve theinteraction of
various properties .The variables inthesystem represent theproperties being studied ,and
theequations describe theinteraction between thevariables .The system iseasiest tostudy
when theequations arealllinear .Often thenumber ofequations isthesame asthenumber
ofvariables ,foronly inthiscase isitlikely thataunique solution willexist.
Notallphysical problems canbereasonably represented using alinear system with the
same number ofequations asunknowns ,butthesolutions tomany problems either have this
form orcanbeapproximated bysuch asystem .Infact,thisisquite often theonly approach
thatcangive quantitative information about aphysical problem .
Inthischapter weconsider direct methods forapproximating thesolution ofasystem
ofnlinear equations innunknowns .Adirect method isonethatgives theexact solution to
thesystem ,ifitisassumed thatallcalculations canbeperformed without round -offenor
effects .However ,wecannot generally avoid round -offerror andweneed toconsider quite
carefully theroleoffinite-digit arithmetic error intheapproximation tothesolution tothe
system ,andhow toarrange thecalculations tominimize itseffect .
6.2 Gaussian Elimination
Ifyou have studied linear algebra ormatrix theory ,you probably have been introduced
toGaussian elimination ,themost elementary method forsystematically determining the
solution ofasystem oflinear equations .Variables areeliminated from theequations until
oneequation involves only onevariable ,asecond equation involves only thatvariable and
oneother,athird hasonly these twoandoneadditional ,andsoon.Thesolution isfound by
solving forthevariable inthesingle equation ,using thistoreduce thesecond equation to
onethatnowcontains asingle variable ,andsoon,until values forallthevariables arefound .
Three operations arcpermitted onasystem ofequations E\,£2,   .£*.
Operations onSystems ofEquations
 Equation £,canbemultiplied byanynonzero constant A.,with theresulting
equation used inplace of£,.This operation isdenoted (A.£,)-(£,).
 Equation E}canbemultiplied byanyconstant A.,andadded toequation £,,with
theresulting equation used inplace of£,-.This operation isdenoted (£;+A.£j)-
(ft).
 Equations E,andEjcanbetransposed inorder.This operation isdenoted (£,)<-*
(Ej).
229
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IllustrationByasequence oftheoperations justgiven ,alinear system canbetransformed toa
more easily -solved linear system with thesame solutions .The sequence ofoperations is
shown inthenext Illustration .
The four equations
Ey.*14-*2 4-3*4=4,
Ey.2*i+*2-*34-*4= 1,
Ey 3*,-*2-*34-2*4=-3,
£4:-*14-2*24-3*3-*4=4,
willbesolved for*j,*2,*3,and*4.Wefirstuseequation £1toeliminate theunknown X\
from equations £2,£3»and£4byperforming (£2— 2£j)-(£2),(£3— 3£j)— (£3),
and(£4+£])-(£4).Forexample ,inthesecond equation
(£2— 2£,)— (£2)
produces
(2*i+X2-*3+*4)-2(x\+x2+3*4)=1-2(4),
which simplifies totheresult shown as£2in
Et:*14-*2 4-3*4= 4,
Ey -*2-*3-5*4=-7,
Ey.-4*2-*u>1-O*II-15,
Ey. 3*24-3*34-2*4= 8.
Forsimplicity ,thenewequations arcagain labeled £1,£2,£3,and£4.
Inthenewsystem ,£2isused toeliminate theunknown JC2from£3and£4byperforming
(£3— 4£2)— (£3)and(£44-3£2)—(£4).This results in
E\ix1+*2 4-3*4
£2:-*2-*3-5*4
£3: 3*3+13*4
£4: — 13*4
The system ofequations (6.2)isnow intriangular (orreduced )form andcanbe
solved fortheunknowns byabackward -substitution process .Since £4implies *4=1,
wecansolve £3for*3togive
x3=*(13-13*4)=
^(13-13)=0.
mJ mJ
Continuing ,£2gives
*2=— (“ 7+5*44-*3)=— (— 74-5+0)=2,4,
“ 7,
13,
-13.(6.2)
and£1gives
*i=4— 3*4— *2=4— 3— 2=— 1.
The solution tosystem (6.2),andconsequently tosystem (6.1),istherefore *1=— 1,
*2=2,*3=0,and*4=1.
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Matrices and Vectors
When performing thecalculations oftheIllustration ,wedidnotneed towrite outthefull
equations ateach steportocarry thevariables X\,X29*3,andx4through thecalculations
because they always remained inthesame column .The only variation from system to
system occurred inthecoefficients oftheunknowns andinthevalues ontheright side of
theequations .Forthisreason ,alinear system isoften replaced byamatrix ,arectangular
array ofelements inwhich notonlyisthevalue ofanelement important ,butalsoitsposition
inthearray.Thematrix contains alltheinformation about thesystem thatisnecessary to
determine itssolution inacompact form.
Thenotation forannxm(nbym)matrix will beacapital letter ,such asA,forthe
matrix ,andlowercase letters with double subscripts ,such asairtorefer totheentry atthe
intersection oftheithrowand jthcolumn ;thatis,
A=[a y)=a\\ d]2
a2ld22a\m
&2m
O/il &n2** *
Example 1Determine thesizeandrespective entries ofthematrix
A2-1 7
3 10
Solution Thematrix hastworows andthree columns ,soitisofsize2x3.Itsentries are
described byan=2,d\2=— l,ai3=7,021=3,d22=1,and 023=0.
The1xnmatrix A=[an d\2 --d\n]iscalled ann-dimensional row vector ,and
annx1matrix
o11
«21A=
1
iscalled an/{-dimensional column vector .Usually theunnecessary subscript isomitted
forvectors andaboldface lowercase letter isused fornotation .So,
*1
*2x=
.x".
denotes acolumn vector ,andy=[yiy2   yn]denotes arow vector.
Asystem ofnlinear equations inthenunknowns x\9*2»   .xnhastheform
011*1+al2x2+  +dlnxn-bu
021*1+022*2H b02nX„=b2.
On1*1+0„ 2*2+**’+Onn^fl— bn.
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Annx{n4-1)matrix canbeused torepresent thislinear system byfirstconstructing
^=laij ]=«11«12'*’«ln'*t‘
«21«22 «2*and b=b2
«nl«n2* * *«nn .b».
Augmented refers tothefactthat
theright-hand side ofthesystem
hasbeen included inthematrix .andthen combining these matrices toform theaugmented matrix :
W,b]«11«12* **«ln b{
«21«22***«2n b2
I !  
ania,,2    ann bn.
where thevertical dotted linebefore thelastcolumn isused toseparate thecoefficients of
theunknowns from thevalues ontheright-hand sideoftheequations .
Repeating theoperations involved intheIllustration onpage 230 with thematrix
notation results infirstconsidering theaugmented matrix :
1 1 0 3 4'
2 1-1 1 1
3-1-1 2-3-1 2 3-1 4
Performing theoperations
(£2-2£,)-»(£2),(£3-3£I)-*(£3),and (£4+£,)-(£4)
produces
1 1 0 3 4'
0-1-1-5-7
0-4-1-7-1 5
0 3 3 2 8
Then
Atechnique similar toGaussian
elimination firstappeared during
theHandynasty inChina inthe
textNine Chapters onthe
Mathematical Art,which was
written inapproximately 200
BCE.Joseph Louis Lagrange
(1736-1813 )described a
technique similar tothis
procedure in1778 forthecase
when thevalue ofeach equation
is0.Gauss gave amore general
description inTheoria Motus
corporum coelestium sectionibus
solem ambientium ,which
described theleast squares
technique heused in1801 to
determine theorbit ofthedwarf
planet Ceres .(£3-4£2)-(£3)and (£4+3£2) (£4),
produces thefinal matrix
110 3
0-1-1-5
0 0 3 1 3
0 0 0 -1 34-7
1 3-1 3
This final matrix canbetransformed into itscorresponding linear system andsolutions
for ,*2,*3,and*4obtained .The procedure involved inthisprocess iscalled Gaussian
Elimination with Backward Substitution .
Thegeneral Gaussian elimination procedure applied tothelinear system
£i:tfn.xi4-a12*2+- - -+a\nXn=bu
£2:«21*1+a22*2H h«2/i*n=bi,
£« «nl*l"fa,j2*24"* * *4“ flnn^n— bfi,
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ishandled inasimilar manner .First form theaugmented matrix A:
an a\2   <*ln
A=[A,b]=aiia22 <*2n  a2,n+l
Qnl an2* **ann
where Adenotes thematrix formed bythecoefficients andtheentries inthe (n+l)st
column arethevalues ofb;thatis,al%n+,=b,foreach i=1,2,...,n.
Suppose thatan#0.Toconvert theentries inthefirstcolumn ,below an,tozero,we
perform theoperations (£*-mk,£,)-(£*)foreach k=2,3...,nforanappropriate
multiplier mk\.Wefirst designate thediagonal element inthecolumn ,a,,asthepivot
element .The multiplier forthek\hrow isdefined bymk\=ak\/an.Performing the
operations (£*— I£i)-(£*)foreach k=2,3,...,neliminates (thatis,changes to
zero)thecoefficient ofx\ineach ofthese rows:
an a\2   a\n:b\ E2— rn2iEi— E2 011 012    Oin:b\
021 022    &2n\b2Ei-mi\E\-£3 0 022** * <*2/1 &2
On1&n2* *’ann ;bn.£„ — mn\E\-£„ 0 On2* * * ann\bn
Although theentries inrows 2,3,...,nareexpected tochange ,forease ofnotation ,we
again denote theentry intheithrowandthejthcolumn byatj.
Ifthepivot element 022#0,weform themultipliers mk2=ak2/a22andperform the
operations (£*— m^Ei)— £*foreach k=3,...,nobtaining
an
0012
022* * *a\n
   a2nbi
biE)-m32E2->£3an
00,2
022***a\n
   a^br
b2
0an2***ann KEn— mn2E2— £„ .0 0 ***ann K
Wethen follow thissequential procedure fortherows i=3...,n— 1.Define themultiplier
mki=aki/aaandperform theoperation
(Ek-m kiEi )-(Ek)
foreach k=i+1,i+2,...,n,provided thepivot element a„ isnonzero .This eliminates
Xiineach rowbelow theithforallvalues ofi=1,2,...,n—1.Theresulting matrix has
theform
0,i0,2  **0,n ai,n+i
0..q
^2  **fl2n a2,n+l
l   
.0 0 Onn an,n+im
where ,except inthefirstrow,thevalues ofa,jarenotexpected toagree with those inthe
original matrix A.Thematrix Arepresents alinear system with thesame solution setasthe
original system ,thatis,
a\\X\-f0,2*2H ha\n*n=01./J+1.
<*22*2H 1-<*2nXn=02.«+!.
&nnXrt=^n.n+1»
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Backward substitution canbeperformed onthissystem .Solving thenthequation for*„ 
gives
®n,n+l
Xn=  
Qnn
Then solving the(n— l)stequation for jcn_iandusing theknown value forxnyields
x n-1&n— l.n+l &n— l,nXn
<*n-\,n-\
Continuing thisprocess ,weobtain
Bj.n+1— (aiJ+lXj+lH bOj.nXn)
anai,n+1^^j=i+\aijXj
an
Theprogram GAUSEL 61
implements Gaussian
Elimination with
Backward Substitution .foreach /=n— 1,n— 2,...,2,1.
The procedure willfailifattheithstep thepivot element a„ -iszero ,forthen either
themultipliers m*,=am/anarenotdefined (thisoccurs ifan=0forsome i<n)orthe
backward substitution cannot beperformed (ifann=0).This does notnecessarily mean
thatthesystem hasnosolution ,butrather thatthetechnique forfinding thesolution must
bealtered byinterchanging rows when apivot is0.
Program GAUSEL 61incorporates row interchanges when required .Anillustration is
given inthefollowing example .
Example 2Represent thelinear system
Ell Xi— *2+2*3-*4=-8,
E2:2*i-2*2+3*3-3*4=-20,
£3:*1+*2+*3 =— 2,
£4:*1-*2+4*3+3*4= 4,
asanaugmented matrix anduseGaussian elimination tofinditssolution .
Solution Theaugmented matrix is
'1-12-1 -8'
2-23-3-20
1 1 1 0-2
1-14 3 4
Performing theoperations
(£2-2£|)-(£2),(£3-Ei) (£3),and (£4-£1)-(£4),
gives thematrix
Thepivot element foraspecific
column istheentry thatisused to
place zeros intheother entries in
thatcolumn .'1-1 2-1-8‘
0 0 -1-1-4
0 2-1 1 6
0 0 2 4 12
The new diagonal entry 022.called thepivot element ,is0,sotheprocedure cannot
continue initspresent form .Butoperations (£,)<+(£;)arepermitted ,soasearch ismade
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oftheelements <232anda42forthefirstnonzero element .Since <232^0,theoperation
(£2)**(£3)isperformed toobtain anew matrix.
'1-1 2-1-8'
0 2-1 1 6
0 0-1-1-4
0 0 2 4 12
Since x2isalready eliminated from£3and£4,thecomputations continue with theoperation
(£4+2£3)-(£4),giving
A(4)='1-1 2-1-8‘ 
0 2-1 1 6
0 0-1-1-4
0 0 0 2 4
Thematrix isnow converted back intoalinear system thathasasolution equivalent tothe
solution oftheoriginal system andthebackward substitution isapplied :
X4=t=2'
1-4-(-1M.*3= : =2,
x2=-1
[6— x4— (-1)*3]
2=3,
[~8—(—l)x4—2X3—(—l)x2]- =
Todefine theinitial augmented matrix inMATLAB ,which wewillcall AA,weenter
thematrix row byrow.Aspace isplaced between each entry inarow,andtherows inAA
areseparated byacolon .So,forthematrix inExample 2wehave
AA=[1-1 2-1-8;2-2 3-3-20;1 1 10-2;1-1434]
MATLAB responds with
1-12-1-8
2-23-3-20
1 11 0-2
1-14 3 4
Toperform theoperation (Ej+m£,)— (£y)inMATLAB weusethecommand
A A(j,:)=A A(j,:)+m*A A(i,:)
The notation A A(k,l)refers toentry inthek\hrow and/thcolumn .The useof:in
MATLAB refers tomultiplying anentire roworcolumn .Forexample ,multiplying thefcth
rowbymisdone with m*A A(k,:).Similarly ,multiplying the/thcolumn bymwould be
done with m*A A(:,1).Sothenext command subtracts twice thefirstrowofAAfrom the
second row.
A A(2,:)=A A(2,:)-2*A A(l,:)
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which gives
1-1 2-1-8
0 0 -1-1-4
1 1 1 0-2
1-1 4 3 4
Wethen subtract thefirstrowofAAfrom thethird row,followed bythesubtraction ofthe
firstrowfrom thefourth rowwith
AA(3,:)=AA(3,:)-AA(1,:)
and
AA(4,:)=A A(4,:)— A A(1,:)
This gives
'1-1 2-1-8'
0 0 -1-1-4A A~0 2-1 1 6
0 0 2 412
Thevariable x\hasnow been eliminated from therows corresponding tothesecond ,
third ,andfourth equations .Since ai2iszero,weneed tointerchange rows tomove anonzero
entry toan.Tointerchange rows 2and3,westore row2inatemporary rowvector B,move
row3torow2,andthen more thetemporary rowvector Btorow3.This isdone with
B=A A(2,:)
AA(2,:)=A A(3,:)
AA(3,:)=B
Theresult is
AA=1-1 2-1-8
0 2-1 1 6
0 0-1-1-4
0 0 2 412
The final operation inGaussian elimination forthismatrix istoadd2times thethird row
tothefourth rowwith
AA(4,:)=A A(4,:)+2*AA(3,:)
This produces
1-1 2-1-8
0 2-1 1 6
0 0 -1-1-4
0 0 0 2 4
Toperform thebackward substitution weneed todefine thevector xthatwillcontain the
solution .Weinitialize avector xasthe0vector andwillreplace these entries asweprogress
through thebackward substitution .
x=[000 0]
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Now wereplace the0inthefourth column ofxwith
x(4)=AA(4,5)/AA(4,4)
which gives
Then
x(3)
gives
x(2)
givesx=0 0 0 2
(AA(3,5)-AA(3,4)*x(4))/AA(3,3)
x=0 0 2 2
(AA(2,5)-(AA(2,3)*x(3)+AA(2,4)*x(4) ))/AA(2,2)
x=0 3 2 2
and
x(l)=(AA(1,5)-(AA(1,2)*x(2)+AA(l,3)*x(3)+AA(l,4)*x(4)))/AA(l,1)
gives thefinal solution
x=-7 3 2 2
which corresponds tox\=-7,X2=3,X3=2,andx4=2.
Example 2illustrates what isdone ifoneofthepivot elements iszero.Iftheithpivot
element iszero,theithcolumn ofthematrix issearched from theithrowdownward for
thefirstnonzero entry ,andarowinterchange isperformed toobtain thenew matrix .Then
theprocedure continues asbefore.Ifnononzero entry isfound theprocedure stops ,and
thelinear system does nothave aunique solution .Itmight have nosolution oraninfinite
number ofsolutions .
Operation Counts
The computations intheprogram areperformed using only one nx(n+1)array for
storage .This isdone byreplacing ,ateach step,theprevious value ofa,jbythenewone.In
addition ,themultipliers arestored inthelocations ofa*«known tohave zero values— that
is,when i<nandk=i+1,1+2,. . .,n.Thus ,theoriginal matrix Aisoverwritten bythe
multipliers below themain diagonal andbythenonzero entries ofthefinal reduced matrix
onandabove themain diagonal .WewillseeinSection 6.5thatthese values canbeused to
solve other linear systems involving theoriginal matrix A.
Both theamount oftime required tocomplete thecalculations andthesubsequent
round -offerror depend onthenumber offloating -point arithmetic operations needed to
solve aroutine problem .Ingeneral ,theamount oftime required toperform amultiplication
ordivision onacomputer isapproximately thesame andisconsiderably greater than
thatrequired toperform anaddition orsubtraction .Even though theactual differences
inexecution time depend ontheparticular computing system being used ,thecount of
theadditions /subtractions arekept separate from thecount ofthemultiplications /divisions
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because ofthetime differential .The total number ofarithmetic operations depends onthe
sizen,asfollows :
Multiplications /divisions :
Additions /subtractions :n3.2-
/i3n2
3+2'3*
5n
6~*
Forlarge n,thetotal number ofmultiplications anddivisions isapproximately n?/3,that
is,0(H3),asisthetotal number ofadditions andsubtractions .The amount ofcomputation ,
andthetime required toperform it,increases with ninapproximate proportion ton3/3,as
shown inTable 6.1.
Table 6.1n Multiplications /Divisions Additions /Subtractions
3 17 11
10 430 375
50 44,150 42,875
100 343,300 338,250
E X E R C I S E S E T 6 . 2
l. Obtain asolution bygraphical methods ofthefollowing linear systems ,ifpossible .
a.x\+2x2=3,
X]— X2=0.
c. x,+2X2=3,
2xi+4*2=6.
e. x,+2X2=0,
2x,4-4X2=0.
g.2xi+x2=-1,
4xi+2X2=-2,
xi— 3x2=5.b.xi+2x2=0,
X]- X2=0.
d. xi+2X2=3,
— 2xi-4X2=6.
f.2x,+x2=-1,
x,+x2=2,
x,— 3X2=5.
h.2xi4-X24-X3=1,
2xi+4X2-*3=-1.
2. UseGaussian elimination andtwo-digit rounding arithmetic tosolve thefollowing linear systems .
Donotreorder theequations .(The exact solution toeach system isxi=1,X2=— 1,X3=3.)
a.4x|— x24-x3=8, b.4xj4-x24-2x3=9,
2xi4-5x24-2x3=3, 2xi4-4x2— X3=-5,
xi4-2x24-4x3=11. xi4-X2-3x3=-9.
3. UseGaussian elimination tosolve thefollowing linear systems ,ifpossible ,anddetermine whether
rowinterchanges arenecessary :
a. X|— x24-3x3=2,
3x,-3x2+x3=-1,
*1+*2 =3.
b.2xj— 1.5X24-3x3=1,-x, +2x3=3,
4xi-4.5X24-5x3=1.
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c.2*,=3,
*i+1.5*2 =4.5,
— 3*2+0.5*3 =— 6.6,
2*i 2*24-*3+*4=0.8.
d.*i-\x2+*3 =4,
2*,-*2-*34-*4=5,
*1+-*2 =2,
*1-2*2+*3+*4=5.
*1+*2 +"*4=2,
2*1+*2~*3+*4=la
4*i*22*3+-2*4=0,
3*i*2*3+-2*4=-3.
*1+*2 +*4=2,
2*i+*2-*3+*4=1,-*,+2*2+-3*3-*4=4,
3*1— *2— *3+2l4=— 3.
4. UseMATLAB with long format andGaussian elimination tosolve thefollowing linear systems .
a.j*i+-J*2+^*3=9,
5*1+|*2+J*3=8,
2*1+*2+2*3=8.
b. 3.333*i+15920 *2-10.333 *3=15913 ,
2.222*i+-16.71*2+-9.612*3=28.544 ,
1.5611 *,+5.1791 *2+1.6852 *3=8.4254 .
C.*l+-+j*3+” 1*4=
5*1+|*2+1*3+"§*4=
5*1+-1*2+-5*3+1*4=§.
1*1+-1*2+-1*3+-5*4=
d.2*,+*2-*3+*4-3*5=7,
*1 +-2*3-*4+-*5=2,
-2*2-*3+*4-*5=-5,
3*i+*2-4*3 +5*5=6,
*1-*2-*3-*4+-*5=3.
5. Given thelinear system
2*t— 6a*:=3,
3a*i-*2=| *
a.Find value (s)ofaforwhich thesystem hasnosolutions .
b.Find value (s)ofaforwhich thesystem hasaninfinite number ofsolutions .
c.Assuming aunique solution exists foragiven a,findthesolution .
6. Given thelinear system
*i-*2+-a*3=-2,-*i+-2*2-a*3=3,
a*i+-*2+-*3=2.
a.Find value (s)ofaforwhich thesystem hasnosolutions .
b.Find value (s)ofaforwhich thesystem hasaninfinite number ofsolutions .
c.Assuming aunique solution exists foragiven a,findthesolution .
7. Suppose thatinabiological system there arenspecies ofanimals andmsources offood.Letx}
represent thepopulation ofthejthspecies foreach j=1 represent theavailable daily
Copyright 2012 Cengagc Learn in*.AIRights Reversed May rotbecopied.scanned .ocimplicated ,inwhole orinpar.Doctocjectronie rlghtv.some third pur.ycontent may bevuppreved rrom theeBook and/orcChaptcnM .Editorial review h*>
deemed Cutanysuppressed content does notmaxrUXy alTcct theoverall learning experience .('engage [.camonreserves theright 10remove additional eonteatatanytime ifsubsequent nghts restrictions require It240 C H A P T E R 6 Direct Methods forSolving Linear Systems
supply oftheithfood;andayrepresent theamount oftheithfood consumed onaverage byamember
ofthey'thspecies .Thelinear system
0n*i+012*2+   +0i„ *„ =bu
021*1+022*2H 4*02«*/»=bl.
0m1*1+0m2*2+***+0mn*n=b„ 
represents anequilibrium where there isadaily supply offood toprecisely meet theaverage daily
consumption ofeach species .
a.Let
A=[au)=12 03
1 0 2 2
0 0 1 1
x=(xj)=[1000 ,500,350,4001 ,andb=fa)=[3500 ,2700 ,900].Isthere sufficient food
tosatisfy theaverage daily consumption ?
b.What isthemaximum number ofanimals ofeach species thatcould beindividually added tothe
system with thesupply offood stillmeeting theconsumption ?
c.Ifspecies 1became extinct ,how much ofanindividual increase ofeach oftheremaining species
could besupported ?
d.Ifspecies 2became extinct ,how much ofanindividual increase ofeach oftheremaining species
could besupported ?
8. AFredholm integral equation ofthesecond kind isanequation oftheform
«(*)=/(*)+r*(*.ouMt ,
where aandbandthefunctions /andKaregiven .Toapproximate thefunction uontheinterval
[a,b],apartition x0=a<x\<   <xm.\<xm=bisselected andtheequations
«(*<)=/(*i)+JK(xltr)u(r)dt,foreach i=0 m,
aresolved foru(jto),w(*i), . ,u(*m).The integrals areapproximated using quadrature formulas
based onthenodes xo,...,xm.Inourproblem ,a=0,b=1,f(x)=x2,andK(x,t)=ex~‘.
a.Show thatthelinear system
0(0)=/(0)+*[A:(0,0)U(0)+tf(0,1)II(1)1,
0(1)=/(1)+*(AT(1,0)0(0)+K{1,1)0(1)]
must besolved when theTrapezoidal ruleisused.
b.Setupandsolve thelinear system thatresults when theComposite Trapezoidal ruleisused with
n=4.
c.Repeat (b)using theComposite Simpson 'srule.
6.3 Pivoting Strategies
Ifallthecalculations could bedone using exact arithmetic ,wecould almost endthechapter
with theprevious section .Wenow know how many calculations areneeded toperform
Gaussian elimination onasystem ,andfrom thisweshould beable todetermine whether
Copyright 2012 Cengagc Learn in*.AIRights Reversed Mayr*xbecopied.scanned .ocimplicated ,inwhole orinpar.Doctocjectronie rights.some third pur.ycontent may besupplied ftem theeBook and/orcChaptcnM .Editorial review h*>
deemed Cutanysuppressed content does notmamlaXy alTcct theioera .1learning experience .('engage [.camonreserves theright 10remove additional eonteat atanytime iisuhvcqjcM nghts restrictions require It