Chapter 4 BMCU 2072

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Gaussian Elimination

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!orward Elimination

=

2.279

2.177

8.106

112144

1864

1525

3

2

1

x

x

x

The (oal of forward elimination is to transform the$oeffi$ient matri) into an upper trian(ular matri)

=

735.0

21.96

8.106

7.000

56.18.40

1525

3

2

1

x

x

x

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!orward Elimination* &tep 1

[ ] [ ]56.28.126456.21525 = [ ]

[ ]

[ ]56.18.40

56.28.1264

1864

112144

1864

1525

112144

56.18.40

1525

+ivide Equation 1 ', "- and

multipl, it ', ./0 56.225

64=

&u'tra$t the result from

Equation "

&u'stitute new equation for

Equation "

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!orward Elimination* &tep 1 $ont2

[ ] [ ]76.58.2814476.51525 =[ ]

[ ]

[ ]76.48.160

76.58.28144

112144

76.48.160

56.18.40

1525

112144

56.18.40

1525


multipl, it ', 1//0 76.525

144=


Equation 3


Equation 3

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!orward Elimination* &tep "

76.48.160

56.18.401525

[ ]( ) [ ]46.58.1605.356.18.40 =[ ]

[ ]

[ ]7.000

46.58.160

76.48.160

7.000

56.18.40

1525

+ivide Equation " ', 4/5

and multipl, it ', 41.50

5.38.4

8.16=


Equation 3


Equation 3

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!orward Elimination

A set of nequations and nun%nowns

11313212111 ... bxaxaxaxa nn =++++

22323222121 ... bxaxaxaxa nn =++++

nnnnnnn bxaxaxaxa =++++ ...332211

n612 steps of forward elimination

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!orward Elimination

Step 1!or Equation "0 divide Equation 1 ', and

multipl, ',

)...( 1131321211111

21 bxaxaxaxaa

ann =++++

111

21

111

21

21211

21

121

... ba

axa

a

axa

a

axa

nn

=+++

11a

21a

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!orward Elimination

1

11

211

11

21212

11

21121 ... b

a

axa

a

axa

a

axa nn =+++

1

11

2121

11

212212

11

2122 ... b

a

abxa

a

aaxa

a

aa nnn =

++

'

2

'

22

'

22 ... bxaxa nn =++

22323222121 ... bxaxaxaxa nn =++++&u'tra$t the result from Equation "

47777777777777777777777777777777777777777777777777

or

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!orward Elimination

8epeat this pro$edure for the remainin(equations to redu$e the set of equations as

11313212111 ... bxaxaxaxa nn =++++'

2

'

23

'

232

'

22 ... bxaxaxa nn =+++'

3

'

33

'

332

'

32 ... bxaxaxa nn =+++

''

3

'

32

'

2 ... nnnnnn bxaxaxa =+++

End of Step 1

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Step 28epeat the same pro$edure for the 3rdterm of

Equation 3

!orward Elimination

11313212111 ... bxaxaxaxa nn =++++

'

2

'

23

'

232

'


"

3

"

33

"

33 ... bxaxa nn =++

""

3

"

3 ... nnnnn bxaxa =++

End of Step 2

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!orward Elimination

At the end of n612 !orward Elimination steps0 the s,stem

of equations will loo% li%e

'

2

'

23

'

232

'

22 ... bxaxaxa nn =+++"

3

"

33

"

33 ... bxaxa nn =++

( ) ( )11 = nnnn

nn bxa

11313212111 ... bxaxaxaxa nn =++++

End of Step (n-1)

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9atri) !orm at End of !orward

Elimination

=

)(n-

n

"

'

n

)(n

nn

"n

"

'

n

''

n

b

b

b

b

x

x

x

x

a

aa

aaa

aaaa

1

3

2

1

3

2

1

1

333

22322

1131211

0000

00

0

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[ ] [ ]56.28.126456.21525 = [ ]

[ ]

[ ]56.18.40

56.28.1264

1864

112144

1864

1525

112144

56.18.40

1525


multipl, it ', ./0 56.225

64=


Equation "


Equation "

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[ ] [ ]76.58.2814476.51525 =[ ]

[ ]

[ ]76.48.160

76.58.28144

112144

76.48.160

56.18.40

1525

112144

56.18.40

1525


multipl, it ', 1//0 76.525

144=


Equation 3


Equation 3

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76.48.160

56.18.401525

[ ]( ) [ ]46.58.1605.356.18.40 =[ ]

[ ]

[ ]7.000

46.58.160

76.48.160

7.000

56.18.40

1525

+ivide Equation " ', 4/5

and multipl, it ', 41.50

5.38.4

8.16=


Equation 3


Equation 3

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#a$% &u'stitution

&olve ea$h equation startin( from the last equation

E)ample of a s,stem of 3 equations

=

735.0

21.96

8.106

7.00056.18.40

1525

3

2

1

xx

x

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#a$% &u'stitution &tartin( Eqns

'

2

'

23

'

232

'


"

3

"

3

"

33

... bxaxann

=++

( ) ( )11 = n

nn

n

nn bxa

11313212111 ... bxaxaxaxa nn =++++

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#a$% &u'stitution

&tart with the last equation 'e$ause it has onl, one un%nown

)1(

)1(

=n

nn

n

n

na

bx

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#a$% &u'stitution

( ) ( )

( ) 1,...,1for

1

1

11

=

=+=

nia

xab

xi

ii

n

ij j

i

ij

i

i

i

)1(

)1(

=

n

nn

nnn

a

bx

( ) ( ) ( ) ( )

( ) 1,...,1for...

1

1

,2

1

2,1

1

1,

1

==

+

++

+

nia

xaxaxabxi

ii

n

i

nii

i

iii

i

ii

i

ii

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E)ample* Thermal Coeffi$ient

A trunnion of diameter 1"3.3:

has to 'e $ooled from a room

temperature of 5;

The equation that (ives the

diametri$ $ontra$tion0 +0 of the

trunnion in dr,6i$e>al$ohol

mi)ture 'oilin( temperature is

41;5

slid throu(h the hu' after$ontra$tin(

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The e)pression for the thermal e)pansion $oeffi$ient0

is o'tained usin( re(ression anal,sis and hen$e solvin( the followin(

simultaneous linear equations*

!ind the values of usin( Nave Gauss Elimination

=

567992

10041621

100571

102435751086472110267

10864721102672860

102672860242

4

3

2

1

1085

85

5

.

.

.

a

a

a

...

..

.

2

321 TaTaa ++=

321 and,, aaa

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?ields

( )=

2860

24

12

RowRow

=

567992

1017972

100571

1024351086472110267

109957910851830

102672860243

4

3

2

1

1085

75

5

.

.

.

a

a

a

...

..

.

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#a$% &u'stitution* &olve for a3usin( the third equation

11

9

2

3

2

3

9

1040661

1053494

1037886

10378861053494

=

=

=

.

.

.a

.a.

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#a$% &u'stitution* &olve for a"usin( the se$ond equation

3

3

7

2

5 101797.2)109957.9(108518.3 =+ aa

( )

( ) ( )

9

5

1173

5

3

73

2

1000872

1085183

1040661109957.91017972

1085183109957.91017972

=

=

=

.

.

..

.a.a

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#a$% &u'stitution*&olve for a1usin( the first equation

( ) 435

21 10057110267286024 =++ .a.aa

( )

( ) ( ) ( )

6

11594

3

5

2

4

1

100690524

10406611026710008722860100571

24

102672860100571

=

=

=

.

....

a.a.a

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=

11

9

6

3

2

1

1040661

1000872

1006905

.

.

.

a

a

a


&olution*The solution ve$tor is

The pol,nomial that passes throu(h the three data

points is then*

( ) 2321 TaTaaT ++=

21196 104066110008721006905 T.T.. +=

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Nave Gauss Elimination

@itfalls

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@itfalls of Nave Gauss Elimination

@ossi'le division ', Bero

ar(e round6off errors

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955

11326

3710

321

321

32

=+

=++=

xxx

xxx

xx

@itfallD1 +ivision ', Bero

=

9

11

3

515

326

7100

3

2

1

x

x

x

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s division ', Bero an issue hereF

955

14356

1571012

321

321

321

=+

=++

=+

xxx

xxx

xxx

=

9

14

15

515

356

71012

3

2

1

x

x

x

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s division ', Bero an issue hereF

?E&

28524

14356

1571012

321

321

321

=+

=++

=+

xxx

xxx

xxx

=

28

14

15

5124

356

71012

3

2

1

x

x

x

=

2

5.6

15

192112

5.600

71012

3

2

1

x

x

x

+ivision ', Bero is a possi'ilit, at an, step

of forward elimination

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@itfallD" ar(e 8ound6off Errors

=

9

751.145

315

7249.23101520

3

2

1

x

x

x

E)a$t &olution

=

1

1

1

3

2

1

x

x

x

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=

9

751.145

315

7249.23101520

3

2

1

x

x

x

&olve it on a $omputer usin( .si(nifi$ant di(its with $hoppin(

=

999995.0

05.1

9625.0

3

2

1

x

x

x

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=

9

751.145

315

7249.23101520

3

2

1

x

x

x

&olve it on a $omputer usin( -si(nifi$ant di(its with $hoppin(

=

99995.0

5.1

625.0

3

2

1

x

x

x

s there a wa, to redu$e the round off errorF

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Avoidin( @itfalls

n$rease the num'er of si(nifi$ant di(its

+e$reases round6off error

+oes not avoid division ', Bero

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Avoidin( @itfalls

Gaussian Elimination with @artial @ivotin(

Avoids division ', Bero

8edu$es round off error

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Gauss Elimination with@artial @ivotin(

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hat is +ifferent A'out @artial

@ivotin(F

pka

At the 'e(innin( of the kthstep of forward elimination0

find the ma)imum of

nkkkkk aaa .......,,........., ,1+

f the ma)imum of the values is

in thep throw0 ,npk then swit$h rowspand k

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9atri) !orm at #e(innin( of "nd

&tep of !orward Elimination

=

'

'3

2

1

3

2

1

''

4

'

32

'3

'3332

22322

1131211

0

0

0

n

'

nnnnn

'

n

n'

'

n

''

n

b

b

b

b

x

x

x

x

aaaa

aaa

aaa

aaaa

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E)ample "ndstep of !E2

=

3

9

8

6

5

431112170

862390

1111240

21670

67.31.5146

5

4

3

2

1

x

x

x

x

x

hi$h two rows would ,ou swit$hF

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E)ample "ndstep of !E2

=

6

9

8

3

5

21670

862390

1111240

431112170

67.31.5146

5

4

3

2

1

x

x

x

x

x

&wit$hed 8ows

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with @artial @ivotin(

A method to solve simultaneous linear

equations of the form [A][X]=[C]

Two steps

1 !orward Elimination

" #a$% &u'stitution

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!orward Elimination

&ame as nave Gauss elimination method

e)$ept that we swit$h rows 'efore eachof

the n612 steps of forward elimination

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E)ample* 9atri) !orm at #e(innin(

of "nd

&tep of !orward Elimination

=

'

'

3

2

1

3

2

1

''4

'32

'

3

'

3332

22322

1131211

0

00

n

'

nnnnn'n

n

'

'

n

''

n

b

bb

b

x

xx

x

aaaa

aaaaaa

aaaa

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9atri) !orm at End of !orward

Elimination

=

)(n-n

"

'

n)(n

nn

"

n

"

'

n

''

n

b

bb

b

x

xx

x

a

aaaaa

aaaa

1

3

2

1

3

2

1

1

333

22322

1131211

0000

000

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#a$% &u'stitution &tartin( Eqns

'

2

'

23

'

232

'


"

3

"

3

"

33 ... bxaxa nn =++

( ) ( )11 = n

nn

n

nn bxa

11313212111 ... bxaxaxaxa nn =++++

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#a$% &u'stitution

( ) ( )

( ) 1,...,1for11

11

=

=

+=

nia

xab

xi

ii

n

ijj

iij

ii

i

)1(

)1(

=

n

nn

nn

na

bx

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Gauss Elimination with

@artial @ivotin(E)ample 1

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E)ample "

=

2279

2177

8106

112144

1864

1525

3

2

1

.

.

.

a

a

a

&olve the followin( set of equations

', Gaussian elimination with partialpivotin(

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E)ample " Cont

=

2279

2177

8106

112144

1864

1525

3

2

1

.

.

.

a

a

a

1 !orward Elimination

" #a$% &u'stitution

2.279112144

2.1771864

8.1061525

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!orward Elimination

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Num'er of &teps of !orward

Elimination

Num'er of steps of forward elimination is

n12=312="

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!orward Elimination* &tep 1 E)amine a'solute values of first $olumn0 first row

and 'elow

144,64,25

ar(est a'solute value is 1// and e)ists in row 3

&wit$h row 1 and row 3

8.1061525

2.1771864

2.279112144

2.279112144

2.1771864

8.1061525

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[ ] [ ]1.1244444.0333.599.634444.02.279112144

=

8.1061525

2.1771864

2.279112144

[ ]

[ ]

[ ]10.53.55560667.20

124.10.44445.33363.99

177.21864

8.1061525

10.535556.0667.20

2.279112144

+ivide Equation 1 ', 1// and

multipl, it ', ./0 4444.0144

64=


Equation "


Equation "

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[ ] [ ]47.481736.0083.200.251736.0279.2112144 =

[ ]

[ ]

[ ]33.588264.0917.20

48.470.17362.08325

106.81525

8.1061525

10.535556.0667.20

2.279112144

33.588264.0917.20

10.535556.0667.20

2.279112144

+ivide Equation 1 ', 1// and

multipl, it ', "-0 1736.0144

25=


Equation 3


Equation 3

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!orward Elimination* &tep " E)amine a'solute values of se$ond $olumn0 se$ond row

and 'elow

2.917,667.2

ar(est a'solute value is "H1I and e)ists in row 3

&wit$h row " and row 3

10.535556.0667.20

33.588264.0917.20

2.279112144

33.588264.0917.20

10.535556.0667.20

2.279112144

! d Eli i ti &t " t 2

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!orward Elimination* &tep " $ont2

[ ] [ ]33.537556.0667.209143.058.330.82642.9170 =

10.535556.0667.20

33.588264.0917.20

2.279112144

[ ]

[ ]

[ ]23.02.000

53.330.75562.6670

53.100.55562.6670

23.02.000

33.588264.0917.20

2.279112144

+ivide Equation " ', "H1I and

multipl, it ', "..I0.9143.0

917.2

667.2=


Equation 3


Equation 3

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#a$% &u'stitution

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#a$% &u'stitution

1.15

2.0

23.0

23.02.0

3

3

=

=

=

a

a

&olvin( for a3

=

2303358

2279

2000

8264091720

112144

23.02.000

33.588264.0917.20

2.279112144

3

2

1

.

.

.

a

a

a

.

..

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#a$% &u'stitution $ont2

&olvin( for a2

6719.

917.2

15.18264.033.58917.2

8264.033.58

33.588264.0917.2

32

32

=

=

=

=+a

a

aa

=

23033582279

2000

8264091720112144

3

2

1

.

..

a

aa

.

..

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#a$% &u'stitution $ont2

&olvin( for a1

2917.0

144

15.167.19122.279144

122.279

2.27912144

321

321

=

=

=

=++

aaa

aaa

=

230

33582279

2000

8264091720112144

3

2

1

.

..

a

aa

.

..

Gaussian Elimination with @artial

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@ivotin( &olution

=

2279

2177

8106

112144

1864

1525

3

2

1

.

.

.

a

a

a

=

15.1

67.19

2917.0

3

2

1

a

a

a

23.02.000

33.588264.0917.20

2.279112144

Gaussian Elimination without

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=

2.279

2.1778.106

112144

18641525

3

2

1

x

xx

=

735.0

21.96

8.106

7.000

56.18.40

1525

3

2

1

x

x

x

@artial @ivotin( &olution

=

05.1

7025.19

1705.8

3

2

1

a

a

a

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Gaussian Elimination withPartial Pivoting Solution


without Partial Pivoting

Solution

=

05.1

7025.19

1705.8

3

2

1

a

a

a

=

15.1

67.19

2917.0

3

2

1

a

a

a

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Gauss Elimination with

@artial @ivotin(E)ample "

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@artial @ivotin(* E)ampleConsider the s,stem of equations

655

901.36099.23

7710

321

321

21

=+

=++

=

xxx

xxx

xx

n matri) form

515

6099.23

0710

3

2

1

x

x

x

6

901.3

7

=

&olve usin( Gaussian Elimination with @artial @ivotin( usin( five

si(nifi$ant di(its with $hoppin(

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@artial @ivotin(* E)ample!orward Elimination* &tep 1

E)aminin( the values of the first $olumn

J1;J0 J63J0 and J-J or 1;0 30 and -

The lar(est a'solute value is 1;0 whi$h means0 to

follow the rules of @artial @ivotin(0 we swit$hrow1 with row1

=

6

901.3

7

515

6099.23

0710

3

2

1

x

x

x

=

5.2

001.6

7

55.20

6001.00

0710

3

2

1

x

x

x

@erformin( !orward Elimination

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@artial @ivotin(* E)ample


E)aminin( the values of the first $olumn

J6;;;1J and J"-J or ;;;;1 and "-

The lar(est a'solute value is "-0 so row " isswit$hed with row 3

=

5.2

001.67

55.20

6001.000710

3

2

1

x

xx

=

001.6

5.27

6001.00

55.200710

3

2

1

x

xx

@erformin( the row swap

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@erformin( the !orward Elimination results in*

=

002.6

5.2

7

002.600

55.20

0710

3

2

1

x

x

x

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#a$% &u'stitution

&olvin( the equations throu(h 'a$% su'stitution

1

002.6

002.63 ==x

15.2

55.2 32 =

=

xx

010

077 321 =

+=

xxx

=

002.6

5.2

7

002.600

55.20

0710

3

2

1

x

x

x

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[ ]

=

=

1

1

0

3

2

1

x

x

x

X exact[ ]

=

=

1

1

0

3

2

1

x

x

x

X calculated

Compare the $al$ulated and e)a$t solution

The fa$t that the, are equal is $oin$iden$e0 'ut it

does illustrate the advanta(e of @artial @ivotin(

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K +e$omposition

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K +e$omposition

K +e$omposition is another method to solve a set of

simultaneous linear equations

hi$h is 'etter0 Gauss Elimination or K +e$ompositionF

To answer this0 a $loser loo% at K de$omposition isneeded

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9ethod

!or most non6sin(ular matri) [A] that one $ould $ondu$t Nave Gauss

Elimination forward elimination steps0 one $an alwa,s write it as

[A] = [L][U]where

[L] = lower trian(ular matri)

[U] = upper trian(ular matri)

K +e$omposition

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Low does K +e$omposition wor%F

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K +e$omposition

Low $an this 'e usedF

Given [A][X] = [C]

1 +e$ompose [A] into [L] and[U]

" &olve [L][] = [C] for []

3 &olve [U][X] = [] for [X]

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9ethod* [A] +e$ompose to [] and [K]

[ ] [ ][ ]

==

33

2322

131211

3231

21

00

0

1

01

001

u

uu

uuu

ULA

[U] is the same as the $oeffi$ient matri) at the end of the forward

elimination step

[L] is o'tained usin( the multipliersthat were used in the forwardelimination pro$ess

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!indin( the [U] matri)Ksin( the !orward Elimination @ro$edure of Gauss Elimination

112144

1864

1525

( )

112144

56.18.40

1525

56.212;56.225

64

== RowRow

( )

76.48.160

56.18.40

1525

76.513;76.525

144

== RowRow

&tep 1*

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!indin( the [K] 9atri)

&tep "*

76.48.160

56.18.40

1525

( )

7.000

56.18.40

1525

5.323;5.38.4

8.16

==

RowRow

[ ]

=

7.000

56.18.40

1525

U

9atri) after &tep 1*

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!indin( the [L] matri)

Ksin( the multipliers used durin( the !orward Elimination @ro$edure

1

01

001

3231

21

56.225

64

11

2121 ===

a

a

76.525

144

11

31

31 ===a

a

!rom the first step

of forward

elimination 1121441864

1525

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!indin( the [] 9atri)

[ ]

= 15.376.50156.2

001

L

!rom the se$ond

step of forward

elimination

76.48.160

56.18.40

1525

5.38.4

8.16

22

3232 =

==

a

a

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+oes [][K] = [A]F

[ ] [ ] =

=7.000

56.18.40

1525

15.376.5

0156.2

001

UL F

E l Th l C ffi i t

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temperature of 5;


diametri$ $ontra$tion + of

the trunnion in dr,6

i$e>al$ohol 'oilin(

temperature is 41;5

throu(h the hu' after

$ontra$tin(

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The e)pression for the thermal e)pansion $oeffi$ient0 a ! a"# a

$T # a

%T$



!ind the values of a"& a

$&and a

%usin( K +e$omposition

=

56799.2

1004162.1

10057.1

1024357.51086472.11026.7

1086472.11026.72860

1026.72860242

4

3

2

1

1085

85

5

a

a

a

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Kse !orward Elimination to find the [K] matri)

1085

85

5

1024357.51086472.11026.7

1086472.11026.72860

1026.7286024

( )

102436.5108647.11026.7

109957.9108518.30

1026.7286024

17.11912;17.11924

2860

1085

75

5

==

RowRow

( )

100474.3109957.90

109957.9108518.30

1026.7286024

3025013;3025024

10267

107

75

5

5

==

RowRow.

&tep 1

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( )

105349.400

109957.9108518.30

1026.7286024

50.25923;50.259108518.3

109957.9

9

75

5

5

7

==

RowRow

[ ]

=9

75

5

105349.400

109957.9108518.30

1026.7286024

U

100474.3109957.90

109957.9108518.30

1026.7286024

107

75

5

This is the matri)

after the 1ststep

&tep "

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Kse the multipliers from !orward Elimination

1

01

001

3231

21

1711924

2860

11

2121 .

a

a=

==

3025024

10267 5

11

3131 =

== .

a

a

!rom the first step of forward elimination

1085

85

5

1024357.51086472.11026.71086472.11026.72860

1026.7286024

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[ ]

=

150.25930250

0117.119

001

L

!rom the se$ond step of forward elimination

502591085183109957.9

5

7

22

3232 .

.aa ===

96

65

5

104742.3010957.990

10957.991085183.30

1026.7286024

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+oes [][K] = [A]F

[ ] [ ] ?

105323.400

109957.9108518.30

1026.7286024

150.25930250

0117.119

001

9

75

5

=

=UL

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&et [][M] = [C]

=

567992

10041621

100571

15025930250

0117119

0012

4

3

2

1

.

.

.

.

.

( ) 56799250259302501004162117119

10057.1

31

221

4

1

.'.'

.''.

'

=++=+

=

&olve for [M]

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4

1 10057.1 ='

( )

( )

00217970

1005711711910041621

1711910041621

42

1

2

2

.

...

'..'

==

=

( )

( )

0637880

002179705025910057130250567992

5025930250567992

4

213

.

....

..

=

=

=

[ ]

=

=

0637880

00217970

100571 4

3

2

1

.

.

.

&olve for [M]

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&et [K][A] = [M]

=

0637880

00217970

100571

105349400

109957.910851830

10267286024 4

3

2

1

9

75

5

.

.

.

a

a

a

.

.

.

( )

( )06378801053484

0021797.0109957.9108518.3

10057.11026.7286024

3

9

3

7

2

5

4

3

5

21

.a.

aa

aaa

=

=+

=++

&olve for A

The 3 equations 'e$ome

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11

93

1040661

1053494

0637880

=

=

.

.

.a

( )

( ) ( )

9

5

117

5

3

7

2

1000872

1085183

1040661109957.900217970

1085183

109957.900217970

=

=

=

.

.

..

.

a.a

&olve for A

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( )

( ) ( )

6

11594

3

5

2

4

1

1006905

24

10406611026710008722860100571

24

102672860100571

=

=

=

.

....

a.a.a

=

11

9

6

3

2

1

1040661

1000872

1006905

.

.

.

a

a

a

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The solution

ve$tor is

The pol,nomial that passes throu(h the three

data points is then*

=

11

9

6

3

2

1

1040661

1000872

1006905

.

.

.

a

a

a

( )21196

2321

104066110008721006905 T.T..

TaTaaT +=

++=

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!indin( the inverse of a square matri)

The inverse [#] of a square matri) [A] is defined as

[A][)] = [*] = [)][A]

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!indin( the inverse of a square matri)

Low $an K +e$omposition 'e used to find the inverseF

Assume the first $olumn of [B] to 'e [b11 b12 bn1]T

Ksin( this and the definition of matri) multipli$ation

!irst $olumn of [B] &e$ond $olumn of [B]

[ ]

=

0

0

1

1

21

11

nb

b

b

A [ ]

=

0

1

0

2

22

12

nb

b

b

A

The remainin( $olumns in [B] $an 'e found in the same manner

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E)ample* nverse of a 9atri)

!ind the inverse of a square matri) [A]

[ ]

=

112144

1864

1525

A

[ ] [ ] [ ]

==

7000

561840

1525

153765

01562

001

.

..

..

.ULA

Ksin( the de$omposition pro$edure0 the [L] and [U] matri$es are found to 'e

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&olvin( for the ea$h $olumn of [B] requires two steps

12&olve [L] [Z] = [C] for [Z]

"2&olve [U] [X] = [Z] for [X]

&tep 1* [ ][ ] [ ]

=

=

0

01

15.376.5

0156.2001

3

2

1

'

''

CL

This (enerates the equations*

05.376.5

056.21

321

21

1

=++

=+ =

'''

'''

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&olvin( for [Z]

( )

( ) ( )23

5625317650

537650

562

15620

5620

1

213

12

1

....

'.'.'

.

.

'.'

'

= =

====

=

[ ]

=

=23

562

1

3

2

1

.

.

'

'

'

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&olvin( [U][X] = [Z] for [X]

=

3.2

2.56

1

7.000

56.18.40

1525

31

21

11

b

b

b

2.37.0

56.256.18.4

1525

31

3121

312111

=

=

=++

b

bb

bbb

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Ksin( #a$%ward &u'stitution

( )

( )04762.0

25

571.49524.05125

51

9524.08.4

571.4560.156.2

8.4

560.156.2

571.47.0

2.3

312111

31

21

31

=

=

=

=

+=

+

=

==

bbb

b

b

b &o the first $olumn ofthe inverse of [A] is*

=

571.4

9524.0

04762.0

31

21

11

b

b

b

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8epeatin( for the se$ond and third $olumns of the inverse

&e$ond Column Third Column

=

0

1

0

112144

1864

1525

32

22

12

b

b

b

=

000.5

417.1

08333.0

32

22

12

b

b

b

=

1

0

0

112144

1864

1525

33

23

13

b

b

b

=

429.1

4643.0

03571.0

33

23

13

b

b

b

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The inverse of [A] is

[ ]

=

429.1000.5571.44643.0417.19524.0

03571.008333.004762.01

A

To $he$% ,our wor% do the followin( operation

[A][A]61= [*] = [A]61[A]

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Gauss6&eidel 9ethod

Gauss6&eidel 9ethod

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Gauss6&eidel 9ethod

An iterativemethod

#asi$ @ro$edure*

6Al(e'rai$all, solve ea$h linear equation for )i

6Assume an initial (uess solution arra,

6&olve for ea$h )iand repeat

6Kse a'solute relative appro)imate error after ea$h iteration

to $he$% if error is within a pre6spe$ified toleran$e

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Gauss6&eidel 9ethod

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Gauss6&eidel 9ethod

Al(orithmA set of nequations and nun%nowns*

11313212111 ... bxaxaxaxa nn =++++

2323222121 ... bxaxaxaxa n$n =++++

nnnnnnn bxaxaxaxa =++++ ...332211

f*the dia(onal elements arenon6Bero

8ewriteea$h equation solvin(for the $orrespondin( un%nown

e)*

!irst equation0 solve for )1

&e$ond equation0 solve for )"

Gauss6&eidel 9ethod

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Gauss6&eidel 9ethod

Al(orithm8ewritin( ea$h equation

11

13132121

1a

xaxaxacx nn

=

nn

nnnnnn

n

nn

nnnnnnnnn

n

nn

a

xaxaxacx

a

xaxaxaxacx

a

xaxaxacx

11,2211

1,1

,122,122,111,11

1

22

232312122

=

=

=

!rom Equation 1

!rom equation "

!rom equation n61

!rom equation n

Gauss6&eidel 9ethod

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Gauss &eidel 9ethod

Al(orithmGeneral !orm of ea$h equation

11

1

1

11

1a

xac

x

n

j

j

jj

=

=

22

21

22

2a

xac

x

j

n

jj

j=

=

1,1

1

1

,11

1

=

=nn

n

nj

j

jjnn

na

xac

x

nn

n

njj

jnjn

na

xac

x

=

=1

Gauss6&eidel 9ethod

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Gauss &eidel 9ethod

Al(orithmGeneral !orm for an, row iO

.,,2,1,1

nia

xac

xii

n

ijj

jiji

i =

==

Low or where $an this equation 'e usedF

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8/13/2019 Chapter 4 BMCU 2072

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temperature of 5;


diametri$ $ontra$tion + of

the trunnion in dr,6i$e>al$ohol

'oilin( temperature is 41;5

is (iven ',*

=108

80

)(363.12 dTTD

Figure 1 Trunnion to 'e slid

throu(h the hu' after$ontra$tin(


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118/140


The e)pression for the thermal e)pansion $oeffi$ient0



!ind the values of a"& a

$&and a

%usin( Gauss6&eidel 9ethod

=

56799.2

1004162.1

10057.1

1024357.51086472.11026.7

1086472.11026.72860

1026.72860242

4

3

2

1

1085

85

5

a

a

a

2321 TaTaa ++=


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The s,stem of equations is*

nitial Guess*

Assume an initial (uess of

=

56799.2

1004162.1

10057.1

1024357.51086472.11026.7

1086472.11026.72860

1026.72860242

4

3

2

1

1085

85

5

a

a

a

=

0

0

0

3

2

1

a

a

a


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8ewritin( ea$h equationteration 1

=

56799.2

1004162.1

10057.1

1024357.51086472.11026.7

1086472.11026.72860

1026.72860242

4

3

2

1

1085

85

5

a

a

a

( ) 654

1 104042424

01026702860100571

=

= ...

a

( ) ( ) 95862

2 100024310267

0108647211040424286010041621

=

= ..

...a

( ) 1210

9865

3 103269110243575

100024310864721104042410267567992

=

= .

.

.....a


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!indin( the a'solute relative appro)imate error

%100100103269.1

0103269.1

%100100100024.3

0100024.3

%100100104042.4

0104042.4

100

12

12

3

9

9

2

6

6

1

=

=

=

=

= =

=

a

a

a

new

i

old

i

new

i

ia x

xx At the end of the first iteration

The ma)imum a'solute

relative appro)imate error is

1;;P

=

12

9

6

3

2

1

1032691

1000243

1040424

.

.

.

a

a

a


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teration "

Ksin( from teration 1 the values of aiare found*

=

12

9

6

3

2

1

1032691

1000243

1040424

.

.

.

a

a

a

( ) ( )

6

12594

1

1080214

24

10326911026710002432860100571

=

=

.

....a

( ) ( ) ( )

9

5

12862

2

102291410267

1032691108647211080214286010041621

=

=

..

....a

( )

12

10

9865

3

1047382

10243575

102291410864721108021410267567992

=

=

.

.

.....a


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!indin( the a'solute relative appro)imate error

At the end of the se$ond iteration The ma)imum a'soluterelative appro)imate error is

/.3.;P

%360.46100104738.2

)103269.1(104738.2

%007.291001022911.4

100024.310221.4

%2864.8100108021.4

104042.4108021.4

12

1212

3

9

99

2

6

66

1

=

=

=

=

=

=

a

a

a

=

12

9

6

3

2

1

1047382

1022914

1080214

.

.

.

a

a

a


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Itrat!on a" a$ a%

1

"

3

/

-

.

//;/"Q1;4.

/5;"1Q1;4.

/H53;Q1;4.

-;.3.Q1;4.

-;H5;Q1;4.

-111"Q1;4.

100

8.2864

3.6300

1.5918

0.6749

0.2593

3;;"/Q1;4H

/""H1Q1;4H

/./I1Q1;4H

/I;"3Q1;4H

/.;33Q1;4H

///1HQ1;4H

100

29.0073

8.9946

1.1922

2.1696

3.6330

413".HQ1;41"

4"/I35Q1;41"

43/H1IQ1;41"

4//;53Q1;41"

4-"3HHQ1;41"

4-HHI"Q1;41"

100

46.3605

29.1527

20.7922

15.8702

12.6290


8epeatin( more iterations0 the followin( values are o'tained

%1a

%2a

%3a

R Noti$e S After si) iterations0 the a'solute relative

appro)imate errors are de$reasin(0 'ut are still hi(h


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Itrat!on a" a$ a%

I-I.

-;.H"Q1;4.

-;.H1Q1;4.""--HQ1;4/

";.3;Q1;4/";13HQ1;4H

";13-Q1;4H;;"/"5;;"""1

41/;/HQ1;411

41/;-1Q1;411;;11"-;;1;"H



%1a

%2a

%3a

The value of $losel, approa$hes the true value of

=

11

9

6

3

2

1

1040511

1001352

1006915

.

.

.

a

a

a

=

11

9

6

3

2

1

1040661

1000872

1006905

.

.

.

a

a

a

from Gauss elimination2


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a p e e a Coe $ e t

The pol,nomial that passes throu(h the three data points is then

( ) 2321 TaTaaT ++=

21196 104051.11000135.21006915.5 TT +=

Gauss6&eidel 9ethod* @itfall

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Even thou(h done $orre$tl,0 the answer ma, not $onver(in( to the

$orre$t answer

This is a pitfall of the Gauss6&iedel method* not all s,stems of

equations will $onver(e

s there a fi)F

ne $lass of s,stem of equations alwa,s $onver(es* ne with a diagonally

dominant$oeffi$ient matri)

+ia(onall, dominant* [A] in [A] [X] = [C] is dia(onall, dominant if*

=

n

jj

ijaa

!1

!! =

>n

ijj

ijii aa1

for all iO and for at least one iO

G & id l 9 th d @itf ll

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Gauss6&eidel 9ethod* @itfall

[ ]

=

116123

14345

3481.52

+ia(onall, dominant* The $oeffi$ient on the dia(onal must 'e at leastequal to the sum of the other $oeffi$ients in that row and at least one row

with a dia(onal $oeffi$ient (reater than the sum of the other $oeffi$ients

in that row

=

1293496

55323

5634124

]#[

hi$h $oeffi$ient matri) is dia(onall, dominantF

9ost ph,si$al s,stems do result in simultaneous linear equations that

have dia(onall, dominant $oeffi$ient matri$es

Gauss6&eidel 9ethod* E)ample "

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p

Given the s,stem of equations

15312 321 x-xx =+

2835 321 xxx =++

761373 321 =++ xxx

=

1

0

1

3

2

1

x

x

x

ith an initial (uess of

The $oeffi$ient matri) is*

[ ]

=

1373

351

5312

A

ill the solution $onver(e usin( the

Gauss6&iedel methodF


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p

[ ]

=

1373

351

5312

A

Che$%in( if the $oeffi$ient matri) is dia(onall, dominant

43155 232122 =+=+== aaa

10731313 323133 =+=+== aaa

8531212 131211 =+=+== aaa

The inequalities are all true and at least one row is stritly(reater than*

Therefore* The solution should $onver(e usin( the Gauss6&iedel 9ethod


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p

=

76

28

1

1373

351

5312

%

$

a

a

a

8ewritin( ea$h equation

12

531 321

xxx

+=

5

328 312

xx

x

=

13

7376 213

xxx

=


=

1

0

1

3

2

1

x

x

x

( ) ( )50000.0

12

150311 =

+=x

( ) ( )9000.4

5

135.028

2

=

=x

( ) ( )0923.3

13

9000.4750000.03763 =

=x

Ga ss &eidel 9ethod E ample "

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The a'solute relative appro)imate error

%00.10010050000.0

0000.150000.01

=

=a

%00.1001009000.4

09000.42a

==

%662.671000923.3

0000.10923.3

3a

=

=

The ma)imum a'solute relative error after the first iteration is 1;;P


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p

=

8118.3

7153.3

14679.0

3

2

1

x

x

x

After teration D1

( ) ( )14679.0

12

0923.359000.4311 =

+=x

( ) ( ) 7153.35

0923.3314679.0282 ==x

( ) ( )8118.3

13

900.4714679.03763 =

=x

&u'stitutin( the ) values into the

equationsAfter teration D"

=

0923.3

9000.4

5000.0

3

2

1

x

x

x


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p

teration D" a'solute relative appro)imate error

%61.24010014679.0

50000.014679.01a

=

=

%889.311007153.3

9000.47153.32a ==

%874.181008118.3

0923.38118.33a

=

=

The ma)imum a'solute relative error after the first iteration is "/;.1P

This is mu$h lar(er than the ma)imum a'solute relative error o'tained in

iteration D1 s this a pro'lemF

Gauss &eidel 9ethod* E)ample "

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Itrat!on a1 a2 a3

12

3456

0.500000.14679

0.742750.946750.991770.99919

100.00240.61

80.23621.5464.53910.74307

4.90003.7153

3.16443.02813.00343.0001

100.0031.889

17.4084.49960.824990.10856

3.09233.8118

3.97083.99714.00014.0001

67.66218.876

4.00420.657720.0743830.00101



%1a

%2a

%3a

=

4

31

3

2

1

x

xx

=

0001.4

0001.399919.0

3

2

1

x

xx

The solution o'tained is $lose to the e)a$t solution of

Gauss &eidel 9ethod* E)ample 3

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Gauss6&eidel 9ethod* E)ample 3

Given the s,stem of equations

761373 321 =++ xxx

2835 321 =++ xxx

15312 321 =+ xxx


=

1

0

1

3

2

1

x

x

x

8ewritin( the equations

3

13776 321

xx

x

=

5

328 312

xxx

=

5

3121 213

=

xxx

Gauss6&eidel 9ethod* E)ample 3

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Itrat!on a1 A2 a3

12

3456

21.000$196.15

$1995.0$20149

2.0364105

$2.0579105

95.238110.71

109.83109.90109.89109.89

0.8000014.421

$116.021204.6$12140

1.2272105

100.0094.453

112.43109.63109.92109.89

50.680$462.30

4718.1$47636

4.8144105

$4.8653106

98.027110.96

109.80109.90109.89109.89

Condu$tin( si) iterations0 the followin( values are o'tained

%1a

%2a

%3a

The values are not $onver(in(

+oes this mean that the Gauss6&eidel method $annot 'e usedF

Gauss6&eidel 9ethod

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The Gauss6&eidel 9ethod $an still 'e used

The $oeffi$ient matri) is not

dia(onall, dominant [ ]

=

5312

351

1373

A

#ut this is the same set of

equations used in e)ample D"0

whi$h did $onver(e [ ]

=

1373

351

5312

A

f a s,stem of linear equations is not dia(onall, dominant0 $he$% to see if

rearran(in( the equations $an form a dia(onall, dominant matri)

Gauss6&eidel 9ethod

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Not ever, s,stem of equations $an 'e rearran(ed to have a

dia(onall, dominant $oeffi$ient matri)

'serve the set of equations

3321 =++ xxx9432 321 =++ xxx

97 321 =++ xxx

hi$h equations2 prevents this set of equation from havin( a

dia(onall, dominant $oeffi$ient matri)F

Gauss6&eidel 9ethod

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&ummar,

6Advanta(es of the Gauss6&eidel 9ethod

6Al(orithm for the Gauss6&eidel 9ethod

6@itfalls of the Gauss6&eidel 9ethod

Chapter 4 BMCU 2072

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Transcript of Chapter 4 BMCU 2072