Persamaan Linier - Eliminasi Gauss

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Persamaan Linier Persamaan Linier Eliminasi GaussEliminasi Gauss

Matrik sederhanaMatrik sederhana

Untuk persamaan linier yang sederhana, maka metode : Graphical Cramer's rule Elimination

nnnn22n11n

2nn2222121

1nn1212111

bxaxaxa

bxaxaxa

b xaxaxa

Graphical MethodGraphical Method

12

12

21

21

x3x3x2x

rearrange 3xx

3xx2

2x1 – x2 = 3x1 + x2 = 3

One solution


2x1 – x2 = 3

2x1 – x2 = – 1

No solution


6x1 – 3x2 = 92x1 – x2 = 3

Infinite many solution


2x1 – x2 = 3

2.1x1 – x2 = 3

Ill conditioned

Cramer’s RuleCramer’s RuleCara menghitung determinan D2 x 2 matrix

3 x 3 matrix

211222112221

1211 aaaaaa

aaD

3231

222113

3331

232112

3332

232211

333231

232221

131211

aa

aaa

aa

aaa

aa

aaa

aaa

aaa

aaa

D

Cramer’s RuleCramer’s RuleTo find xk for the following system

Replace kth column of as with bs (i.e., aik bi ) nnnn22n11n

2nn2222121

1nn1212111

bxa...xaxa

bxa...xaxabxa...xaxa

)

)(

ijk D(a

matrix newDx

ExampleExample3 x 3 matrix

333231

232221

131211

aaa

aaa

aaa

D

33231

22221

11211

33

33331

23221

13111

22

33323

23222

13121

11

baa

baa

baa

D

1

D

Dx

aba

aba

aba

D

1

D

Dx

aab

aab

aab

D

1

D

Dx

Ill-Conditioned SystemIll-Conditioned System

What happen if the determinant D is very small or zero?

Divided by zero (linearly dependent system)Divided by a small number: Round-off errorLoss of significant digits

0AdetD

Eliminate x2

Subtract to get

2222121

1212111

bxaxa

bxaxa

2122221212112

1222122211122

baxaaxaa

baxaaxaa

aaaa

babax

aaaa

babax

babaxaaxaa

21122211

1212112

21121122

2121221

2121221211211122

Elimination MethodElimination Method

Not very practical for large number (> 4) of equations

Gauss EliminationGauss Elimination

Manipulate equations to eliminate one of the unknownsDevelop algorithm to do this repeatedlyThe goal is to set up upper triangular matrix

Back substitution to find solution (root)

nn

n333

n22322

n1131211

a

aa

aaa

aaaa

U

Basic Gauss EliminationBasic Gauss Elimination

Direct Method (no iteration required)Forward eliminationColumn-by-column elimination of the

below-diagonal elementsReduce to upper triangular matrixBack-substitution

Naive Gauss EliminationNaive Gauss Elimination

Begin with

Multiply the first equation by a21 / a11 and subtract from

second equation

nnnn22n11n

2nn2222121

1nn1212111

bxa...xaxa

bxa...xaxa

bxa...xaxa

nnnn22n11n

111

212nn1

11

21n2212

11

2122111

11

2121

1nn212111

bxa...xaxa

ba

abxa

a

aa...xa

a

aaxa

a

aa

bxa...xaxa


Reduce to

Repeat the forward elimination to get

nnnn22n11n

2nn2222

1nn1212111

bxa...xaxa

bxa...xa

bxa...xaxa

nnnn22n

2nn2222

1nn1212111

bxa...xa

bxa...xa

bxa...xaxa


First equation is pivot equationa11 is pivot elementNow multiply second equation by a'32 /a'22

and subtract from third equation

222

323nn2

22

32n3323

22

3233

2nn2323222

1nn1313212111

ba

abxa

a

aaxa

a

aa

bxaxaxa

bxaxaxaxa

Gauss EliminationGauss EliminationRepeat the

elimination of

ai2 and get

Continue and get nnnn33n

3nn3333

2nn2323222

1nn1313212111

bxaxa

bxaxa

bxaxaxa

bxaxaxaxa

)()( 1nnn

1nnn

3nn3333

2nn2323222

1nn1313212111

bxa

bxaxa

bxaxaxa

bxaxaxaxa

Now we can perform back substitution to get {x}By simple division

Substitute this into (n-1)th equation

Solve for xn-1

Repeat the process to solve for xn-2 , xn-3 , …. x2, x1

)()(,

)(,

2n1nn

2nn1n1n

2n1n1n bxaxa

Back SubstitutionBack Substitution

)(

)(

1nnn

1nn

n a

bx

Back substitution: starting with xn

Solve for xn1 , xn2 , … , 3, 2, 1

Back SubstitutionBack Substitution

a

xab

x

a

bx

1iii

n

1ijj

1iij

1ii

i

1nnn

1nn

n

)(

)()(

)(

)(

for i = n1, n2, …, 1

0a 1iii )(

Naive Gauss EliminationNaive Gauss Elimination

Elimination of first columnElimination of first column

)()(

)()(

)()(

1f4

1f3

1f2

baaa0

baaa0

baaa0

baaaa

41

31

21

4444342

3343332

2242322

114131211

114141

113131

112121

444434241

334333231

224232221

114131211

aaf

aaf

aaf

baaaa

baaaa

baaaa

baaaa

/

/

/

Elimination of second columnElimination of second column

)()(

)()(

/

/

2f4

2f3

baa00

baa00

baaa0

baaaa

aaf

aaf

baaa0

baaa0

baaa0

baaaa

42

32

44443

33433

2242322

114131211

224242

223232

4444342

3343332

2242322

114131211

Elimination of third columnElimination of third column

Upper triangular matrix

)()(

/

3f4

ba000

baa00

baaa0

baaaa

aaf

aaa00

aaa00

baaa0

baaaa

43444

33433

2242322

114131211

33434344443

33433

2242322

114131211

1141431321211

2242432322

3343433

4444

a/)xaxaxab(x

a/)xaxab(x

a/)xab(x

a/bx

Back-SubstitutionBack-Substitution

Upper triangular matrix

0a,a,a,a 44332211

ba000

baa00

baaa0

baaaa

444

33433

2242322

114131211

ExampleExample

41

31

21

41

31

21

f14

f13

f12

5141020

24110

00420

13201

6f

0f

1f

14226

24110

13221

13201

)()(

)()(

)()(

Forward EliminationForward Elimination

42

32

42

32

f24

f23

5141400

24100

00420

13201

1f

1/2f

5141020

24110

00420

13201

)()(

)()(

Upper Triangular MatrixUpper Triangular Matrix

43

43

f34

3370000

24100

00420

13201

14f

5141420

24110

00420

13201

)()(


13/70x3x21x

8/35x2x

4/352x4x

33/707033/x

431

32

43

4

33/70

4/35

8/35

13/70

x

3370000

24100

00420

13201

MATLAB Script File: MATLAB Script File: GaussNaiveGaussNaive

Print all factor and Aug

(do not suppress output)Eliminate first column

Eliminate second column

Eliminate third column Back-substitution

Aug = [A, b]

>> format short>> x = GaussNaive(A,b)m = 4n = 4Aug = 1 0 2 3 1 -1 2 2 -3 -1 0 1 1 4 2 6 2 2 4 1factor = -1Aug = 1 0 2 3 1 0 2 4 0 0 0 1 1 4 2 6 2 2 4 1factor = 0Aug = 1 0 2 3 1 0 2 4 0 0 0 1 1 4 2 6 2 2 4 1factor = 6Aug = 1 0 2 3 1 0 2 4 0 0 0 1 1 4 2 0 2 -10 -14 -5

factor = 0.5000Aug = 1 0 2 3 1 0 2 4 0 0 0 0 -1 4 2 0 2 -10 -14 -5factor = 1Aug = 1 0 2 3 1 0 2 4 0 0 0 0 -1 4 2 0 0 -14 -14 -5

factor = 14Aug = 1 0 2 3 1 0 2 4 0 0 0 0 -1 4 2 0 0 0 -70 -33

x4

x3

x2

x1

x = 0 0 0 0.4714x = 0 0 -0.1143 0.4714x = 0 0.2286 -0.1143 0.4714x = -0.1857 0.2286 -0.1143 0.4714

Algorithm for Gauss eliminationAlgorithm for Gauss elimination1. Forward elimination for each equation j, j = 1 to n-1

for all equations k greater than j

(a) multiply equation j by akj /ajj

(b) subtract the result from equation kThis leads to an upper triangular matrix

2. Back-Substitution (a) determine xn from

(b) put xn into (n-1)th equation, solve for xn-1

(c) repeat from (b), moving back to n-2, n-3, etc. until all equations are solved

)1n(nn

)1n(nn a/bx

Operation CountOperation Count

Important as matrix gets large

For Gauss elimination

Elimination routine uses on the order of O(n3/3) operations

Back-substitution uses O(n2/2)

Operation CountOperation Count

))(())((,

))(())((,

))(())((,

))(())((,

3121nn1n

2knkn1knknn1kk

n2n1n2nn32

1n1nn1nn21

flops

iontion/DivisMultiplica

flops

ubtractionAddition/S

i

Loop Inner

k

Loop Outer

Total operation counts for elimination stage = 2n3/3 + O(n2)

Total operation counts for back substitution stage = n2 + O(n)

Operation CountOperation Count Number of flops (floating-point operations) for

Naive Gauss elimination

%....

%.

%.

859910676106861000000106761000

539866666768155010000671550100

588766780510070510

nEliminatio to

Due Percentage

3

2nFlops

Total

onSbustituti

Back

nElimination

868

3

Computation time increase rapidly with n Most of the effort is incurred in the elimination step Improve efficiency by reducing the elimination effort

Partial PivotingPartial Pivoting

Problems with Gauss elimination division by zero

round off errors

ill conditioned systems

Use “Pivoting” to avoid this Find the row with largest absolute coefficient below the

pivot element

Switch rows (“partial pivoting”)

complete pivoting switch columns also (rarely used)

Round-off ErrorsRound-off Errors

A lot of chopping with more than n3/3 operations

More important - error is propagated

For large systems (more than 100 equations), round-off error can be very important (machine dependent)

Ill conditioned systems - small changes in coefficients lead to large changes in solution

Round-off errors are especially important for ill-conditioned systems

Ill-conditioned SystemIll-conditioned System

2x1 – x2 = 3

2.1x1 – x2 = 3

Consider

Since slopes are almost equal

22

21

22

212

12

11

12

112

2222121

1212111

a

bx

a

ax

a

bx

a

ax

bxaxa

bxaxa

22

21

12

11

a

a

a

a

Ill-Conditioned SystemIll-Conditioned System

0aa

aaD

2221

1211 Divided by small number

)1n(nn

n333

n22322

n1131211

a

aa

aaa

aaaa

U

)(detdet 1nnn332211 aaaaUA

DeterminantDeterminant

Calculate determinant using Gauss elimination

Gauss Elimination with Partial PivotingGauss Elimination with Partial Pivoting

Forward elimination

for each equation j, j = 1 to n-1

first scale each equation k greater than j then pivot (switch rows) Now perform the elimination

(a) multiply equation j by akj /ajj

(b) subtract the result from equation

Partial (Row) PivotingPartial (Row) Pivoting

1x4x2x2x6

2x4xx

1x3x2x2x

1 x3x2x

4321

432

4321

431

14226

24110

13221

13201

bA


Interchange rows 1 & 4

41

31

21

41

31

21

f14

f13

f12

5/67/35/31/30

24110

5/67/37/37/30

14226

1/6f

0f

1/6f

13201

24110

13221

14226

)()(

)()(

)()(


No interchange required

42

32

42

32

f24

f23

5/72200

33/145000

5/67/37/37/30

14226

1/7f

3/7f

5/67/35/31/30

24110

5/67/37/37/30

14226

)()(

)()(


Interchange rows 3 & 4

13/70/6x 2x 2x 41x

8/357/3/x 7/3x 7/35/6x

4/35/2x 25/7x

33/70/533/14x

2341

342

43

4

)(

)()(

)(

)(

33/70

4/35

8/35

13/70

x

0f

33/145000

5/72200

5/67/37/37/30

14226

43

MATLAB M-File: MATLAB M-File: GaussPivotGaussPivot

Partial Pivoting (switch rows)

[big,i] = max(x)

largest element in {x}

index of the largest element

Partial Pivoting

>> format short>> x=GaussPivot0(A,b)Aug = 1 0 2 3 1 -1 2 2 -3 -1 0 1 1 4 2 6 2 2 4 1big = 6i = 4ipr = 4Aug = 6 2 2 4 1 -1 2 2 -3 -1 0 1 1 4 2 1 0 2 3 1factor = -0.1667Aug = 6.0000 2.0000 2.0000 4.0000 1.0000 0 2.3333 2.3333 -2.3333 -0.8333 0 1.0000 1.0000 4.0000 2.0000 1.0000 0 2.0000 3.0000 1.0000factor = 0Aug = 6.0000 2.0000 2.0000 4.0000 1.0000 0 2.3333 2.3333 -2.3333 -0.8333 0 1.0000 1.0000 4.0000 2.0000 1.0000 0 2.0000 3.0000 1.0000factor = 0.1667Aug = 6.0000 2.0000 2.0000 4.0000 1.0000 0 2.3333 2.3333 -2.3333 -0.8333 0 1.0000 1.0000 4.0000 2.0000 0 -0.3333 1.6667 2.3333 0.8333

Aug = [A b]

Interchange rows 1 and 4

Find the first pivot element and its index

Eliminate first columnNo need to interchange

big = 2.3333i = 1ipr = 2factor = 0.4286Aug = 6.0000 2.0000 2.0000 4.0000 1.0000 0 2.3333 2.3333 -2.3333 -0.8333 0 0 0 5.0000 2.3571 0 -0.3333 1.6667 2.3333 0.8333factor = -0.1429Aug = 6.0000 2.0000 2.0000 4.0000 1.0000 0 2.3333 2.3333 -2.3333 -0.8333 0 0 0 5.0000 2.3571 0 0 2.0000 2.0000 0.7143big = 2i = 2ipr = 4Aug = 6.0000 2.0000 2.0000 4.0000 1.0000 0 2.3333 2.3333 -2.3333 -0.8333 0 0 2.0000 2.0000 0.7143 0 0 0 5.0000 2.3571factor = 0Aug = 6.0000 2.0000 2.0000 4.0000 1.0000 0 2.3333 2.3333 -2.3333 -0.8333 0 0 2.0000 2.0000 0.7143 0 0 0 5.0000 2.3571

x = 0 0 0 0.4714x = 0 0 -0.1143 0.4714x = 0 0.2286 -0.1143 0.4714x =

-0.1857 0.2286 -0.1143 0.4714

Back substitutionSecond pivot element and index

Third pivot element and index

Eliminate second column

Eliminate third column

Interchange rows 3 and 4

No need to interchange

Save factors fij for

LU Decomposition

1

4

2 3

5

F14

F23F12

F24

F45

H1

F35F25

V3V1

TRUSSTRUSS

W = 100 kg

Statics: Force BalanceStatics: Force Balance

0FFFF

0FFF

0FFFF

0FFF

0FFF

0FVF

0FFFFF

100FFF

0FFHF

0FVF

4535255x

35255y

4524144x

24144y

35233x

3533y

252423122x

25242y

141211x

1411y

coscos

sinsin

coscos

sinsin

cos

sin

coscos

sinsin

sin

sin

,

,

,

,

,

,

,

,

,

,

Node 1

Node 2

Node 3

Node 4

Node 5

Example: Forces in a Simple TrussExample: Forces in a Simple Truss

0

0

0

0

0

0

0

100

0

0

F

F

F

F

F

F

F

V

H

V

1coscos0000000

0sinsin0000000

100cos0cos0000

000sin0sin0000

0cos00100000

0sin00000100

00coscos101000

00sinsin000000

00000cos1010

00000sin0001

45

35

25

24

23

14

12

3

1

1

function [A,b]=Truss(alpha,beta,gamma,delta)

A=zeros(10,10);A(1,1)=1; A(1,5)=sin(alpha);A(2,2)=1; A(2,4)=1; A(2,5)=cos(alpha);A(3,7)=sin(beta); A(3,8)=sin(gamma);A(4,4)=-1; A(4,6)=1; A(4,7)=-cos(beta); A(4,8)=cos(gamma);A(5,3)=1; A(5,9)=sin(gamma);A(6,6)=-1; A(6,9)=-cos(delta);A(7,5)=-sin(alpha); A(7,7)=-sin(beta);A(8,5)=-cos(alpha); A(8,7)=cos(beta); A(8,10)=1;A(9,8)=-sin(gamma); A(9,9)=-sin(delta);A(10,8)=-cos(gamma); A(10,9)=cos(delta); A(10,10)=-1;

b=zeros(10,1); b(3,1)=100;

Define Matrices A and b in script fileDefine Matrices A and b in script file

>> alpha=pi/6; beta=pi/3; gamma=pi/4; delta=pi/3;>> [A,b] = Truss(alpha,beta,gamma,delta)A = 1.0000 0 0 0 0.5000 0 0 0 0 0 0 1.0000 0 1.0000 0.8660 0 0 0 0 0 0 0 0 0 0 0 0.8660 0.7071 0 0 0 0 0 -1.0000 0 1.0000 -0.5000 0.7071 0 0 0 0 1.0000 0 0 0 0 0 0.7071 0 0 0 0 0 0 -1.0000 0 0 -0.5000 0 0 0 0 0 -0.5000 0 -0.8660 0 0 0 0 0 0 0 -0.8660 0 0.5000 0 0 1.0000 0 0 0 0 0 0 0 -0.7071 -0.8660 0 0 0 0 0 0 0 0 -0.7071 0.5000 -1.0000b = 0 0 100 0 0 0 0 0 0 0>> x = GaussPivot(A,b)x = 40.5827 0 48.5140 70.2914 -81.1655 34.3046 46.8609 84.0287 -68.6091 -93.7218

Gauss Elimination with Partial PivotingGauss Elimination with Partial Pivoting

Simple truss

Banded matrix

Banded MatrixBanded Matrix

HBW: Half Band Width

Tridiagonal MatrixTridiagonal Matrix

Only three nonzero elements in each equation (3n instead of n2 elements)

Subdiagonal, diagonal, superdiagonal

Row Scaling (not implemented in textbook)

-- scale the diagonal element to aii = 1

Solve by Gauss elimination

n

1n

i

3

2

1

n

1n

i

3

2

1

nn

1n1n1n

iii

333

222

11

r

r

r

r

r

r

x

x

x

x

x

x

fe

gfe

gfe

gfe

gfe

gf


Special case of banded matrix with bandwidth = 3

Save storage, 3 n instead of n n


Forward elimination

Back substitution

n32k

rf

err

gf

eff

1k1k

kkk

1k1k

kkk

,,,

1232n1nk f

xgrx

f

rx

k

1kkkk

n

nn

,,,,,

Use factor = ek / fk1

to eliminate subdiagonal element

Apply the same matrix operations to right hand side

Hand Calculations: Tridiagonal MatrixHand Calculations: Tridiagonal Matrix

53

2

5

3

x

x

x

x

2515000

50210

0152

0021

4

3

2

1

...

.

22 2 1

1

22 2 1

1

33 3 2

2

33 3 2

2

44 4 3

3

44 4 3

3

e 2f f g 5 2 1

f 1

e 2r r r 5 3 1

f 1

e 1f f g 2 1 1

f 1

e 1r r r 2 1 1

f 1

e 0 5f f g 1 25 0 5 1

f 1

e 0 5r r r 3 5 1 4

f 1

( )

( )

( )

( )

.. ( . )

.. ( )

11

223

f

xgrx

21

311

f

xgrx

31

4501

f

xgrx

41

4

f

rx

1

2111

2

3222

3

4333

4

44

))((

))((

))(.(

(a) Forward elimination (b) Back substitution

MATLAB M-file:MATLAB M-file: Tridiag Tridiag

» [e,f,g,r] = example

e = 0 -2.0000 4.0000 -0.5000 1.5000 -3.0000f = 1.0000 6.0000 9.0000 3.2500 1.7500 13.0000g = -2.0000 4.0000 -0.5000 1.5000 -3.0000 0r = -3.0000 22.0000 35.5000 -7.7500 4.0000 -33.0000

» x = Tridiag (e, f, g, r)x = 1 2 3 -1 -2 -3

Example: Tridiagonal matrixExample: Tridiagonal matrixfunction [e,f,g,r] = example

e=[ 0 -2 4 -0.5 1.5 -3];

f=[ 1 6 9 3.25 1.75 13];

g=[-2 4 -0.5 1.5 -3 0];

r=[-3 22 35.5 -7.75 4 -33];

33

4

75.7

5.35

22

3

x

x

x

x

x

x

133

375.15.1

5.125.35.0

5.094

462

21

6

5

4

3

2

1

Note: e(1) = 0 and g(n) = 0

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Persamaan Linier - Eliminasi Gauss