gaussian lu eliminations seidel

7/28/2019 Gaussian LU Eliminations Seidel

1/100

ES 84 Numerical Methods

Stephen H. Haim

Computer Engineering Dept./EECE

Gaussian Elimination,

LU Decomposition &

Gauss-Seidel


2/100

Gaussian Elimination

2


3/100

Nave Gaussian Elimination

One of the most popular techniques for

solving simultaneous linear equations of the

form

Consists of 2 steps

1. Forward Elimination of Unknowns.

2. Back Substitution

CXA


4/100

Forward Elimination

The goal of Forward Elimination is to transform

the coefficient matrix into an Upper Triangular

Matrix

7.000

56.18.40

1525

112144

1864

1525


5/100

Forward Elimination

Linear Equations

A set ofn equations and n unknowns

11313212111 ... bxaxaxaxa nn 22323222121 ... bxaxaxaxa nn

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .


6/100

Forward Elimination

Transform to an Upper Triangular Matrix

Step 1: Eliminate x1 in 2nd equation using equation 1 as

the pivot equation

)(1

21

11

aa

Eqn

Which will yield

1

11

211

11

21212

11

21121 ... b

a

axa

a

axa

a

axa nn


7/100

Forward Elimination

Zeroing out the coefficient of x1 in the 2nd equation.

Subtract this equation from 2nd equation

1

11

2121

11

212212

11

2122 ... ba

abxaa

aaxaa

aa nnn

'

2

'

22

'

22 ... bxaxa nn

nnna

a

aaa

aa

a

aa

1

11

21

2

'

2

12

11

2122'22

Or Where


8/100

Forward Elimination

Repeat this procedure for the remaining

equations to reduce 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

. . .

. . .

. . .


9/100

Forward Elimination

Step 2: Eliminate x2 in the 3rd equation.

Equivalent to eliminating x1 in the 2nd equation

using equation 2 as the pivot equation.

)(2

332

22

aa

EqnEqn


10/100

Forward Elimination

This procedure is repeated for the remaining

equations to reduce the set of equations as

11313212111 ... bxaxaxaxa nn '

2

'

23

'

232

'

22 ... bxaxaxa nn "

3

"

33

"

33 ... bxaxa nn

""

3

"

3 ... nnnnn bxaxa

. .

. .

. .


11/100

Forward Elimination

Continue this procedure by using the third equation as the pivot

equation and so on.

At the end of (n-1) Forward Elimination steps, the system of

equations will look like:

'

2

'

23

'

232

'

22 ... bxaxaxa nn

"3

"3

"33 ... bxaxa nn

11 nnnn

nn bxa

. .

. .

. .

11313212111 ... bxaxaxaxa nn


12/100

Forward Elimination

At the end of the Forward Elimination steps

)-(nnn

3

2

1

n

nn

n

n

n

b

b

b

b

x

x

x

x

a

aa

aaa

aaaa

1

"

3

'2

1

)1(

"

3

"

33

'2

'23

'22

1131211


13/100

Back Substitution

The goal of Back Substitution is to solve each of

the equations using the upper triangular matrix.

3

2

1

3

2

1

33

2322

131211

x

xx

00

0

b

b

b

a

aa

aaa

Example of a system of 3 equations


14/100

Back Substitution

Start with the last equation because it has only

one unknown

)1(

)1(

n

nn

n

n

nabx

Solve the second from last equation (n-1)th

using xn solved for previously.

This solves for xn-1.


15/100

Back Substitution

Representing Back Substitution for all equations

by formula

1

1

11

i

ii

n

ijjiijii

ia

xabx Fori=n-1, n-2,.,1

and

)1(

)1(

n

nn

nn

na

bx


16/100

Example: Rocket Velocity

The upward velocity of a rocket

is given at three different times

Time, t Velocity, v

s m/s

5 106.8

8 177.2

12 279.2

The velocity data is approximated by a polynomial as:

12.t5,322

1 atatatv

Find: The Velocity at t=6,7.5,9, and 11 seconds.


17/100


Assume

12.t5,atatatv 32

2

1

3

2

1

3

2

3

2

2

2

1

2

1

1

1

1

v

v

v

a

a

a

tt

tt

tt

3

2

1

Results in a matrix template of the form:

Using date from the time / velocity table, the matrix becomes:

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a


18/100


Forward Elimination: Step 1

)64(

25

12

RowRow

Yields

2.27921.96

81.106

aa

a

11214456.18.40

1525

3

2

1


19/100


)144(

25

13

RowRow

0.336

21.96

8.106

a

a

a

76.48.160

56.18.40

1525

3

2

1

Yields



20/100


Yields

)8.16(8.4

23

RowRow

735.0

21.96

8.106

a

a

a

7.000

56.18.40

1525

3

2

1

This is now ready for Back Substitution



21/100


Back Substitution: Solve fora3 using the third equation

735.07.0 3 a

70

7350

.

.a 3

0501.a 3


22/100


Back Substitution: Solve fora2 using the second equation

21.9656.18.4 32 aa

8.4

56.121.96 32

aa

84

05015612196

.-

...-

a 2

7019.a 2


23/100


Back Substitution:Solve fora1 using the first equation

8.106525 321 aaa

25

58.106 321

aaa

25

050.170.1958.106

1

a

2900.01 a


24/100


Solution:The solution vector is

050.1

70.19

2900.0

3

2

1

a

a

a

The polynomial that passes through the three data points is

then:

3221 atatatv

125,050.170.192900.0 2 ttt


25/100


Solution:

Substitute each value of t to find the corresponding velocity

.s/m1.165

050.15.770.195.72900.05.7v2

.s/m8.201

050.1970.1992900.09v2

.s/m8.252

050.11170.19112900.011v2

./69.129

050.1670.1962900.062

sm

v


26/100

Pitfalls

Two Potential Pitfalls-Division by zero: May occur in the forward elimination

steps. Consider the set of equations:

655

901.33099.26

7710

321

321

32

xxx

xxx

xx

- Round-off error: Prone to round-off errors.


27/100

Pitfalls: Example

Consider the system of equations:

Use five significant figures with chopping

515

6099.23

0710

3

2

1

x

x

x

6

901.3

7

=

At the end of Forward Elimination

1500500

6001.00

0710

3

2

1

x

x

x

15004

001.6

7

=


28/100

Pitfalls: Example

Back Substitution

99993.015005

150043 x

5.1001.0

6001.6 32

xx

350.010

077 321

xx

x

15004

001.67

1500500

6001.000710

3

2

1

x

xx


29/100

Pitfalls: Example

Compare the calculated values with the exact solution

99993.0

5.1

35.0

3

2

1

x

x

x

X calculated

1

1

0

3

2

1

x

x

x

X exact


30/100

Improvements

Increase the number of significant digits

Decreases round off error

Does not avoid division by zero

Gaussian Elimination with Partial Pivoting

Avoids division by zero

Reduces round off error


31/100

Partial Pivoting

pka

Gaussian Elimination with partial pivoting applies row switching to

normal Gaussian Elimination.

How?

At the beginning of the kth step of forward elimination, find the maximum of

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

If the maximum of the values is In the pth row, ,npk

then switch rows p and k.


32/100

Partial Pivoting

What does it Mean?

Gaussian Elimination with Partial Pivoting ensures that

each step of Forward Elimination is performed with the

pivoting element |akk| having the largest absolute value.


33/100

Partial Pivoting: Example

Consider the system of equations

6x5xx5

901.3x3x099.2x3

7x7x10

321

321

21

In matrix form

5156099.23

0710

3

2

1

x

x

x

6901.3

7

=

Solve using Gaussian Elimination with Partial Pivoting using five

significant digits with chopping


34/100



Examining the values of the first column

|10|, |-3|, and |5| or 10, 3, and 5

The largest absolute value is 10, which means, to follow therules of Partial Pivoting, we switch row1 with row1.

6

901.37

515

6099.230710

3

2

1

x

xx

5.2

001.67

55.20

6001.000710

3

2

1

x

xx

Performing Forward Elimination


35/100



Examining the values of the first column

|-0.001| and |2.5| or 0.0001 and 2.5

The largest absolute value is 2.5, so row 2 is switched withrow 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

Performing the row swap


36/100



Performing the Forward Elimination results in:

002.6

5.2

7

002.600

55.20

0710

3

2

1

x

x

x


37/100


Back Substitution

Solving the equations through back substitution

1002.6

002.6

3x

15.2

55.2 22

xx

010

077 321

xxx

002.6

5.2

7

002.600

55.20

0710

3

2

1

x

x

x


38/100


1

1

0

3

2

1

x

x

x

X exact

1

1

0

3

2

1

x

x

x

Xcalculated

Compare the calculated and exact solution

The fact that they are equal is coincidence, but it does

illustrate the advantage of Partial Pivoting


39/100

Summary

-Forward Elimination

-Back Substitution-Pitfalls

-Improvements

-Partial Pivoting


40/100

LU Decomposition

40


41/100

LU Decomposition

LU Decomposition is another method to solve a set of

simultaneous linear equations

Which is better, Gauss Elimination or LU Decomposition?

To answer this, a closer look at LU decomposition is

needed.


42/100

LU Decomposition

L

Method

A

ULA

U

For most non-singular matrix that one could conduct Nave Gauss

Elimination forward elimination steps, one can always write it as

Where

= lower triangular martix

= upper triangular martix


43/100

LU Decomposition

CXUL

CXA

Proof

If solving a set of linear equations

If Then

Multiply by

Which gives

Remember which leads to

Now, if then

Now, let

Which ends with

and

1L

CLXULL 11

ILL 1 CLXUI 1

UUI CLXU1

ULA

ZCL 1

CL Z

ZU X

(1)

(2)


44/100

LU Decomposition

CZL Z

ZU X

How can this be used?

Given

Decompose into and U L

Then solve for

And then solve for X

CXA

A


45/100

LU Decomposition

CZL

How is this better or faster than Gauss

Elimination?

Lets look at computational time.

n = number of equations

To decompose [A], time is proportional to

To solve and

time proportional to

3

3n

CXU

2

2n


46/100

LU Decomposition

Therefore, total computational time for LU Decomposition is

proportional to

23

3n

n)

2(2

3

23nn

or

Gauss Elimination computation time is proportional to

23

23nn

How is this better?


47/100

LU Decomposition

)2

n

3

n(m

23

)n(m3

n 23

51033.8

What about a situation where the [C] vector changes?

In LU Decomposition, LU decomposition of [A] is independent

of the [C] vector, therefore it only needs to be done once.

Let m = the number of times the [C] vector changes

The computational times are proportional to

LU decomposition = Gauss Elimination=

Consider a 100 equation set with 50 right hand side vectors

LU Decomposition = Gauss Elimination =71069.1


48/100

LU Decomposition

Another Advantage

Finding the Inverse of a Matrix

LU Decomposition Gauss Elimination

3

4)(

3

32

3 nnn

n

2323

3423 nnnnn

For large values of n

3

4

23

334 nnn


49/100

LU Decomposition

Method: [A] Decompose to [L] and [U]

33

2322

131211

3231

21

00

0

1

01

001

u

uu

uuu

ULA

[U] is the same as the coefficient matrix at the end of the forward

elimination step.

[L] is obtained using the multipliers that were used in the forwardelimination process


50/100

LU Decomposition

Finding the [U] matrix

Using the Forward Elimination Procedure of Gauss Elimination

112144

1864

1525

112144

56.18.40

1525

)64(25

1Row2Row

76.48.160

56.18.40

1525

)144(25

1Row3Row


51/100

LU Decomposition

Finding the [U] matrix

Using the Forward Elimination Procedure of Gauss Elimination

76.48.160

56.18.40

1525

7.000

56.18.40

1525

)8.16(8.4

2Row3Row

7.000

56.18.40

1525

U


52/100

LU Decomposition

Finding the [L] matrix

Using the multipliers used during the Forward Elimination Procedure

101

001

3231

21

56.2

25

64

11

21

21 a

a

76.525

144

11

31

31 a

a

From the first step

of forwardelimination

From the secondstep of forward

elimination

76.48.160

56.18.401525

5.38.4

8.16

22

32

32

a

a


53/100

LU Decomposition

15.376.5

0156.2

001

L

Does ? AUL

7.00056.18.40

1525

15.376.50156.2

001

UL


54/100

LU Decomposition

Example: Solving simultaneous linear equations using LU Decomposition

Solve the following set of

linear equations using LU

Decomposition

2.279

2.177

8.106

a

a

a

112144

1864

1525

3

2

1

Using the procedure for finding the [L] and [U] matrices

7.000

56.18.40

1525

15.376.5

0156.2

001

ULA


55/100

LU Decomposition


Set

Solve for

CZL

Z

2.279

2.177

8.106

15.376.5

0156.2

001

3

2

1

z

z

z

2.2795.376.5

2.17756.2

10

321

21

1

zzz

zz

z


56/100

LU Decomposition


Complete the forward substitution to solve for Z

735.0

)21.96(5.3)8.106(76.52.279

5.376.52.279

2.96

)8.106(56.22.177

56.22.177

8.106

213

12

1

zzz

zz

z

735.0

21.96

8.106

3

2

1

z

z

z

Z


57/100

LU Decomposition

ZXU

X


Set

Solve for

0.735

96.21-

106.8

7.000

56.18.40

1525

3

2

1

a

a

a

The 3 equations become

735.07.0

21.9656.18.4

8.106525

3

32

321

a

aa

aaa


58/100

LU Decomposition


From the 3rd equation

1.050

0.7

0.735a

a

3

735.07.0 3

Substituting in a3 and using the

second equation

7019.

4.8-1.0501.5696.21-

8.4

56.121.96

21.9656.18.4

3

2

32

aa

aa


59/100

LU Decomposition


Substituting in a3 and a2

using the first equation

2900.0

25

050.170.1958.106

25

aa58.106

8.106525

321

321

a

aaa

Hence the Solution Vector is:

050.1

70.19

2900.0

3

2

1

a

a

a


60/100

LU Decomposition

3

4

23

334 nnn

Finding the inverse of a square matrix

Remember, the relative computational time comparison

of LU decomposition and Gauss elimination is:

Review: The inverse [B] of a square matrix [A] is defined as

ABIBA


61/100

LU Decomposition

Finding the inverse of a square matrix

How can LU Decomposition be used to find the inverse?

Assume the first column of [B] to be

Using this and the definition of matrix multiplication

First column of [B] Second column of [B]

Tn112 bbb 11

0

0

1

b

b

b

A

1n

21

11

0

1

0

b

b

b

A

2n

22

12

The remaining columns in [B] can be found in the same manner


62/100

LU Decomposition

Example: Finding the inverse of a square matrix

112144

1864

1525

A

0.700

1.56-4.8-0

1525

15.376.5

0156.2

001

ULA

Find the inverse of [A]

Using the Decomposition procedure, the [L] and [U] matrices are found to be


63/100

LU Decomposition


Solving for the each column of [B] requires to steps

1) Solve [L] [Z] = [C] for [Z] and 2) Solve [U] [X] = [Z] for [X]

Step 1: CZL

0

0

1

15.376.5

0156.2

001

3

2

1

z

z

z

This generates the equations: 11

z056.2 21 zz

05.376.5 321 zzz


64/100

LU Decomposition


Solving for [Z]

2.3

56.25.3176.50

z5.3z76.50z

56.2

156.20

213

12

1

2.56z-0z

1z

2.3

56.2

1

z

z

z

Z

3

2

1


65/100

LU Decomposition


Solving for [U] [X] = [Z] for [X]

3.2

2.56-

1

7.000

56.18.40

1525

31

21

11

b

b

b

1525 312111 bbb

56.256.18.4 3121 bb

2.37.0 31 b


66/100

LU Decomposition


Using Backward Substitution

04762.0=25

571.4)9524.0(51=

2551=

9524.0=8.4

)571.4(560.1+56.2=

8.4

560.1+56.2=

571.4=

7.0

2.3=

3 121

1 1

3 1

2 1

3 1

bbb

bb

bSo the first column of

the inverse of [A] is:

571.4

9524.0

04762.0

31

21

11

b

b

b


67/100

LU Decomposition


Repeating for the second and third columns of the inverse

Second 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

b

b

b

112144

1864

1525

33

23

13

429.1

4643.0

03571.0

33

23

13

b

b

b


68/100

LU Decomposition


The inverse of [A] is

429.1050.5571.4

4643.0417.19524.00357.008333.04762.0

1A

To check your work do the following operation

AAIAA 11


69/100

Gauss-Seidel Method

69


70/100

Gauss-Seidel Method

An iterative method.

Basic Procedure:

-Algebraically solve each linear equation for xi

-Assume an initial guess solution array

-Solve for each xi and repeat

-Use absolute relative approximate error after each iteration

to check if error is within a pre-specified tolerance.


71/100

Gauss-Seidel Method

Why?

The Gauss-Seidel Method allows the user to control round-off error.

Elimination methods such as Gaussian Elimination and LU

Decomposition are prone to prone to round-off error.

Also: If the physics of the problem are understood, a close initialguess can be made, decreasing the number of iterations needed.


72/100

Gauss-Seidel Method

AlgorithmA set ofn equations and n unknowns:

11313212111 ... bxaxaxaxa nn

2323222121 ... bxaxaxaxa n2n

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .

If: the diagonal elements arenon-zero

Rewrite each equation solvingfor the corresponding unknown

ex:

First equation, solve for x1

Second equation, solve for x2


73/100

Gauss-Seidel Method

AlgorithmRewriting each equation

11

13132121

1a

xaxaxacx nn

nn

nnnnnn

n

nn

nnnnnnnnn

n

nn

a

xaxaxacx

a

xaxaxaxac

x

a

xaxaxacx

11,2211

1,1

,122,122,111,11

1

22

232312122

From Equation 1

From equation 2

From equation n-1

From equation n


74/100

Gauss-Seidel Method

AlgorithmGeneral Form of each equation

11

11

11

1a

xac

x

n

jj

jj

22

21

22

2a

xac

x

j

n

jj

j

1,1

11

,11

1

nn

n

njj

jjnn

na

xac

x

nn

n

njj

jnjn

na

xac

x

1


75/100

Gauss-Seidel Method

AlgorithmGeneral Form for any row i

.,,2,1,1

nia

xac

xii

n

ijj jiji

i

How or where can this equation be used?


76/100

Gauss-Seidel Method

Solve for the unknowns

Assume an initial guess for [X]

n

-n

2

x

x

x

x

1

1

Use rewritten equations to solve for

each value of xi.

Important: Remember to use themost recent value of xi. Which

means to apply values calculated to

the calculations remaining in the

current iteration.


77/100

Gauss-Seidel Method

Calculate the Absolute Relative Approximate Error

100x

xxnew

i

old

i

new

i

ia

So when has the answer been found?

The iterations are stopped when the absolute relativeapproximate error is less than a prespecified tolerance for all

unknowns.


78/100

Gauss-Seidel Method: Example


79/100

Gauss Seidel Method: Example

1

3

2

1

3

2

3

2

2

2

1

2

1

1

1

1

v

v

v

a

a

a

tt

tt

tt

3

2

1

Using a Matrix template of the form

The system of equations becomes

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a

Initial Guess: Assume an initial guess of

5

2

1

3

2

1

a

a

a



80/100


1

Rewriting each equation

2.279

2.177

8.106

112144

1864

1525

3

2

1

a

a

a

25

58.106 321

aaa

8

642.177 312

aaa

1

121442.279 213

aaa



81/100


1

Applying the initial guess and solving for ai

5

2

1

3

2

1

a

a

a 6720.325

)5()2(58.106a1

8510.7

8

56720.3642.177a2

36.155

1

8510.7126720.31442.279a3

Initial Guess

When solving for a2, how many of the initial guess values were used?



82/100


1

%76.72100x6720.3

0000.16720.31

a

%47.125100x8510.7

0000.28510.7

2

a

%22.103100x36.155

0000.536.1553

a

Finding the absolute relative approximate error

100x

xxnew

i

old

i

new

i

ia

At the end of the first iteration

The maximum absolute

relative approximate error is

125.47%

36.155

8510.7

6720.3

3

2

1

a

a

a



83/100


1

Iteration #2Using

36.155

8510.7

6720.3

3

2

1

a

a

a

056.12

25

36.1558510.758.1061

a

882.54

8

36.155056.12642.1772

a

34.798

1

882.5412056.121442.2793

a

from iteration #1

the values of ai are found:



84/100


1

Finding the absolute relative approximate error

%542.69100x056.12

6720.3056.121

a

%695.85100x

882.54

8510.7882.542

a

%54.80100x34.798

36.15534.7983

a

At the end of the second iteration

34.798882.54

056.12

3

2

1

a

a

a

The maximum absolute

relative approximate error is

85.695%



85/100


1

0858.1

690.19

29048.0

3

2

1

a

a

a

Repeating more iterations, the following values are obtained

Iteration a1 a2 a3

1

2

3

4

5

6

3.672

12.056

47.182

193.33

800.53

3322.6

72.767

67.542

74.448

75.595

75.850

75.907

-7.8510

-54.882

-255.51

-1093.4

-4577.2

-19049

125.47

85.695

78.521

76.632

76.112

75.971

-155.36

-798.34

-3448.9

-14440

-60072

-249580

103.22

80.540

76.852

76.116

75.962

75.931

%1a

%2a

%3a

! Notice The relative errors are not decreasing at any significant rate

Also, the solution is not converging to the true solution of

G S id l M th d Pitf ll


86/100

Gauss-Seidel Method: Pitfall

What went wrong?

Even though done correctly, the answer is not converging to the

correct answer

This example illustrates a pitfall of the Gauss-Siedel method: not all

systems of equations will converge.

Is there a fix?

One class of system of equations always converges: One with a diagonally

dominantcoefficient matrix.

Diagonally dominant: [A] in [A] [X] = [C] is diagonally dominant if:

n

jj

ijaa

i1

ii

n

ijj

ijii aa1

for all i and for at least one i

G S id l M th d Pitf ll


87/100

Gauss-Seidel Method: Pitfall

116123

14345

3481.52

A

Diagonally dominant: The coefficient on the diagonal must be at leastequal to the sum of the other coefficients in that row and at least one row

with a diagonal coefficient greater than the sum of the other coefficients

in that row.

1293496

55323

5634124

]B[

Which coefficient matrix is diagonally dominant?

Most physical systems do result in simultaneous linear equations that

have diagonally dominant coefficient matrices.



88/100

p

2

Given the system of equations

15x-3x12x 321

283x5xx 321

7613x7x3x 321

1

0

1

3

2

1

x

x

x

With an initial guess of

The coefficient matrix is:

1373

351

5312

A

Will the solution converge using the

Gauss-Siedel method?



89/100

p

2

1373

351

5312

A

Checking if the coefficient matrix is diagonally dominant

43155 232122 aaa

10731313 323133 aaa

8531212 131211 aaa

The inequalities are all true and at least one row is strictlygreater than:

Therefore: The solution should converge using the Gauss-Siedel Method



90/100

p

2

76

28

1

1373

351

5312

3

2

1

a

a

a

Rewriting each equation

12

531 321

xxx

5

328 312

xxx

13

7376 213

xxx


1

0

1

3

2

1

x

x

x

50000.0

12

150311

x

9000.4

5

135.0282

x

0923.3

13

9000.4750000.03763

x



91/100

p

2

The absolute relative approximate error

%662.6710050000.0

0000.150000.01a

%00.1001009000.4

09000.42a

%662.671000923.3

0000.10923.33a

The maximum absolute relative error after the first iteration is 100%



92/100

p

2

8118.3

7153.3

14679.0

3

2

1

x

x

x

After Iteration #1

14679.0

12

0923.359000.4311

x

7153.35

0923.3314679.0282

x

8118.3

13

900.4714679.03763

x

Substituting the x values into the equations After Iteration #2

0923.3

9000.4

5000.0

3

2

1

x

x

x



93/100

p

2

Iteration #2 absolute relative approximate error

%62.24010014679.0

50000.014679.01

a

%887.311007153.3

9000.47153.32

a

%876.181008118.3

0923.38118.33

a

The maximum absolute relative error after the first iteration is 240.62%

This is much larger than the maximum absolute relative error obtained in

iteration #1. Is this a problem?



94/100

p

2

Repeating more iterations, the following values are obtained

1a

2a

3a

Iteration a1 a

2 a

3

1

2

3

4

5

6

0.50000

0.14679

0.74275

0.94675

0.99177

0.99919

67.662

240.62

80.23

21.547

4.5394

0.74260

4.900

3.7153

3.1644

3.0281

3.0034

3.0001

100.00

31.887

17.409

4.5012

0.82240

0.11000

3.0923

3.8118

3.9708

3.9971

4.0001

4.0001

67.662

18.876

4.0042

0.65798

0.07499

0.00000

4

3

1

3

2

1

x

x

x

0001.40001.3

99919.0

3

2

1

x

x

xThe solution obtained

is close to the exact solution of



95/100

p

3

Given the system of equations

7613x7x3x 321

283x5xx 321

15x-3x12x 321


1

0

1

3

2

1

x

x

x

Rewriting the equations

3

13776 321

xxx

5

328 312

xxx

53121 21

3

xxx



96/100

p

3

Conducting six iterations

1a

2a

3aIteration a1 a2 a3

1

2

34

5

6

21.000

-196.15

-1995.0-20149

2.0364x105

-2.0579x105

110.71

109.83

109.90109.89

109.90

1.0990

0.80000

14.421

-116.021204.6

-12140

1.2272x10

5

100.00

94.453

112.43109.63

109.92

109.89

5.0680

-462.30

4718.1-47636

4.8144x105

-4.8653x106

98.027

110.96

109.80109.90

109.89

109.89

The values are not converging.Does this mean that the Gauss-Seidel method cannot be used?

Gauss Seidel Method


97/100

Gauss-Seidel Method

The Gauss-Seidel Method can still be used

The coefficient matrix is not

diagonally dominant

5312

351

1373

A

But this is the same set of

equations used in example #2,

which did converge.

1373

351

5312

A

If a system of linear equations is not diagonally dominant, check to see if

rearranging the equations can form a diagonally dominant matrix.

Gauss Seidel Method


98/100

Gauss-Seidel Method

Not every system of equations can be rearranged to have adiagonally dominant coefficient matrix.

Observe the set of equations

3321 xxx9432 321 xxx

97 321 xxx

Which equation(s) prevents this set of equation from having a

diagonally dominant coefficient matrix?

Gauss Seidel Method


99/100

Gauss-Seidel Method

Summary

-Advantages of the Gauss-Seidel Method

-Algorithm for the Gauss-Seidel Method

-Pitfalls of the Gauss-Seidel Method


100/100

gaussian lu eliminations seidel

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