C3_L3 Special Matrices and Gauss Seidal (1)

8/12/2019 C3_L3 Special Matrices and Gauss Seidal (1)

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CHAPTER 3

Special Matrices


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SPECIAL MATRICESBanded Matrix:

A banded matrix is a square matrix that has all elements equal to zero, with the exception of a band centered on the main diagonal.

The dimensions of a banded system can be quantified by two

parameters:

a) The bandwidth BW, and

b) The half bandwidth HBW.

These two values are related by


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A banded matrix with a bandwidth of 3 is called a tridiagonal

system. It can be expressed generally as

TRIDIAGONAL SYSTEM

n

n

n

n

nn

nnn

r

r

r

r

r

x

x

x

x

x

f e

g f e

g f e

g f e

g f

1

3

2

1

1

3

2

1

111

333

222

11

Notice that we have changed our notation for the

coefficients from a and b to e, f , g and r .


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By LU decomposition, the tridiagonal system is transformed to

TRIDIAGONAL SYSTEM

n

nn

n

n

nn

nnn

f

g f

g f

g f

g f

e

e

e

e

f e

g f e

g f e

g f e

g f

~

~

~

~

1~1~

1~1~

1

11

33

22

11

1

2

2

111

333

222

11


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A tridiagonal system can be solved efficiently by using Thomas

algorithm. The algorithm consists of three steps:

1. Decomposition:

where k = 2,3,.,n

THOMAS ALGORITHM

1

1

~~~

k k k k

k

k k

g e f f

f e

e


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2. Forward Substitution:

where k = 2,3 ,.,n

3. Back Substitution:

where k = n- 1 ,n- 2,..,2,1

THOMAS ALGORITHM

1~~

k k k k r er r

k

k k k k

n

nn

f x g r x

f

r x

~~

~~

1


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Solve the following tridiagonal system:

EXAMPLE 1

8.200

8.0

8.0

8.40

04.2100

104.210

0104.21

00104.2

4

3

2

1

x

x

x

x


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LU decomposition

SOLUTION

323.1000

1395.100

01550.10

00104.2)(04.2100

1395.100

01550.10

00104.2

)(04.2100104.210

01550.10

00104.2

)(

04.2100

104.210

0104.21

00104.2

3395.1

1

4

2550.1 13

104.21

2

R R

R R

R R


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The factors employed to obtain the upper triangular matrix are

LU decomposition

SOLUTION

323.1000

1395.100

01550.10

00104.2

1717.000

01645.00

00149.0

0001

]][[][ U L A

717.0~

645.0~

49.0~

395.11

4

550.11

3

04.21

2

f

f

f


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Forward decomposition

Thus, the right-hand-side vector has been modified to

SOLUTION

996.210)221.14)(717.0(8.200

221.14)8.20)(645.0(8.0

8.20)8.40)(49.0(8.0

4

3

2

r

r

r

996.210

221.14

8.20

8.40


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Back decomposition

SOLUTION

970.65

778.93

538.124

480.159

04.2)778.93)(1(8.40

1

550.1)538.124)(1(.20

2

395.1)480.159)(1(221.14

3

323.1996.210

4

1

211

2

322

3

433

4

4

f T g r

f T g r

f T g r

f r

x

x

x

x


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a) Solve the following tridiagonal system with the Thomas algorithm

b) Determine the matrix inverse for [A] based on the LU

decomposition and unit vectors.

EXAMPLE 2

105

25

41

8.04.00

4.08.04.0

04.08.0

3

2

1

x

x

x


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a)

We have

Decomposition:

For k = 2,

SOLUTION

10525

41

8.04.004.08.04.0

04.08.0

3

2

1

x x

x

105,8.0,4.0

25,4.0,8.0,4.0

41,4.0,8.0

333

2222

111

r f e

r g f e

r g f

6.0)4.0)(5.0(8.0~~

5.08.04.0~

1222

1

22

g e f f

f e

e


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For k = 3,

Thus, the LU decomposition is

SOLUTION

53333.0)4.0)(66667.0(8.0~~

66667.0

6.0

4.0~

~

2333

2

33

g e f f f

ee

53333.000

4.06.00

04.08.0

166667.00

015.0

001

]][[][ U L A


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Forward substitution:

For k = 2,

For k = 3,

Thus, the right-hand-side becomes

SOLUTION

5.45)41)(5.0(25~ 1222 r er r

53333.000

4.06.00

04.08.0

166667.00

015.0

001

]][[][ U L A

3333.135)5.45)(66667.0(105~~ 2333 r er r

3333.135

5.45

41


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Back substitution:

For k = 3,

For k = 2,

For k = 1,

Thus, the required solution is

SOLUTION

53333.000

4.06.00

04.08.0

166667.00

015.0

001

]][[][ U L A

75.25353333.0

3333.135

3

33

f

r x

2456.0

)]75.253)(4.0(5.45[

2

3222

f

x g r x

75.1738.0)]245)(4.0(41[

1

2111 f

x g r x

75.253,245,75.173 321 x x x


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b) The LU decomposition is

To compute the first column of the inverse,

SOLUTION

53333.000

4.06.00

04.08.0

][,

166667.00

015.0

001

][ U L

333335.0,5.0,1

066667.0

05.0

1

0

0

1

166667.00

015.0

001

}{ where ,

0

01

}]{[

321

32

21

1

3

2

1

3

2

1

d d d

d d

d d

d

d

d

d

d

d d

D D L


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And

By back substitution, we have

The first column of the inverse is

SOLUTION

333335.053333.0

5.04.06.0

14.08.0

333335.0

5.01

53333.000

4.06.0004.08.0

}{}]{[

31

3121

2111

31

21

11

1

A

A A

A A

A

A A

D AU

875.1,25.1,625.0 112131 A A A

625.0

25.1

875.1


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For the second column,

SOLUTION

66667.0,1,0

066667.0

15.0

00

1

0

166667.00

015.0

001

}{ where ,

0

10

}]{[

321

32

21

1

3

2

1

3

2

1

d d d

d d

d d

d d

d

d

d

d d

D D L


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And


The second column of the inverse is

SOLUTION

66667.053333.0

14.06.0

04.08.0

66667.0

10

53333.000

4.06.0004.08.0

}{}]{[

32

3222

2212

32

22

12

1

A

A A

A A

A

A A

D AU

25.1,5.2,25.1 122232 A A A

25.1

5.2

25.1


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For the third column,

SOLUTION

1,0,0

166667.0

05.0

0

1

0

0

166667.00

015.0

001

,

1

0

0

}]{[

321

32

21

1

3

2

1

d d d

d d

d d

d

d

d

d

D L


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And


The third column of the inverse is

SOLUTION

153333.0

04.06.0

04.08.0

1

00

53333.000

4.06.0004.08.0

}{}]{[

33

3323

2313

33

23

13

1

A

A A

A A

A

A A

D AU

625.0,25.1,875.1 132333 A A A

875.1

25.1

625.0


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The matrix inverse is

SOLUTION

875.125.1625.0

25.15.225.1

625.025.1875.1

][ 1 A

C3_L3 Special Matrices and Gauss Seidal (1)

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