chapter 5f. simultaneous linear equations · chapter 5f. simultaneous linear equations '...

1

� A. J. Clark School of Engineering �Department of Civil and Environmental Engineering

by

Dr. Ibrahim A. AssakkafSpring 2001

ENCE 203 - Computation Methods in Civil Engineering IIDepartment of Civil and Environmental Engineering

University of Maryland, College Park

CHAPTER 5f.SIMULTANEOUS LINEAR EQUATIONS

© Assakkaf

Slide No. 171

� A. J. Clark School of Engineering � Department of Civil and Environmental Engineering

ENCE 203 � CHAPTER 5f. SIMULTANEOUS LINEAR EQUATIONS

Banded Matrices

� A banded matrix was defined earlier as square matrix with elements of zero except on the principal diagonal and the values in the positions adjacent to the principal diagonal.

� A tridiagonal matrix is a special case of a banded matrix.

2

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Slide No. 172



Banded Matrices

� Tridiagonal Matrix� A triadigonal matrix is a special case of a

banded matrix that has zeros except in the three diagonals:

5554

454443

343332

322221

1211

00000

0000000

aaaaa

aaaaaa

aa

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Slide No. 173



Banded Matrices

� Tridiagonal Matrix� The tridigonal matrix can be described as

having a band width of 3.� It can also be described as having a half-

band width of 1, in reference to the number of nonzero diagonals on one side of the principal diagonal, that is, not including the diagonal where i = j for aij.

3

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Slide No. 174



Banded Matrices

� Tridiagonal Matrix� The following relationship between the

band width bw1 and half-band width bw2 can be obtained:

12 2`1 += ww bb

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Slide No. 175



Banded Matrices

� Tridiagonal Matrix� Banded matrices such as a tridigonal

matrix have so many applications in engineering.

� The can be stored more efficiently by storing the banded elements only, thereby reducing storage requirements for a solution.

4

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Slide No. 176



Symmetric Matrices

� Definition� A symmetric matrix is defined as a matrix

where aij = aji.� Examples:

=

221031082321

A

−

−−−

=

1542047532541

0313

B

−

−=

20552

C

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Slide No. 177



Symmetric Matrices

� Applications� In engineering, it is common to deal with

symmetric matrices.� For example, the stiffness properties of a

structural element can be described using symmetric stiffness matrix.

� Also, correlations among the structural variables can be described using symmetric correlation matrix.

5

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Slide No. 178



Symmetric Matrices

� Applications� Example: Stiffness Matrix for a Beam

EI = constant

P

L

−

−

=

LEI

LEI

LEI

LEI

S46

612

2

23

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Slide No. 179



Symmetric Matrices

� Properties� A symmetric matrix has the following

properties:

� The decomposition of A can be expressed as

TAA =

TLLA =

6

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Slide No. 180



Symmetric Matrices

� Cholesky Decomposition Method� This method uses a recurrence procedure

to decompose a symmetric matrix into upper and lower triangular matrices.

� As a result, the LU decomposition can be computed more effectively.

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Slide No. 181



Symmetric Matrices

� Cholesky Decomposition Method

ijijl

llal

nilal

al

jj

j

kjkikij

ij

i

kikiiii

<−=−

=

=−=

=

∑

∑−

=

−

=

and ,1,,3,2,1for

,,3,2for

1

1

1

1

2

1111

L

L

(1a)

(1b)

(1c)

7

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Slide No. 182



Symmetric Matrices

� Example:Decompose the following matrix into its lower and upper triangular matrices using the Cholesky method:

=

221031082321

A

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Slide No. 183



Symmetric Matrices

� Example (cont�d)Using Eq. 1a:

Using Eq. 1c:

111111 === al

=

221031082321

A

22108321

33

3223

22

3113

2112

11

===

=====

=

aaa

aaaaa

a

jj

j

kjkikij

ij l

llal

ji

∑−

=

−=

==1

1

1,2

212

11

21

11

11

11221

21 ===−

=∑

−

=

la

l

llal k

kk

8

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Slide No. 184



Symmetric Matrices

� Example (cont�d)Using Eq. 1b:

=

221031082321

A

222108321

21

33

3223

22

3113

2112

11

==

===

====

=

la

aaa

aaaa

a

∑−

=−=

=1

1

2

: 2i

kikiiii lal

i

( ) 228 222122

12

1

222222 =−=−=−= ∑

−

=

lalalk

k

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Slide No. 185



Symmetric Matrices

� Example (cont�d)Using Eq. 1c:

=

221031082321

A

jj

j

kjkikij

ij l

llal

ji

∑−

=

−=

==1

1

2 and 1,3

22

)2(310

313

22

213132

22

12

12332

32

11

31

11

11

11331

31

=−=−=−

=

===−

=

∑

∑

−

=

−

=

llla

l

llal

la

l

llal

kkk

kkk

2222108321

22

21

33

3223

22

3113

2112

11

===

===

====

=

lla

aaa

aaaa

a

9

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Slide No. 186



Symmetric Matrices

� Example (cont�d)Using Eq. 1b:

=

221031082321

A

∑−

=−=

=1

1

2

: 3i

kikiiii lal

i

( ) ( ) 32322 22232

23133

13

1

233333 =−−=−−=−= ∑

−

=

llalalk

k

232222108321

32

31

22

21

33

3223

22

3113

2112

11

=====

===

====

=

lllla

aaa

aaaa

a

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Slide No. 187



Symmetric Matrices

� Example (cont�d)� Therefore

=

=

300220321

323022001

TL

L

323221

33

32

31

22

21

11

======

llllll

10

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Slide No. 188



Symmetric Matrices

� Example (cont�d)� Note that the validity of LLT = A

=

==

221031062321

300220321

323022001

ALLT

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Slide No. 189



Iterative Equation-Solving Methods

� Simultaneous equations can also be solved using trial-and error procedure.

� In this procedure, a solution can be assumed, that is, a set of estimates for the unknowns.

� Then, these estimates can be revised through some set of rules.

11

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Slide No. 190




� This approach is the basis for iterative methods for solving simultaneous equations.

� Among these iterative methods, two procedures are considered:� Jacobi Iteration, and� Gauss-Seidel Iteration

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� Jacobi Iteration� Consider the following general set of

simultaneous equation:

nnnnnnn

nn

nn

nn

CXaXaXaXa

CXaXaXaXaCXaXaXaXa

CXaXaXaXa

=++++

=++++=++++

=++++

L

MMMMM

L

L

L

332211

33333232131

22323222121

11313212111

(2)

12

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Slide No. 192




� Jacobi Iteration� The first step in this method is rearrange

each equation in Eq. 2 to produce an expression for a single unknown.

� To start the iterative calculations, an initial solution estimate for Xi unknowns is required.

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Slide No. 193




� Jacobi Iteration

nn

nnnnnnn

nn

nn

aXaXaXaC

X

aXaXaXaCX

aXaXaXaCX

1,13221

22

232312122

11

131321211

−−−−−−=

−−−−=

−−−−=

L

M

L

L

(3)

13

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Slide No. 194



Iterative Equation-Solving Methods� Jacobi Iteration

� The initial estimates for all the Xi are substituted into the right sides of Eq. 3 to obtain a new set of calculated (left side) values for the Xi�s.

� These new values are substituted into the right side of Eq. 3 and a new set of values for the Xi�s is obtained.

� This iterative process continues until the calculated values for Xi�s converge to an acceptable solution.

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Slide No. 195




� Example: Jacobi IterationSolve the following set of equations using the Jacobi iterative method:

14 834

92 3

321

32

321

=+−−=−+−

=−+

XXXXXXXXX

14

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Slide No. 196




� Example (cont�d): Jacobi IterationThe first step is to rearrange each equation as

follows:

14 834

92 3

321

32

321

=+−−=−+−

=−+

XXXXXXXXX

41

438

329

213

312

321

XXX

XXX

XXX

+−=

++−=

+−=

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Slide No. 197




� Example (cont�d): Jacobi IterationLet an estimate of the solution be X1 = X2 = X3 = 1,

therefore,

250.041

4111

41

14

)1(3184

38

333.33

103

)1(2193

29

213

312

321

==+−=+−

=

−=++−=++−

=

==+−=+−

=

XXX

XXX

XXX

15

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Slide No. 198




� Example (cont�d): Jacobi Iteration

25.01333.3

:for valuesRevised

3

2

1

=−=

=

XXX

X i

8333.04

)1()3333.3(14

1

9792.04

)25.0(33333.384

38

5.33

)25.0(2)1(93

29

213

312

321

−=−+−=+−

=

−=++−=++−

=

=+−−=+−

=

XXX

XXX

XXX

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Slide No. 199




� Example (cont�d): Jacobi Iteration

8333.09792.0

5.3:for valuesRevised

3

2

1

−=−=

=

XXX

X i

8698.04

)9792.0()5.3(14

1

7500.14

)8333.0(35.384

38

7709.23

)8333.0(2)9792.0(93

29

213

312

321

−=−+−=+−

=

−=−++−=++−

=

=−+−−=+−

=

XXX

XXX

XXX

16

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Slide No. 200




� Example (cont�d): Jacobi IterationIteration X 1 | ∆X 1| X 2 | ∆X 1| X 3 | ∆X 1|

0 1 - 1 - 1 -1 3.3333 2.3333 -1.0000 2.0000 0.2500 0.75002 3.5000 0.1667 -0.9792 0.0208 -0.8333 1.08333 2.7708 0.7292 -1.7500 0.7708 -0.8698 0.03654 3.0035 0.2326 -1.9596 0.2096 -0.8802 0.01045 3.0664 0.0629 -1.9093 0.0503 -0.9908 0.11066 2.9759 0.0905 -1.9765 0.0672 -0.9939 0.00317 2.9962 0.0203 -2.0015 0.0250 -0.9881 0.00588 3.0084 0.0122 -1.9920 0.0094 -0.9994 0.01139 2.9977 0.0107 -1.9975 0.0054 -1.0001 0.000710 2.9991 0.0014 -2.0007 0.0032 -0.9988 0.001311 3.0010 0.0019 -1.9993 0.0013 -0.9999 0.001112 2.9998 0.0012 -1.9997 0.0004 -1.0001 0.000213 2.9998 0.0000 -2.0001 0.0004 -0.9999 0.000214 3.0001 0.0003 -1.9999 0.0002 -1.0000 0.000115 3.0000 0.0001 -2.0000 0.0000 -1.0000 0.0000

The solution convergesto:X1 = 3X2 = -2X3 = -1

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Slide No. 201




� Gauss-Seidel Iteration� Like in the Jacobi iteration, the first step in

this method is rearrange each equation inEq. 2 to produce an expression for a single unknown.

� To start the iterative calculations, an initial solution estimate for Xi unknowns is required.

17

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Slide No. 202




� Gauss-Seidel Iteration

nn

nnnnnnn

nn

nn

aXaXaXaC

X

aXaXaXaCX

aXaXaXaC

X

1,13221

22

232312122

11

131321211

−−−−−−=

−−−−=

−−−−=

L

M

L

L

(4)

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Slide No. 203



Iterative Equation-Solving Methods� Gauss-Seidel Iteration

� In Jacobi iteration a full cycle is completed over all the equations before updating the solutions estimates.

� In Gauss-Seidel procedure, each unknown is updated as soon as a new estimate of that unknown is completed.

� The notion here is that the most recent estimate is the best estimate, and therefore, should be used as soon it is available.

18

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Slide No. 204




� Example: Gauss-Seidel IterationSolve the following system of equations using the Gauss-Seidel Iteration procedure with initial estimates of X1 = X2 = X3 = 1:

2.43 3 1.14 42

7.15324

−=−+−=−+−

=+−

ZYXZYXZYX

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Slide No. 205



Iterative Equation-Solving Methods� Example (cont�d): Gauss-Seidel Iteration

332.4421.14

4327.15

−−−−=

++−=

−+=

YXZ

ZXY

ZYX

2.43 3 1.14 42

7.15324

−=−+−=−+−

=+−

ZYXZYXZYX

19

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Slide No. 206




� Example (cont�d): Gauss-Seidel IterationAn estimate of the solution is X1 = X2 = X3 = 1,

therefore,

5958.43

)4375.1()6750.3(32.4332.4

4375.14

)1()6750.3(21.14421.14

6750.34

)1(3)1(27.154

327.15

=−

−−−−=−

−−−=

−=++−=++−=

=−+=−+=

YXZ

ZXY

ZYX

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Slide No. 207




� Example (cont�d): Gauss-Seidel Iteration

5958.44375.1

6750.3:for valuesRevised

=−=

=

ZYX

X i

3273.03

)4964.2()2406.0(32.4332.4

4964.24

)5958.4)2406.0(21.14421.14

2406.04

)5958.4(3)4375.1(27.15

=−

−−−−−=−

−−−=

−=+−+−=++−=

−=−−+=

YXZ

ZXY

X

20

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Slide No. 208




� Example (cont�d): Gauss-Seidel Iteration

3273.04964.22406.0

:for valuesRevised

=−=−=

ZYX

X i

0888.33

)2275.2()4313.2(32.4332.4

2275.24

3273.0)4313.2(21.14421.14

4313.24

)3273.0(3)4964.2(27.154

327.15

=−

−−−−=−

−−−=

−=++−=++−=

=−−+=−+=

YXZ

ZXY

ZYX

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Slide No. 209



Iterative Equation-Solving Methods� Example (cont�d): Gauss-Seidel Iteration

The solution convergesto:X = 1.3Y = -2.4Z = 1.9

Iteration X | ∆X | Y | ∆Y | Z | ∆Z |0 1 - 1 - 1 -1 3.6750 2.6750 -1.4375 2.4375 4.5958 3.59582 -0.2406 3.9156 -2.4964 1.0589 0.3273 4.26863 2.4314 2.6720 -2.2275 0.2689 3.0889 2.76164 0.4946 1.9368 -2.5055 0.2780 1.0594 2.02955 1.8777 1.3831 -2.3213 0.1842 2.5039 1.44456 0.8864 0.9913 -2.4558 0.1345 1.4678 1.0361. . . . . . .

30 1.2999 0.0003 -2.4000 0.0000 1.8999 0.000331 1.3001 0.0002 -2.4000 0.0000 1.9001 0.000232 1.2999 0.0002 -2.4000 0.0000 1.8999 0.000233 1.3001 0.0001 -2.4000 0.0000 1.9001 0.000134 1.3000 0.0001 -2.4000 0.0000 1.9000 0.000135 1.3000 0.0001 -2.4000 0.0000 1.9000 0.000136 1.3000 0.0000 -2.4000 0.0000 1.9000 0.0000

21

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Slide No. 210



Iterative Equation-Solving Methods� Convergence Consideration for the Iterative

Methods� For these iterative techniques of Jacobi and

Gauss-Seidel to work, certain additional conditions must be considered:1. The set of equations must possess a strong

diagonal.2. A sufficient condition for a solution to be

found is that the absolute value of the diagonal coefficient in any equation must greater than the sum of the absolute values of all other coefficients appearing in that equation.

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Slide No. 211




� Convergence Consideration for the Iterative Methods� Before solving a set of equations using the

iterative methods, do the following:� Rearrange the set of n × n equations so that the

diagonal coefficient is the largest in any equation.

22

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Slide No. 212




� Example: ConvergenceCheck the following set of equations for convergence. If they do not meet the condition for convergence, try to rearrange the equation so that they meet the requirement.

1742 4 25324

321

321

321

=++=+−=−+

XXXXXXXXX

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Slide No. 213




� Example (cont�d): Convergence

equations. therearrangeTry N.G 61241 <⇒−+<

1742 4 25324

321

321

321

=++=+−=−+

XXXXXXXXX

1742 324 4 25

321

321

321

=++=−+=+−

XXXXXXXXX

methods. iterativein set above theuse Therefore,O.K. 34214

O.K. 34214

O.K 35125

>⇒+>

>⇒−+>

>⇒+−>

23

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Slide No. 214



Use of Determinants

� As was noted earlier, a determinant is a unique number that can be used to represent a square matrix.

� For the matrix A, the determinant is denoted as |A|.

� Recall that the system of equations is given by [A]{X} = {C}

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Slide No. 215



Use of Determinants

� Or in equivalent general matrix form as

=

nnnnnn

n

n

n

C

CCC

X

XXX

aaaa

aaaaaaaaaaaa

MM

L

MMMMM

L

L

L

3

2

1

3

2

1

321

3333231

2232221

1131211

(5)

24

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Slide No. 216



Use of Determinants

� In which, the coefficient matrix A is given by

=

=

=

nnnnnn

n

n

n

C

CCC

C

X

XXX

X

aaaa

aaaaaaaaaaaa

AMM

L

MMMMM

L

L

L

3

2

1

3

2

1

321

3333231

2232221

1131211

and , ,

X = vector of unknowns, and C = vector of constants

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Slide No. 217



Use of Determinants

� Cramer�s RuleThe value of Xi is obtained using Cramer�s

rule as

AA

X ii =

Where |A| is the determinant of A and |Ai| is the determinantof a matrix formed by replacing column I of A with thecolumn vector of constant of Eq.5.

25

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Slide No. 218



Use of Determinants

� Cramer�s Rule� For example, |A1| and |A2| would be given

by

nnnn

n

n

nnnnn

n

n

aaCa

aaCaaaCa

A

aaaC

aaaCaaaC

A

L

MOMMM

L

L

L

MOMMM

L

L

331

223221

113111

2

32

223222

113121

1 and ==

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Slide No. 219



Use of Determinants

� Example: Use of DeterminantSolve the following set of simultaneous linear equations using the method of the determinants:

2.43 3 1.14 42

7.15324

−=−+−=−+−

=+−

ZYXZYXZYX

26

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Slide No. 220



Use of Determinants

� Example (cont�d): Use of Determinant[ ]{ } { }

−−=

−−−

−=

2.41.14

7.15

313142

324

ZYX

CXA

−−=

−−−

−=

2.41.14

7.15 and

313142

324CA

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Slide No. 221



Use of Determinants

� Example (cont�d): Use of Determinant

[ ] [ ] [ ] 68)4(3)1(23)1(3)3(22)1(1)3(44313142

324−=−−+−−−−+−−−=

−−−

−=A

3.168

4.8868

312.4141.14

327.15

1 =−

−=−

−−−−

−

==AA

X

−−=

−−−

−=

2.41.14

7.15 and

313142

324CA

27

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Slide No. 222



Use of Determinants


4.268

2.16368

32.4311.142

37.154

2 −=−

=−

−−−−−

==AA

Y

−−=

−−−

−=

2.41.14

7.15 and

313142

324CA

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Slide No. 223



Use of Determinants


9.168

2.12968

2.4131.1442

7.1524

3 =−

−=−

−−−

−

==AA

Z

−−=

−−−

−=

2.41.14

7.15 and

313142

324CA

28

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Slide No. 224



Use of Determinants

� Example (cont�d): Use of DeterminantTherefore, the solution for set of equation

is

2.43 3 1.14 42

7.15324

−=−+−=−+−

=+−

ZYXZYXZYX

−=

9.14.2

3.1

ZYX

chapter 5f. simultaneous linear equations · chapter 5f. simultaneous linear equations '...

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