Lecture01 - Gaussian Elimination

8/9/2019 Lecture01 - Gaussian Elimination

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Solution of Linear System of Equations

Lecture 1:

Gaussian Elimination

MTH2212 Computational Methods and Statistics


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Dr. M. HrairiDr. M. Hrairi MTH2212MTH2212 -- Computational Methods and StatisticsComputational Methods and Statistics 22

Objectives

Introduction


Pitfalls in Gauss Elimination Method

Pivoting

Scaling


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Introduction

Well see how to solve a set ofn linearequations with nunknowns.

If n 3, the small set of linearequations can be solved

analytically. Simultaneous linearequations arise in a variety of

engineering problems context i.e. trusses, reactors, electric

circuits,

Broadly there are two types of methods to solve the linear

equations: Direct methods and Iterative methods.


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Solving a set ofn linearequations:

Solution consists of two phases:1. Forward elimination of unknowns

2. Back substitution

!

!

!

nnnnnnn

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

...

.

.

.

...

...

332211

22323222121

11313212111(1a)

(1b)

(1c)


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1. Forward elimination of the unknowns Eliminate the first unknown, x1, from the second through the n

th equations.Multiply equation (1) by a21/a11

Subtract equation (2) from equation (1b) to get

or

where the prime indicates the new values for the elements.

1

11

211

11

21313

11

21212

11

21121 ... ba

a

xaa

a

xaa

a

xaa

a

xa nn !

1

11

2121

11

212313

11

2123212

11

2122 )(...)()( b

a

abxa

a

aaxa

a

aaxa

a

aa

nnn!

(2)

(3)

'

2

'

23

'

232

'

22... bxaxaxa

nn!


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Dr. M. HrairiDr. M. Hrairi MTHMTH22122212 -- Computational Methods and StatisticsComputational Methods and Statistics 66


Similarly equation (1) can be multiplied by a 1/a11 and the result subtracted

from the third equation.

Repeating the procedure for the remaining equations give the following

modified forms:

''

3

'

32

'

2

'

2

'

23

'

232

'

22

11313212111

...

.

.

.

...

...

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxaxa(4a)

(4b)

(4c)


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The second unknown x2 is eliminated by multiplying through the next

equation by a 2/a22 and subtracting the result from the following

equation.

The procedure is continued using the remaining pivot equations. At the

end of the elimination stage, the original system of equations istransformed to an upper triangular system:

!

!

!

)1()1(

'

2

'

23

'

232

'

22

11313212111

.

.

.

...

...

n

nn

n

nn

nn

nn

bxa

bxaxaxa

bxaxaxaxa

(5a)

(5b)

(5c)


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2. Back Substitution

The last unknown xn is now given by:

This value of xn is then back-substituted into equation (n-1) to find xn-1.

This procedure can be repeated to evaluate the values of the remaining xs.

)1(

)1(

!n

nn

n

nn

a

bx (6)


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Example 1

Use Gauss elimination to solve:

3x1 0.1x2 0.2x3 = 7.85 (a)

0.1x1 + 7x2 0.3x3 = -19.3 (b)

0.3x1 0.2x2 + 10x3 = 71.4 (c)

use 6 significant figures in your computation.


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Solution of Example 1

1. Forward elimination Multiply 1st equation (a) by 0.1/ and subtract from 2nd equation (b) to eliminate

x1 from the latter equation:

7.00 x2 0.29 x = -19.5617

Then multiply 1st equation (a) by 0. / and subtract from rd equation (c) toeliminate x1 from the latter equation:

-0.190000x2 + 10.0200x = 70.6150

The system of equations is now:

x1 0.1x2 0.2x = 7.85 (a)

7.00 x2 0.29 x = -19.5617 (b)

-0.190000x2 + 10.0200x = 70.6150 (c)


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Now lets eliminate x2 from latestrd equation (c) by multiplying equation (b)

by -0.190000/7.00 and subtract the result from equation (c):

10.0200x = 70.084

The system of equations is now reduced to an upper triangular form:

x1 0.1 x2 0.2 x = 7.85 (a)

7.00 x2 0.29 x = -19.5617 (b)

10.0200 x = 70.084 (c)


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2. Back Substitution

Solve last equation (c) to find x :

x = 70.084 /10.0200 = 7.0000

Substitute x in equation (b) and solve for x2 :

7.00 x2 0.29 (7.0000 ) = -19.5617

x2 = -2.50000

Finally substitute values of x2 and x in equation (a) to find x1 :

x1 0.1(-2.5000) 0.2(7.0000 ) = 7.85

x1 = .00000


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3. Check yourresults

Substitute the values of x1 , x2 and x in the original system of equations.

( .00000) 0.1 (-2.50000) 0.2 (7.0000 ) = 7.84999

0.1 ( .00000) + 7 (-2.50000) 0. (7.0000 ) = -19. 0000

0. ( .00000) 0.2 (-2.50000) + 10 (7.0000 ) = 71.400 0


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Some Pitfalls of Gauss Elimination Method

1. Division by zero

Caused by coefficient value equals zero orvery close to zero

This can be solved by using pivoting technique.


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Some Pitfalls of Gauss Elimination Method

2. Round-off errors This may be due to some of the following:

Large numberof equations to be solved due to the fact that everyresult is dependent on previous results.

Errors in early steps will tend to propagate.

Round-off errors may be solved by:

Using more significant figures.

Using fractions instead of decimals.

Always substitute youranswers back into the original system ofequations to check foroccurrence of substantial errors.


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Pivoting

This involves the following steps:

1. Determine the largest coefficient available in the column below thepivot element.

2. Switch the rows so that the largest element is the pivot element.This is known as partial pivoting.

3. If columns as well as rows are searched for the largest element

and then switched, the process is called complete pivoting

.


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Example 2

Solve the following system of equations

0.0003 x1 + 3.0000 x2 = 2.0001 (1)

1.0000 x1 + 1.0000 x2 = 1.0000 (2)

1. Using nave Gauss elimination

2. Using partial pivoting


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Solution of example 2

1. Firstly, lets solve these equations the same way we did in example 1:

Forward elimination:

Multiply equation (1) by 1/0.0003 to get

x1 + 10,000x2 = 6667

Subtract the resulting equation from equation (2) to get

-9999x2 = -6666 x2 = 2/3

Back substitution:0.0003x1 + 3.0000x2 = 2.0001 x1 = 2.0001 3(2/3)

0.0003


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Now lets check the effects of numberof significant digits

on the result of x1 due to x2:

Significant

figures x2 x1 x1(%)

3 0.667 -3.33 1099

4 0.6667 0.0000 100

5 0.66667 0.30000 10

6 0.666667 0.330000 1

7 0.6666667 0.3330000 0.1


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2. Secondly, lets solve these equations by applying partial pivoting:

1.0000 x1 + 1.0000 x2 = 1.0000 (1)

0.0003 x1 + 3.0000 x2 = 2.0001 (2)

Forward elimination:

Multiply equation (1) by 0.0003 to get

0.0003 x1 + 0.0003 x2 = 0.0003

Subtract the resulting equation from equation (2) to get

2.9997x2

= 1.9998 x2

= 2/3

Back substitution:

x1 + x2 = 1 x1 = 1 (2/3)

1


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This case is much less sensitive to the numberof significant

figures on the result of x1 due to x2:

Significant

figures x2 x1 x1(%)

3 0.667 0.333 0.1

4 0.6667 0.3333 0.01

5 0.66667 0.33333 0.001

6 0.666667 0.333333 0.0001

7 0.6666667 0.3333333 0.00001


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Scaling

Scaling is necessary when different units are used in the

same system of equations.

Scaling is used to:1. Standardize the size of the coefficients in the system of

equations.

2. Minimize the round-off errors caused by having much larger

coefficients than others.


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Example 3

Use three significant figures to solve

2 x1 + 100000 x2 = 10000 (1)

x1 + x2 = 2. (2)

1. Nave Gauss elimination

2. Gauss elimination with scaling and pivoting

3. Gauss elimination without scaling and with pivoting


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1. Gauss elimination:

Forward elimination

2 x1 + 100000 x2 = 10000 (1)

- 50000 x2 = -5000 (2)

Back substitution

x2 = 0.10 and x1 = 0.00

2. Gauss elimination with scaling and pivoting:

Scaling0.00002 x1 + x2 = 0.1 (1)

x1 + x2 = 2 (2)


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Pivoting

x1 + x2 = 2 (1)

0.00002 x1 + x2 = 0.1 (2)

Forward elimination

x1 + x2 = 2

x2 = 0.1

Back substitution

x1 = 1.90 and x2 = 0.10


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3. Gauss elimination without scaling and with pivoting of the

original equations:

Pivoting

x1 + x2 = 2.

2 x1 + 100000 x2 = 10000 (1)

Forward elimination

x1 + x2 = 2

100000 x2 = 10000

Back substitutionx1 = 1.90 and x2 = 0.10

Lecture01 - Gaussian Elimination

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