lec 10 gauss elimination1

7/30/2019 Lec 10 Gauss Elimination1

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Simultaneous Linear Equations

Topic: Gaussian Elimination

Dr. Nasir M Mirza

Numerical Methods

Email: [email protected]


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Gaussian Elimination


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Basic Principles

The general description of a set of linear equations in the matrixform: [A][X] = [B]

Where, [A] is ( m x n ) matrix, the [X] is a ( n x 1 ) vector, andthe [B] is a (m x 1 ) vector.

How we solve such a system:

Write the equations in natural form

Identify unknowns and order them

Isolate unknowns

Write equations in matrix form


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Types of Matrix Formulation

If m = n The solution of [A]{x} ={b} with n unknowns and m(n) equations

If m > n The system is overdetermined system (Least Square Problems)

If m < n The system is underdetermined system (Optimization Problems)

mnmnmmm

nn

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

332211

33333232131

22323222121

11313212111

Suppose we have an ( m x n ) Array


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Matrix Representation

nnnnnnn

n

n

n

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaa

aaaa

3

2

1

3

2

1

321

3333231

2232221

1131211

mnmnmmm

nn

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

332211

33333232131

22323222121

11313212111The set of n linear equationswith n unknowns:

The matix form:


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Consistency

nnnnnnn

n

n

n

b

b

b

b

x

x

x

x

aaaa

aaaa

aaaaaaaa

bXA

3

2

1

3

2

1

321

3333231

2232221

1131211

The problem is consistent, if a solution exists for the problem.

The problem is inconsistent, if there is no solution for the problem.


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Rank of Matrix

Rank of a matrix is the number of linearlyindependent column vectors in the matrix.

For n x n matrix, A:

If rank(A) = n and is consistent, A has an uniquesolution exists

If rank(A) = n and is inconsistent, A has no solution

exists If rank(A) < n and system is consistent, A has an

infinite number of solutions


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Matrix

For an n x n system, rank(A) = n automaticallyguarantees

that the system is consistent.

The columns of A are linearly independent The rows of A are linearly independent

rank(A) = n

det(A) ~= 0

A-1 exists;

The solution to [A]{x} ={b} exist and is unique.


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Matrix Definition

1

6

15.0

12

y

x

5

6

12

12

y

x

Considery = -2.0 x + 6

y = 0.5 x + 1

There are 2 unknowns x , y

and rank is 2

Considery = -2 x + 6

y = -2 x + 5

There are 2 unknowns x , y and

rank is 1 and is inconsistent


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Gaussian Elimination

Gaussian Elimination is one of the most popular techniques for

solving simultaneous linear equations of the form: [A][X] =[b]

The method consists of 2 steps

1. Forward Elimination of Unknowns.2. Back Substitution

Let us learn the method first by examples:

mnmnmmm

nn

nn

nn

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

bxaxaxaxa

332211

33333232131

22323222121

11313212111


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Example 1:Consider following three linear equations with three unknowns:

)3(74.506.113.569.3

)2(69.475.378.046.1

)1(76.128.406.337.2

321

321

321

xxx

xxx

xxx

)4(7426.08059.12911.1 321 xxx

)5(6058.3366.66650.20

69.475.3780.046.1

0842.16366.2885.146.1

32

321

321

xx

xxx

xxx

First step is to divide by 2.37 so that the coefficient of x1 is one. Thus

Now multiply this by -1.46 and adding it with Eq2.

)6(4802.86038.58942.90 32 xx

Now multiply Eq.4 with 3.69 and adding it to Eq3.


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Example 1:Re-writing Equations 4, 5 and 6:

)6(4802.86038.58942.9)0(

)5(6058.33866.66650.2)0(

)4(7426.08059.12911.1

321

321

321

xxx

xxx

xxx

)7(3530.13965.20 32 xx

8671.211077.18)0(0 32 xx

The second step is to divide by -2.6650 so that the coefficient of x2 is one. Thus

Now multiply this by -9.8942 and adding it to Eq 6.

dividing it by 18.1077, we get

)8(2076.1)0(0 32 xx


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Example 1:Final form of Eq. 4, 7 and 8 is

)8(2076.100

)7(3530.13965.20

)4(7426.08059.12911.1

3

32

321

x

xx

xxx

5410.12076.13965.23530.13965.23530.1 32

xx

Now x3 is known and we can back-substitute it into Eq. 7 to find x2.

Similarly, we can find x1 using values of x2 and x3 from Eq. 4

Therefore, the solution is given as

x1= 0.9338 ; x

2= 1.5410 ; x

3= 1.2076

9338.0

2076.13965.25410.12911.17426.0

3965.22911.17426.0321

xxx


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Forward Elimination

The goal of Forward Elimination is to transform the coefficientmatrix into an Upper Triangular Matrix

7.000

56.18.401525

112144

1864

1525


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Forward Elimination

Linear Equations:A set ofn equations and n unknowns

11313212111 ... bxaxaxaxa nn

22323222121 ... bxaxaxaxa nn

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .

( Eq.1 )

( Eq.2 )


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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 change the Eq. 1 as following:

1

11

21

1

11

21

212

11

21

121 ... ba

axa

a

axa

a

axa nn

11313212111 ... bxaxaxaxa nn (Eq.1)


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Forward Elimination

Zeroing out the coefficient of x1

in the 2nd equation.

Subtract Equation 1 from Eq.2

1

11

2121

11

212212

11

2122 ... b

a

abxa

a

aaxa

a

aa nnn

'

2

'

22

'

22 ... bxaxa nn

nnn aa

aaa

aaaaa

1

11

212

'

2

12

11

2122

'22

Or

1

11

21

1

11

21

212

11

21

121... b

a

axa

a

axa

a

axa

nn

Where

22323222121 ... bxaxaxaxa nn ( Eq.2 )

( Eq.1 )


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Forward Elimination

Repeat this procedure for the remaining equations to reduce the set ofequations as

11313212111 ... bxaxaxaxa nn '

2

'

23

'

232

'

22 ... bxaxaxa nn

'

3

'

33

'

332

'

32 ... bxaxaxa nn

''

3

'

32

'

2 ... nnnnnn bxaxaxa

. . .

. . .

. . .


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Forward Elimination

Step 2: Eliminate x2 in the 3rd equation.Equivalent to eliminating x1 in the 2

nd equation using equation 2 as the pivot

equation.

)(23 3222

aa

EqnEqn


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Forward Elimination

This procedure is repeated for the remaining equations to reduce theset of equations as

11313212111 ... bxaxaxaxa nn

'2

'23

'232

'22 ... bxaxaxa nn

"

3

"

33

"

33 ... bxaxa nn

""

3

"

3 ... nnnnn bxaxa

. .

. .

. .


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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 n

nn

n

nn bxa

. .

. .

. .

11313212111 ... bxaxaxaxa nn


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Forward Elimination

At the end of the Forward Elimination steps:

)-(n

nn

3

2

1

n

nn

n

n

n

b

bb

b

x

xx

x

a

aaaaa

aaaa

1

"

3

'

2

1

)1(

"

3

"

33

'

2

'

23

'

22

1131211


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Back Substitution

The goal of Back Substitution is to solve each of the equations usingthe upper triangular matrix.

3

2

1

3

2

1

33

2322

131211

x

x

x

00

0

b

b

b

a

aa

aaa

Example of a system of 3 equations


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Back Substitution

Start with the last equation because it has only

one unknown

)1(

)1(

n

nn

n

n

n

a

bx

Solve the second from last equation (n-1)th using xn solved for

previously.

This solves for xn-1.


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Back Substitution

Representing Back Substitution for all equations by formula

1

1

11

iii

n

ijj

i

ij

i

i

i

a

xab

x

Fori=n-1,

n-2,.,1

and)1(

)1(

n

nn

n

n

n

a

bx


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Computer Programfunction x = gaussE(A,b,ptol)

% GEdemo Show steps in Gauss elimination and back substitution% No pivoting is used.

%

% Synopsis: x = GEdemo(A,b)

% x = GEdemo(A,b,ptol)

%

% Input: A,b = coefficient matrix and right hand side vector

% ptol = (optional) tolerance for detection of zero pivot

% Default: ptol = 50 * eps

%

% Output: x = solution vector, if solution exists

A=[25 5 1; 64 8 1; 144 12 1]

b=[106.8; 177.2; 279.2]

if nargin


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Computer Program (continued)% program continued

for i =1:n-1pivot = Ab(i,i);

if abs(pivot)

lec 10 gauss elimination1

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