LINEAR EQUATIONS SYSTEMS 2

8/9/2019 LINEAR EQUATIONS SYSTEMS 2

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NUMERICAL METHODS

From University of Michigan , department of atmospheric,

oceanic and spaces sciences


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LINEAR EQUATIONS SYSTEMSSOLUTION OF LINEAR EQUATIONS SYSTEMS

DIRECT METHODS

(1)Crammers Rules

Given a system of linear equations, Cramer's Rule is a

handy way to solve for just one of the variables withouthaving to solve the whole system of equations. They don't

usually teach Cramer's Rule this way, but this is supposed

to be the point of the Rule: instead of solving the entire

system of equations, you can use Cramer's to solve for justone single variable.

(1)http://www.purplemath.com/modules/cramers.htm


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(1)Crammers RulesExample: Use the crammers rules to solve:We have the left-hand side of the system with the variables (the

"coefficient matrix") and the right-hand side with the answer values.

Let D be the determinant of the coefficient matrix of the abovesystem, and let Dx be the determinant formed by replacing the x-

column values with the answer-column values:

2x+ y+ z = 3

x y z = 0

x+ 2y+ z = 0

System of

equations

Coefficient matrix's

determinant

Answer column Dx:coefficient

determinant with answer-column values in x-column

2 x + 1 y + 1 z = 3

1x 1y 1 z = 0

1x + 2y + 1z = 0


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(1)Crammers RulesExample: (Cont)Similarly, Dy and Dz would then be:

Evaluating each determinant, we get:


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(1)Crammers RulesExample: (Cont)Cramer's Rule says thatx = Dx D, y = Dy D, and z = Dz D.

That is:x = 3/3 = 1, y = 6/3 = 2, and z = 9/3 = 3

(2)For more than three equations, Cramer's rule becomes

impractical because, as the number of equations increases, thedeterminants are time consuming to evaluate by hand (or

computer). So using other more efficient alternatives.

(2)Steven c. Chapra, Numerical Methods for Engineers. Fifth Edition Chapter 9.


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(3)Elimination of UnknownsThe elimination of unknowns by combining equations is an algebraic

approach that can be illustrated for a set of two equations:

a11 x1 + a12x2 = b1 (1)

a21 x1 + a22x2 = b2 (2)

The basic strategy is to multiply the equations by constants so that

will be eliminated when two equations are combined. The result is

a single equation that can be solved for remaining unknown. Thisvalue can be substituted into either of the original equations to

compute the other variable.

(3)Steven c. Chapra, Numerical Methods for Engineers. Fifth Edition Chapter 9.


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(3)Elimination of UnknownsFor example, equation (1) might be multiplied by a21 and Equation

(2) by a11 to give

a21 a11 x1 + a12 a12x2 = a21 b1 (3)

a11a21 x1 + a11 a22x2 = a11 b2 (4)

Subtracting equation (3) from (4) will, therefore eliminate the X1term from the equations to yield

a22 a11 x2 - a12 a21x2 = a21 b2 - b1a21 (5)

Which can be solved for

(6)

21122211

121211

2

aaaa

babax

Equation (6) can be substituted intoeq (1) which can be solved for

21122211

212122

1

aaaa

babax


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(3)Elimination of UnknownsExample: Use the Elimination of Unknowns to solve:

Solution. Using equations (6) and (7)

22

1823

21

21

xx

xx

4)1(2)2(3)2(2)18(2

1

x 3)1(2)2(3

18)1()2(32

x


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(4)Gauss MethodA method to solve simultaneous linear equations of the form

[A][X]=[C]

Two steps

1. Forward Elimination2. Back Substitution

(4)http://numericalmethods.eng.usf.edu


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(4)Gauss Method

1. Forward Elimination: The goal of forward elimination is

to transform the coefficient matrix into an upper

triangular matrix

2.279

2.177

8.106

112144

1864

1525

3

2

1

x

x

x

735.0

21.96

8.106

7.000

56.18.40

1525

3

2

1

x

x

x


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(4)Gauss Method1. Forward Elimination: A set of n equations and n

unknowns

(1)

(2)

(n)

11313212111 ... bxaxaxaxa nn

22323222121... bxaxaxaxa nn

nnnnnnn bxaxaxaxa ...332211

. .

. .

. .

(n-1) steps of forward elimination


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(4)Gauss Method1. Forward Elimination:

Step 1. For Equation 2, divide Equation 1 by and

multiply by .

)...(11313212111

11

21 bxaxaxaxaa

ann

1

11

21

1

11

21

212

11

21

121... b

aaxa

aaxa

aaxa

nn


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Subtract the result from Equation 2.

1

11

21

21

11

21

2212

11

21

22 ... ba

a

bxaa

a

axaa

a

a nnn

'

2

'

22

'

22

... bxaxann

22323222121 ... bxaxaxaxa nn

_________________________________________________

or

1

11

21

1

11

21

212

11

21

121... b

aaxa

aaxa

aaxa nn


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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

End of Step 1

. . .

. . .

. . .


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Step 2. Repeat the same procedure for the 3rd term of

Equation 3.

11313212111 ... bxaxaxaxa nn '

2

'

23

'

232

'

22... bxaxaxa nn

"

3

"

33

"

33...

bxaxa nn

""

3

"

3 ... nnnnn bxaxa

. .

. .

. .

End of Step 2

Q O S S S S


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

of equations will look like

'

2

'

23

'

232

'

22 ... bxaxaxa nn

"

3

"

33

"

33... bxaxa nn

11 n

nn

n

nn bxa

. .. .. .

11313212111 ... bxaxaxaxa nn

End of Step (n-1)

LINEAR EQUATIONS SYSTEMS


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Matrix Form at End of Forward Elimination

)(n-n

"

'

n

)(n

nn

"

n

"

'

n

''

n

b

b

bb

x

x

xx

a

aa

aaaaaaa

1

3

2

1

3

2

1

1

333

22322

1131211

0000

00

0


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(4)Gauss Method2. Back Substitution:

Back Substitution Starting Eqns

'

2

'

23

'

232

'

22 ... bxaxaxa nn

"

3

"

3

"

33 ... bxaxa nn

11 n

nn

n

nn bxa

11313212111 ... bxaxaxaxa nn



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(4)Gauss Method2. Back Substitution:

Start with the last equation because it has only one

unknown

1,...,1for

1

1

11

nia

xab

xi

ii

n

ijj

i

ij

i

i

i

)1(

)1(

n

nn

n

n

n a

b

x

1,...,1for...

1

1

,2

1

2,1

1

1,

1

ni

a

xaxaxabx

i

ii

n

i

nii

i

iii

i

ii

i

i

i



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(5) Gauss-Jordan MethodIs a variant of Gaussian Elimination. Again, we are

transforming the coefficient matrix into another matrix

that is much easier to solve, and the system represented

by the new augmented matrix has the same solution setas the original system of linear equations. In Gauss-Jordan

Elimination, the goal is to transform the coefficient matrix

into a diagonal matrix, and the zeros are introduced into

the matrix one column at a time. We work to eliminate

the elements both above and below the diagonal element

of a given column in one pass through the matrix.

(5)http://ceee.rice.edu/Books/CS/chapter2/linear44.html



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(5) Gauss-Jordan MethodSteps:

Step1. Write the augmented matrix for the system of

linear equations.



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(5) Gauss-Jordan MethodStep 2.Use elementary row operations on the augmented

matrix [A|b] to transformA into diagonal form.


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LU Decomposition(6)The LU-decomposition method first "decomposes"

matrix A into A = L.U, where L and U are lower triangular

and upper triangular matrices, respectively. More

precisely, if A is a nn matrix, L and U are also nnmatrices with forms like the following:

33

2322

131211

3231

21

00

0

1

01

001

u

uu

uuu

ULA

(6) http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/INT-APP/CURVE-linear-

system.html



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LU Decomposition(7)How does LU Decomposition work?

If solving a set of linear equations

If [A] = [L][U] then

Multiply by

Which gives

Remember [L]-1[L] = [I] which leads to

Now, if [I][U] = [U] then

Now, let

Which ends with

and

[A][X] = [C]

[L][U][X] = [C]

[L]-1

[L]-1[L][U][X] = [L]-1[C]

[I][U][X] = [L]-1[C]

[U][X] = [L]

-1

[C][L]-1[C]=[Z]

[L][Z] = [C] (1)

[U][X] = [Z] (2)

(7)http://numericalmethods.eng.usf.edu



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LU Decomposition(7)Steps:

Step 1. Finding the [U] matrix: Using the Forward EliminationProcedure of Gauss Elimination

2279

21778106

112144

18641525

3

2

1

.

.

.

x

xx

112144

56.18.40

1525

56.212;56.225

64

RowRow

76.48.160

56.18.40

1525

76.513;76.525

144

RowRow

Step a.



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Step 1. Finding the [U] matrix:

Step b:

76.48.160

56.18.40

1525

7.000

56.18.40

1525

5.323;5.38.4

8.16

RowRow

7.000

56.18.40

1525

U

Matrix after Step a:



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Step 2. Finding the [L] matrix: Using the multipliers usedduring the Forward Elimination Procedure

From the step a of

forward elimination

56.225

64

11

21

21

a

a

76.525

144

11

31

31

a

a

112144

1864

1525



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Step 2. Finding the [L] matrix:

15.376.5

0156.2

001

L

From the step b offorward elimination

76.48.160

56.18.40

15255.3

8.4

8.16

22

32

32

a

a



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Step 3.

[L][Z] = [C]

Solve for [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



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Step 3.

735.0

21.965.38.10676.52.279

5.376.52.279

2.96

8.10656.22.177

56.22.177

8.106

213

12

1

zzz

zz

z

735.0

21.96

8.106

3

2

1

z

z

z

Z



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Step 4.

[U][X] = [Z]

Solve for [X] The 3 equations become

7350

2196

8106

7.00056.18.40

1525

3

2

1

..

.

xx

x

735.07.0

21.9656.18.4

8.106525

3

32

321

a

aa

aaa



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Step 4.

From the 3rd equation

050170

7350

735070

3

3

3

.a.

.a

.a.

Substituting in a3 and using the

second equation

21965618432 .a.a.

7019

84

05015612196

84

5612196

2

2

3

2

.a

.

...a

.

a..a



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Step 4.

Substituting in a3

and a2using the first equation

8106525321 .aaa

Hence the SolutionVector is:

050.170.19

2900.0

3

2

1

aa

a

29000

25

0501701958106

2558106 32

1

.

...

aa.a

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LINEAR EQUATIONS SYSTEMS 2