MDB3053 Chap9 LinearAlgebra Part 1 May16

7/26/2019 MDB3053 Chap9 LinearAlgebra Part 1 May16

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RECAP ON CHAPTER 6

1. Bracketing Method:

Bisection Method

False-Position Method

2. Open Method:

Fixed Point Iteration

Newton-Raphson Method

Secant Method


2/27

2x

1x

How to solve a system of

simultaneous algebraic

equations using CM ?

Cokes example


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4/27

LINEAR ALGEBRAIC EQUATIONS

Example of a system of L.A.E

Why coupled? Each equation has terms in common with the

others, that is x1, x2 and x3.

Why linear? It means the effects areproportionalto their cause.

Each equation contains only first order terms of x1, x2 and x3. No

terms like x12, or log(x3) or 1/x2x3.

132

132

8253

321

321

321

xxx

xxx

xxx

Three coupled,

linear equations

bxA


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GENERAL FORM OF L.A.E

The general form of a system of L.A.E forn number ofequations

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111coefficients

constants

unknown

variables


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MATRIX FORM OF L.A.E

Linear Algebraic Equations in matrix form

54

132

21

21

xx

xx

nnmnm

n

b

b

b

x

x

x

aa

aa

A

11

1

111

,,

bxA

m x n

coefficient

matrix, A

n x 1 column vector of

unknowns

n x 1 column vector

of constants

5

1

14

32

5

1

14

32

2

1

x

x

or

Example: augmented formbxA


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TRY THIS

Write the matrix representation for the L.A.Eshown below:

132

132

8253

321

321

321

xxx

xxx

xxx

1

1

8

321

132

253


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SPECIAL TYPES OF MATRICES

33

2322

131211

00

0

a

aa

aaa

333231

2221

11

0

00

aaa

aa

a

33

22

11

00

00

00

a

a

a

100

010

001

Diagonal matrix

Identity matrix, I

Lowertriangular matrix, L

Uppertriangular matrix, U


9/27

WAYS TO SOLVE L.A.E

Graphical Methodby plottingCramers Ruleby determinant concept

Direct Method Gaussian Elimination(Exact Sol) Gauss-Jordan

LU Factorization

Iterative Method Jacobis Method

(Approx. Sol) Gauss-Seidel Method


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GRAPHICAL METHOD

The SOLUTION isshown graphically

They are in the forms of

straight linebecause of

linear equations

22

1823

21

21

xx

xx

)2(..........1

2

)1.......(2

39

12

12

xx

xx


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BASIC CONCEPTS OF SOLUTION

Unique Solution (Nonhomogeneous equation)

Without Solution

2x

1x

63

593

21

21

xx

xx

1x

2x

parallel lines

864

432

21

21

xx

xxLinearly

dependentequations


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Infinite # of Solutions (No unique solutions)

ill-conditioned slopes are close to each

other, very hard to determine the solution

BASIC CONCEPTS OF SOLUTION (CONT)

864

432

21

21

xx

xx

2x

1x

linearly

dependant

5.522

7.52.22

21

21

xx

xx

1x

2x

identical lines


13/27

GAUSSIAN ELIMINATION

Gaussian = Forward + Back

Elimination Elimination Substitution

To transform the original

matrix, A into an upper, U

triangular matrix

Unknowns are solved

through Back Substitution

in the upper triangular

matrix

Most frequently used Direct Method

Principle: to reduce coefficient matrix, A into

equivalent upper triangular form

Consists ofTWO stages:


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Fig. 9.3

upper

triangular

matrix

Augmented

form


15/27

FORWARD ELIMINATION

Forward Elimination stage reduces the augmented

matrix into an uppertriangular matrix.

It requires Elementary Row Transformations

(3 Basic approaches). You can:

1. Interchange any two row in position

2. Multiply (or divide) a row by a

nonzero scalar , k=2 (eg. 2 R2)

3. Multiply (or divide) a row by ascalar and adding it (or subtracting

it from) another row (eg R1 + 2 R2)

1

5

32

14

5

1

14

32

2

5

64

14

7

5

70

14


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Example: Reducing matrix to upper matrix

1 2 2 3

Consider the matrix 4 4 2 6

4 6 4 10

Lets perform the transformation at second row R2:

new R2

It means To get the second row of the new matrix,

multiply every element of the first row by 4 and subtract

it from the element in the second row.

12

'

2

1

4RRR

R1

R2

R3

1

4 Elimination factor or

multiplier


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Thus, the new 2nd row becomes :

R2 = [4 4 2 | 6] 4 [1 2 2 | 3]

= [0 4 6 |6]

EXAMPLE (cont)

1221

4RRR

133

1

4RRR

You can perform one

more transformation to

obtain U matrix

Original Matrix:

1 2 2 3

4 4 2 6

4 6 4 10

Transformation:

1 2 2 3

0 -4 -6 -6

0 -2 -4 -2

Next, transformation at 3rd row

Thus, the new 3rd row becomes:

R3 = [4 6 4 | 10] 4 [1 2 2 | 3]

= [ 0 2 4 |2]R

3

'' = [0 -2 -4 | -2] 1/2 [0 -4 -6 | -6]

= [ 0 0 1 | 1]


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BACK SUBSTITUTION

Suppose that the next forward elimination step results in

the following augmented matrix:

1 2 2 3

0 4 6 6

0 0 1 1

All the entries

below the

diagonal are

ZEROS

By using Back Substitution, SOLUTIONS are:

R3: x3 = 1 x3 =1

R2: 4x26x3 =6 x2 = 3

R1 : x1 + 2x2 + 2x3 = 3 x1 =1

Direct method

yields exact sol


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CLASS ACTIVITY #1

Use fraction number to reduce round-off errors.

3232

14

232

321

321

321

xxx

xxx

xxx

Try this in EXCEL: (p296)

To inverse, highlight the

range, enter minverse(..)

and press Ctrl+Shift+Enter

To multiply,enter mmult(..)

Try this in MATLAB: (p298)

>> A=[2 1 3;4 1 1;-2 3 2]

>> b=[2 ; 1 ; 3 ]

>> x = A\b


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CLASS ACTIVITY #1 (cont)

Step 1: Augmented Matrix Form:

3232

14

232

321

321

321

xxx

xxx

xxx

3

1

2

232

114

312

462423

5

2

232

530

312

2 112

RRR

Step 2: Forward Elimination by performing

elementary row transformation


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Pitfalls of Gaussian Elimination

Division by zero. It is possible that during bothelimination and back-substitution phases a division by

zero can occur. can be solved by (Partial) Pivoting

Round-off errors. Because computer can only store afixed no. of digits to use more S.F. orfractions

Ill-conditioned systems. Systems where small changes

in coefficients result in large changes in the solution. Ithappens when two or more equations are nearly

identical.

1x

2x


22/27

GAUSS-JORDAN

A variation of Gauss elimination but 50% more

computational work than Gaussian Elimination

Elimination step results in an identity matrix (not

upper triangular matrix) due to normalization step

No back substitution required

3

2

1

332331

232221

131211

c

c

c

aaa

aaa

aaa

3

2

1

100

010

001

h

h

h

Elimination +Normalization

Identity

matrix


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Fig 9.9

IdentityMatrix

No back

substitution !

Elimination +

Normalization


24/27

EXAMPLE

Suppose that the forward elimination and normalizing

steps result in the following augmented matrix:

1 0 0 5

0 1 0 3

0 0 1 7

The SOLUTION will be : x1 = 5 x2 = 3 and x3 = 7

Hence there is no need for back substitution


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CLASS ACTIVITY #2

Solve the system of equations below using

Gauss-Jordan method

123

6252

72

321

321

321

xxx

xxx

xxx

Ans: x1 = 2 ; x2 = 8; x3 = 21


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MATRIX INVERSION

Inverse Matrix A-1

can be found through Gauss-Jordan Method

STEP 1 : Form

STEP 2 : Transform to using theElementary Row Transformations.

STEP 3 : Finally, B = A -1, where B is the inversematrix

IA

IA BI


27/27

CLASS ACTIVITY #3

MATRIX INVERSION

Find the inverse of the matrix below usingthe Gauss-Jordan Method

612

131

221

A

135

124

410171

A

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MDB3053 Chap9 LinearAlgebra Part 1 May16