MATHEMATICS FOR ENGINEERING STUDENTS: NUMERICAL METHOD

MATHEMATICS FOR
ENGINEERING STUDENTS

1

Norhafizah binti Arshad

MATHEMATICS FOR
ENGINEERING STUDENT

NUMERICAL METHOD
First Edition

AUTHOR:
NORHAFIZAH BINTI ARSHAD

POLITEKNIK SULTAN IDRIS SHAH

EDITOR:
NORSHAFIZA BINTI MOHAMAD KHAIRUDDIN

POLITEKNIK SULTAN IDRIS SHAH
2

Copyright © Politeknik Sultan Idris Shah 2022
First Edition 2022

All Rights Reserved.

No part of this publication may be reproduced, stored in a retrieval
system or transmitted in any form or by any means, electronics,

mechanical, photocopying, recording or otherwise, without the prior
permission in writing from the Publisher.

National Library of Malaysia Cataloguing-in-Publication Data

Norhafizah Arshad, 1984-
MATHEMATICS FOR ENGINEERING STUDENTS : NUMERICAL
METHOD /
AUTHOR: NORHAFIZAH BINTI ARSHAD ; EDITOR:
NORSHAFIZA BINTI MOHAMAD KHAIRUDDIN. – First Edition.
Mode of Access: Internet
eISBN 978-967-2860-33-4
1. Engineering mathematics.
2. Mathematics.
3. Numerical analysis.
4. Government publicatio3 ns--Malaysia.
5. Electronic books.
I. Norshafiza Mohamad Khairuddin, 1987-.
II. Title.
510.2462

Published by:
Politeknik Sultan Idris Shah
Sungai Lang, 45100, Sungai Air Tawar,
Selangor Darul Ehsan

Phone: 03-32806200
Fax: 03-32806400
Website: https://psis.mypolycc.edu.my/

PREFACE

Thanks to Allah S.W.T for granting the strength and time to
accomplish Mathematics For Engineering Student: Numerical
Method e-book.
First of all, i would like to take immense pleasure in thanking
my family and my facilitator Puan Norafiza Akma binti
Shamsudin for their greatest support along the way and for
their continuous encouragement so that I can gain inner
strength to complete this e-book.
I also would like to thank those who were involved directly or
indirectly in making this e-book success.
Thanks for all support and encouragement. Any suggestions
and comments from teachers and students for improvement in
the future edition of this e-book will be appreciated.

4

Norhafizah Binti Arshad
Politeknik Sultan Idris Shah

iii

ABSTRACT

Mathematics For Engineering Student : Numerical Method is
specially written for diploma students studying Electrical
Engineering Mathematics 3 (DBM 30043) courses in polytechnic
institutions. Based on the Malaysian polytechnic syllabus, this e-
book presents the fundamentals of advanced engineering
mathematics and its application in an easy-to-understand
manner. This e-book is designed to cover the Numerical Method
topic.

The purpose of this e-book is to strengthen student
understanding about the concept of mathematics in Numerical
methods. This e-book will help the student develop the insight
and intuition necessary to master Electrical Engineering
Mathematics 3 (DBM 30043) for Numerical Method. It was
written with the following main objectives:
1. to help the students identify mathematical methods in solving
the mathematical problems.
2. to help the students solve the mathematical problems by using
appropriate techniques and solutions.
3. to give the students the5 ability to practice mathematical
knowledge and skills in different mathematics problem.

With absolute trust in our work, we hope, our holistic efforts
towards making this e-book are paid off if students understand
mathematics conceptually rather than just focusing on the
problem solving part.

iv

TABLE OF CONTENTS

SUBTOPIC PAGE
Preface iii
Abstract iv
Table of Content v
1.0 Solving Linear Equation
3
1.1 Gaussian Elimination Method 18
1.2 LU Decomposition : Doolittle Method 28
1.3 LU Decomposition6 : Crout Method
2.0 Solving Polynomial Equation 39
2.1 Newton-Raphson Method 49
2.2 Simple Fixed Point Iteration Method

v

NUMERICAL
METHOD

Solving Solving
Linear Polynomial
Equation Equation

Gaussian LU Decomposition Simple Newton-
Elimination Fixed Raphson
Point Method
Method Iteration
Doolittle Method
Crout
Method 7 Method

1

Linear Equation:

11 1 + 12 2 + 13 3 + ⋯ + 1 = 1

21 1 + 22 2 + 23 3 + ⋯ + 2 = 2

. .. .

. .. .

. .. .

1 1 + 2 2 + 3 3 + ⋯ + =

In Matrix Form: Ax=B

11 12 … 1 1 1
21 22 … 2 2 2
⋮ = ⋮
⋮ ⋮⋮ ⋮

1 2 … 8

In Augmented Matrix Form:

11 12 … 1 1
21 22 … 2 2

⋮ ⋮ ⋮ ⋮⋮
1 2 …

2

STEP 1  Change into matrix form =
STEP 2  Change into augmented matrix [ | ].

STEP 3  Performing Row Operation on a matrix
to convert the matrix to row-echelon
1 12 13
form 0 1 23
00 1

 Re-arrange the row to obtaining row-
echelon form:

• Interchange between two rows
(Notation: ↔ )

• Ad9d/minus a multiple of one row to
another row (Notation: + ) @
(Notation: − )

• Multiply/divide a row by any
number except 0 (Notation: )

STEP 4  Use back-substitution to solve the system
of equations by equating the LHS and RHS
from the same row [start from bottom]

3



11 1 + 12 2 + 13 3 = 1
21 1 + 22 2 + 23 3 = 2
31 1 + 32 2 + 33 3 = 3

Gaussian elimination method are used to obtain row-echelon form of
matrix and then used Backward substitution to solve Linear Equation

STEP 1 Matrix Form: Ax=B 1
2
11 12 13 1 3
21 22 23 2 =
31 32 33 3

STEP 2 STEP 3

Augmented Matrix Form: Row Row-echelon form matrix

11 12 13 1 operation 1 12 13 1
21 22 23 2 0 1 23 2
31 32 33 3 0 0 1 3

STEP 4

Backward substitution

3= 3 1 + 12 2 + 13 3= 1

2= 2 − 23 3 2 + 23 3= 2
1= 1 − 12 2+ 12 23 3 − 13 3 3= 3

4

Solve the following system using Gaussian method for:
+ − 2 = 1
2 − + = 4
3 + − = 2

01 02
Matrix form: Ax=B Augmented matrix:

1 1 −2 1 1 1 −2 1
2 −1 1 = 4 2 −1 1 ቮ4
3 1 −1 2 3 1 −1 2

03 Performing Row Operation
row-echelon form
1 12 13
0 1 23
00 1

(i) The first row already 1 in row 1, column 1.

1 1 −2 1
2 −1 1 ቮ4
3 1 −1 2

(ii) Subtract the second row to 2 times the first row.
2′ = 2 − 2 1

1 1 −2 1 1 1 −2 1

2 −1 1 ቮ4 → 0 −3 5 ቮ2

3 1 −1 2 3 1 −1 2

5

03 (iii) Subtract the third row to 3 times the first row.
′ = −

1 1 −2 1 1 1 −2 1

0 −3 5 ቮ2 → 0 −3 5 ቮ 2

3 1 −1 2 0 −2 5 −1

03 (iv) Divide the second row by -3

′ → −

1 1 −2 1 1 1 −2 1

0 −3 5 ቮ 2 → 0 1 −5/3 ቮ−2/3

0 −2 5 −1 0 −2 5 −1

03 (v) Add the third row to 2 times the second row.
′ = +

1 1 −2 1 1 1 −2 1

0 1 −5/3 ቮ−2/3 → 0 1 −5/3 ቮ−2/3

0 −2 5 −1 0 0 5/3 −7/3

03 (vi)Divide the third row by -5/3
′ →

−


1 1 −2 1 1 1 −2 1

0 1 −5/3 ቮ−2/3 → 0 1 −5/3 ቮ−2/3

0 0 5/3 −7/3 0 0 1 −7/5

6

04 04
The first row of the matrix
Use back-substitution. represents:
+ − 2 = 1
1 1 −2 1
0 1 −5/3 ቮ−2/3 Substitute = − 7 , = −3 into
0 0 1 −7/5
5
Equating the LHS and RHS from
the same row. Start from bottom the first row equation.
which is the third row.
7
7 + −3 − 2 − 5 = 1
= − 5
14
The second row of the matrix − 3 + 5 = 1
represents:
1
52 − 5 = 1
− 3 = − 3
1
Back-substitute = − 7 into the = 1 + 5

5 6
= 5
second equation.

57 2
− 3 − 5 = − 3

72 Therefore,
+ 3 = − 3
= 6 , = −3 and = − 7
5 5

27
= − 3 − 3

= −3

7

MODE BUTTON

+ − 2 = 1
2 − + = 4
3 + − = 2

Press
3x

14

67
= 5 , = −3 , = − 5

8

Solve the following system using Gaussian method for:
+ 2 = 4

− 2 − = −8
3 − − 4 = −15

01 Matrix form: Ax=B 02
Augmented matrix:
01 2 4
1 −2 −1 = −8 01 2 4
3 −1 −4 −15 1 −2 −1 ቮ −8
3 −1 −4 −15

03 Performing Row Operation
row-echelon form
1 12 13
0 1 23
00 1

(i) Interchange first row and second row.

01 2 4 1 −2 −1 −8

1 −2 −1 ቮ −8 = 0 1 2 ቮ 4

3 −1 −4 −15 3 −1 −4 −15

(ii) The second row already has a 0 in row 2, column 1.

1 −2 −1 −8
0 1 2ቮ 4
3 −1 −4 −15

9

03 (iii) Subtract the third row to 3 times the first row.
3′ = 3 − 3 1

1 −2 −1 −8 1 −2 −1 −8

0 1 2 ቮ 4 → 0 1 2 ቮ4

3 −1 −4 −15 0 5 −1 9

03 (iv) The second row already 1
in row 2, column 2.

1 −2 −1 −8
0 1 2 ቮ4
0 5 −1 9

03 (v) Subtract the third row to 5 times the second row.
′ = −

1 −2 −1 −8 1 −2 −1 −8

0 1 2 ቮ4 → 0 1 2 ቮ 4

0 5 −1 9 0 0 −11 −11

03 (vi)Divide the third row by -11

′ → −

1 −2 −1 −8 1 −2 −1 −8

0 1 2 ቮ 4 → 0 1 2 ቮ4

0 0 −11 −11 0 0 11

10

04 04
The first row of the matrix
Use back-substitution. represents:
− 2 − = −8
1 −2 −1 −8 Substitute = 1 , = 2 into the
0 1 2 ቮ4 first row equation.
0 0 11
− 2 2 − 1 = −8
Equating the LHS and RHS from
the same row. Start from bottom − 4 − 1 = −8
which is the third row.
− 5 = −8
= 1 = −8 + 5
= −3
The second row of the matrix
represents: Therefore,

+ 2 = 4 = −3, = 2 and = 1

Substitute = 1 into the second
row equation.

+ 2 1 = 4
+ 2 = 4
= 4 − 2
= 2

11

MODE BUTTON

+ 2 = 4
− 2 − = −8
3 − − 4 = −15

Press
3x

18

= −3, = 2, = 1

12

Solve the following system using Gaussian method for:
−3 + 2 + = 8

9 − 5 − 2 = −21
−5 + 3 + = 17

01 02
Matrix form: Ax=B
Augmented matrix:
−3 2 1 8
9 −5 −2 = −21 −3 2 1 8
−5 3 1 17 9 −5 −2 ቮ−21
−5 3 1 17

03 Performing Row Operation row-echelon form
1 12 13
0 1 23
00 1

(i) Divide the first row by -3

′ →
−

−3 2 1 8 1 −2/3 −1/3 −8/3

9 −5 −2 ቮ−21 → 9 −5 −2 ቮ −21

−5 3 1 17 −5 3 1 17

(ii) Subtract the second row to 9 times the first row.
′ = −

1 −2/3 −1/3 −8/3 1 −2/3 −1/3 −8/3

9 −5 −2 ቮ −21 → 0 1 1 ቮ3

−5 3 1 17 −5 3 1 17

13

03 (iii) Add the third row to 5 times the first row.
3′ = 3 + 5 1

1 −2/3 −1/3 −8/3 1 −2/3 −1/3 −8/3

01 1 ቮ3 →0 1 1 ቮ3

−5 3 1 17 0 −1/3 −2/3 11/3

03 (iv) The second row already 1 in row 2, column 2.

1 −2/3 −1/3 −8/3

01 1 ቮ3

0 −1/3 −2/3 11/3

03 (v) Add the third row to (1/3) times the second row.

′ = +

1 −2/3 −1/3 −8/3 1 −2/3 −1/3 −8/3

01 1 ቮ3 →0 1 1 ቮ3

0 −1/3 −2/3 11/3 0 0 −1/3 14/3

03 (vi)Divide the third row by -1/3

′ →
−


1 −2/3 −1/3 −8/3 1 −2/3 −1/3 −8/3

01 1 ቮ3 →0 1 1 ቮ3

0 0 −1/3 14/3 00 1 −14

14

04

Use back-substitution.

1 −2/3 −1/3 −8/3 04

01 1 ቮ3 The first row of the matrix

00 1 −14 represents:

21 8
− 3 − 3 = − 3
Equating the LHS and RHS from
the same row. Start from bottom Substitute = −14, = 17 into
which is the third row. the first row equation.

= −14

21 8
− 3 17 − 3 −14 = − 3
The second row of the matrix
represents: 34 14 8
− 3 + 3 = − 3
+ = 3
20 8
Substitute = −14 into the − 3 = − 3
second row equation.
8 20
+ −14 = 3 = − 3 + 3
− 14 = 3
= 3 + 14 = 4
= 17
Therefore,
= 4, = 17 and = −14

15

MODE BUTTON

-3 + 2 + = 8
9 − 5 − 2 = −21
−5 + 3 + = 17

Press
3x

22

= 4, = 17, = −14

16

Solve each of the following system
by using Gaussian method:

+ 2 = 5 + + = 2
3 + 2 + = 10 + 2 + 3 = 1
2 + 4 + = 13 2 − − 3 = 6

= 1, = 2 , = 3 = 2, = 1 , = −1

2 + 3 + 4 = 7 2 + 9 − 7 = −1
+ 2 + 3 = 2 −8 − 39 + 28 = 2
+ 5 + 7 = 1 −3 − 12 + 9 = 1

= 4, = 5 , = −4 = 0, = 2/3 , = 1

23

−a + 3b − 2c = −5 2 − + 3 = −25
3a − 3b + c = 3 5 + 4 − 3 = −1
−2a + b = −1 3 − 2 − = −17

= 2, = 3 , = 6 = −5, = 3 , = −4

17

LU DECOMPOSITION is method of solving a system of
linear equations of the form , = . Matrix A is

decomposed into a product of a lower triangular matrix, L
and an upper triangular matrix, U, that is = .

A LU

11 12 13 1 0 0 11 12 13
21 22 23 = 21 1 0 0 22 23
31 32 33 31 32 1 0 0 33

lower triangular upper triangular
matrix
matrix

18

STEP 1  Change into matrix form =

 Apply matrix A to LU Decomposition which

is the lower triangular matrix, L and upper

STEP 2 triangular matrix, U by using Doolittle

Method A=LU

A LU

. 1 0 0 11 12 13
11 12 13

21 22 23 = 21 1 0 0 22 23
31 32 33 31 32 1 0 0 33

STEP 3  Construct the lower triangular matrix, L and

upper triangular matrix, U by find all the

elements in th25 e matrices LHS = RHS.

STEP 4  Form matrix L and U

STEP 5  Find Ly = B by using forward substitution

STEP 6  Find Ux = y by using backward substitution.

19

Solve the following system using Doolittle method

+ + = 2
+ 2 + 3 = 1
2 − − 3 = 6

01 Matrix form: Ax=B

11 1 2
12 3 = 1
2 −1 −3 6

02 Apply matrix A to LU Decomposition using
Doolittle method

26

=

A LU
1 0 0 11 12 13
11 1
1 2 3 = 21 1 0 0 22 23
31 32 1 0 0 33
2 −1 −3

20

03 Construct matrix L and U by finding all the elements in
the matrices

11 1 11 12 13

1 2 3 = 21 11 21 12 + 22 21 13 + 23

2 −1 −3 31 11 31 12 + 32 22 31 13 + 32 23 + 33

11 = 1 12 = 1 13 = 1

21 11 = 1 21 12 + 22 = 2 21 13 + 23 = 3
21(1) = 1 1 (1) + 22 = 2 1 (1) + 23 = 3

21 = 1 22 = 1 23 = 2

31 11 = 2 31 12 + 32 22 = −1 31 13 + 32 23 + 33 = −3
31(1) = 2 2 (1) + 32(1) = −1 2 1 + −3 2 + 33 = −3

31 = 2 32 = −3 −4 + 33 = −3
33 = 1

04 Form matrix L and U

11 = 1 27 12 = 1 13 = 1

21 = 1 22 = 1 23 = 2

31 = 2 32 = −3 33 = 1

=
L U1 0 0 11 12 13
A

11 1
1 2 3 = 21 1 0 0 22 23
31 32 1 0 0 33
2 −1 −3

11 1 1 0 0111

1 2 3 = 127 1 0 0 1 2

2 −1 −3 2 −3 1 0 0 1

21

05 Find Ly = B by using 06 Find Ux = y by using
forward substitution. backward substitution.

1 0 0 1 2 1 1 1 2
1 1 0 2 = 1
2 −3 1 3 6 0 1 2 = −1

0 0 1 −1

Equating the LHS and RHS Equating the LHS and RHS
from the same row. Start from from the same row. Start from
first row. third row.

1 = 2 = −1

The second row of the The second row of the matrix
matrix represents: represents:

1 + 2 = 1 + 2 = −1

Substitute 1 = 2 into the Substitute = −1 into the
second row equation. second row equation.

2 + 2 = 1 + 2(−1) = −1
2 = −1 − 2 = −1
= 1
28
The first row of the matrix
The third row of the matrix represents:
represents:

2 1 − 3 2 + 3 = 6 + + = 2
Substitute 1 = 2 , 2 = −1
into the third equation. Substitute = −1, = 1 into
the first equation.
2 2 − 3 −1 + 3 = 6
7 + 3 = 6 x + 1 + (−1) = 2
3 = 6 − 7 = 2
3 = −1
Therefore,
Therefore, = 2, = 1 = −1
1 = 2, 2 = −1, 3 = −1

22

MODE BUTTON

+ + = 2
+ 2 + 3 = 1
2 − − 3 = 6

Press
3x

29

= 2, = 1, = −1

23

Solve the following system using Doolittle method

5 − + 7 = −1
4 + = 1

7 + + 2 = 1

01 Matrix form: Ax=B

5 −1 7 −1
410 = 1
712 1

02 Apply matrix A to LU Decomposition using
Doolittle method

30

=

A LU
1 0 0 11 12 13
5 −1 7
4 1 0 = 21 1 0 0 22 23
31 32 1 0 0 33
712

24

03 Construct matrix L and U by finding all
the elements in the matrices

5 −1 7 11 12 13

4 1 0 = 21 11 21 12 + 22 21 13 + 23

712 31 11 31 12 + 32 22 31 13 + 32 23 + 33

11 = 5 12 = −1 13 = 7

21 11 = 4 21 12 + 22 = 1 21 13 + 23 = 0
21(5) = 4 4/5 (−1) + 22 = 1 4/5 (7) + 23 = 0
21 = 4/5
22 = 9/5 23 = −28/5
31 11 = 7
31(5) = 7 31 12 + 32 22 = 1 31 13 + 32 23 + 33 = 2
31 = 7/5 7/5 (−1) + 32 9/5 = 1 7/5 7 + 4/3 −28/5 + 33 = 2
− 7/5 + 9/5 32 = 1
32 = 1 + 7/5 ÷ 9/5 33 = 2 − 49/5 + 112/15
33 = −1/3
32 = 4/3

04 Form matrix L and U

11 = 5 31 12 = −1 13 = 7
21 = 4/5 22 = 9/5 23 = −28/5

31 = 7/5 32 = 4/3 33 = −1/3

=
A LU
5 −1 7 1 0 0 11 12 13

4 1 0 = 21 1 0 0 22 23
31 32 1 0 0 33
712

5 −1 7 1 0 0 5 −1 7
31
4 1 0 = 4/5 1 0 0 9/5 −28/5

7 1 2 7/5 4/3 1 0 0 −1/3

25

05 Find Ly = B by using 06 Find Ux = y by using
forward substitution. backward substitution.

1 0 0 1 −1 5 −1 7 −1
4/5 1 0 2 = 1 0 9/5 −28/5 = 9

7/5 4/3 1 3 1 0 0 −1/3 5
0

Equating the LHS and RHS from Equating the LHS and RHS from
the same row. Start from first row. the same row. Start from third row.

1 = −1 1
− 3 = 0
The second row of the matrix
represents: = 0

4 The second row of the matrix
5 1 + 2 = 1 represents:

Substitute 1 = −1 into the 9 28 9
second row equation. 5 − 5 = 5
Substitute = 0 into the second
4 row equation.
5 −1 + 2 = 1
9 28 9
2 = 9/5 5 − 5 (0) = 5

The third row of the matrix 32 99
represents: 5 = 5

7 4 = 1
5 3
1 + 2 + 3 = 1 The first row of the matrix

Substitute 1 = −1 , 2 = 9/5 represents:
into the third equation.
5 − + 7 = −1

7 −1 +4 9 + 3 = 1 Substitute = 0, = 1 into the
5 5 first row equation.
3
5 − (1) + 7(0) = −1
− 7 + 36 + 3 = 1 5 = −1 + 1
5 15 = 0

3 = 1 + 7- 36
15
5

3 = 0 Therefore,
= 0, = 1 = 0
Therefore,

1 = −1, 2 = 9/5, 3 = 0

26

MODE BUTTON

5 − + 7 = −1
4 + = 1

7 + + 2 = 1

Press
3x

33

= 0, = 1, = 0

27

LU DECOMPOSITION is method of solving a system of
linear equations of the form , = . Matrix A is

decomposed into a product of a lower triangular matrix, L
and an upper triangular matrix, U, that is = .

A L U

11 12 13 11 0 0 1 12 13
21 22 23 = 21 22 0 0 1 23
31 32 33 31 32 33 00 1

lower triangular upper triangular
matrix matrix

34

28

STEP 1  Change into matrix form =

 Apply matrix A to LU Decomposition which

is the lower triangular matrix, L and upper

STEP 2 triangular matrix, U by using Crout Method

A=LU

A LU
. 11 12 13 11 0 0 1 12 13

21 22 23 = 21 22 0 0 1 23
31 32 33 31 32 33 0 0 1

STEP 3  Construct the lower triangular matrix, L and

upper triangular matrix, U by find all the

elements in th35 e matrices LHS = RHS.

STEP 4  Form matrix L and U

STEP 5  Find Ly = B by using forward substitution

STEP 6  Find Ux = y by using backward substitution.

29

Solve the following system using Crout method

4 − 4 + 3 = 5
1 − 2 + = 3
2 − 1 + 5 = 12

01 Matrix form: Ax=B

4 −4 3 5
1 −2 1 = 3
2 −1 5 12

02 Apply matrix A to LU Decomposition using
Crout method

36

=

A LU
11 0 0 1 12 13
4 −4 3

1 −2 1 = 21 22 0 0 1 23
31 32 33 0 0 1
2 −1 5

30

03 Construct matrix L and U by finding all
the elements in the matrices

4 −4 3 11 11 12 11 13

1 −2 1 = 21 21 12 + 22 21 13 + 22 23

2 −1 5 31 31 12 + 32 31 13 + 32 23 + 33

11 = 4 11 12 = −4 11 13 = 3
21 = 1 (4) 12 = −4 (4) 13 = 3
31 = 2 12 = −1 13 = 3/4

04 21 12 + 22 = −2 21 13 + 22 23 = 1
1 −1 + 22 = −2 3
22 = −1 1 4 − 1 23 = 1

31 12 + 32 = −1 23 = −1/4
2(−1) + 32 = −1
32 = 1 31 13 + 32 23 + 33 = 5

2(3/4) + (1)(−1/4) + 33 = 5
5
4 + 33 = 5

33 = 15/4

Form matrix L and U

11 = 4 37 12 = −1 13 = 3/4

21 = 1 22 = −1 23 = −1/4

31 = 2 32 = 1 33 = 15/4

=
A LU
11 0 0 1 12 13
4 −4 3

1 −2 1 = 21 22 0 0 1 23
31 32 33 0 0 1
2 −1 5

4 −4 3 40 0 1 −1 3/4
37

1 −2 1 = 1 −1 0 0 1 −1/4

2 −1 5 2 1 15/4 0 0 1

31

05 Find Ly = B by using 06 Find Ux = y by using
forward substitution. backward substitution.

4 0 0 1 5 1 −1 3/4 5/4

1 −1 0 2 = 3 0 1 −1/4 = −7/4

2 1 15/4 3 12 0 0 1 3

Equating the LHS and RHS from Equating the LHS and RHS from
the same row. Start from first row. the same row. Start from third row.

4 1 = 5 = 3
1 = 5/4
The second row of the matrix

The second row of the matrix represents:
represents:
− 1 = − 7
4 4

1 − 2 = 3 Substitute = 3 into the second
row equation.
Substitute 1 = 5 into the second
4
row equation.
− 1 3 = − 7
5 38
4 − 2 = 3 44

2 = −7/4 − 3 = − 7

The third row of the matrix 44
represents:
= − 7 + 3
2 1 + 2 + 15 3 = 12
4 44

Substitute 1 = 5, 2 = − 7 into = −1
4
4 The first row of the matrix
represents:

− + 3 = 5

44

the third row equation. Substitute = 3, = −1 into the
first row equation.
2 5 + −7 + 15 3 = 12
4 4
4

3+ 15 3 = 12 − (−1) + 3 (3) = 5
4
4 44

15 3 = 45 = 5 − 9 − 1
4 4
44
3 = 3
= −2

Therefore, Therefore,
= −2, = −1, = 3
1 = 5, 2 = − 7 , 3 = 3
4
4

32

MODE BUTTON

4 − 4 + 3 = 5
1 − 2 + = 3
2 − 1 + 5 = 12

Press
3x

39

= −2, = −1, = 3

33

Solve the following system using Crout method

7 − 3 − 3 = 2
− + = −1
− + = 1

01 Matrix form: Ax=B

7 −3 −3 2
−1 1 0 = −1
−1 0 1 1

02 Apply matrix A to LU Decomposition using
Crout method

40

=

A LU
11 0 0 1 12 13
7 −3 −3

−1 1 0 = 21 22 0 0 1 23
31 32 33 0 0 1
−1 0 1

34

03 Construct matrix L and U by finding all
the elements in the matrices

7 −3 −3 11 11 12 11 13

−1 1 0 = 21 21 12 + 22 21 13 + 22 23

−1 0 1 31 31 12 + 32 31 13 + 32 23 + 33

11 = 7 11 12 = −3 11 13 = −3
21 = −1 7 12 = −3 7 13 = −3
31 = −1 12 = −3/7 13 = −3/7

21 12 + 22 = 1 21 13 + 22 23 = 0
(−1) −3/7 + 22 = 1 −1 −3/7 + 4/7 23 = 0

22 = 4/7 23 = −3/7 ÷ 4/7
23 = −3/4
31 12 + 32 = 0
−1 −3/7 + 32 = 0 31 13 + 32 23 + 33 = 1
(−1) −3/7 + −3/7 −3/4 + 33 = 1
32 = −3/7
33 = 1 − 3/7 − 9/28
33 = 1/4

04 11 = 7 Form matrix L and U 13 = −3/7

41 12 = −3/7

21 = −1 22 = 4/7 23 = −3/4

31 = −1 32 = −3/7 33 = 1/4

=

A LU
7 −3 −3 11 0 0 1 12 13

−1 1 0 = 21 22 0 0 1 23
31 32 33 0 0 1
−1 0 1

7 −3 −3 7 0 0 1 −3/7 −3/7

−1 1 0 = −1 4/7 0 0 1 −3/4

−1 0 1 −1 −3/7 1/4 0 0 1

35

05 Find Ly = B by using 06 Find Ux = y by using
forward substitution. backward substitution.

70 0 1 2 1 −3/7 −3/7 2/7

−1 4/7 0 2 = −1 0 1 −3/4 = −5/4

−1 −3/7 1/4 3 1 00 1 3

Equating the LHS and RHS from Equating the LHS and RHS from
the same row. Start from first row. the same row. Start from third row.

7 1 = 2 = 3

1 = 2/7

The second row of the matrix The second row of the matrix
represents:
represents:
35
4 − 4 = − 4
− 1 + 7 2 = −1
2 Substitute = 3 into the second
Substitute 1 = 7 into the second row equation.

row equation.

24 − 3 3 = − 5
− 7 + 7 2 = −1
44
9 5
2 = −1 + 2 ÷ 4 − 4 = − 4
7 7
= − 5 + 9
2 = −5/4 4 4
42
The third row of the matrix = 1

represents: The first row of the matrix
represents:
− 1 − 3 2 + 1 3 = 1
7 4 332
− 7 − 7 = 7
Substitute 1 = 2, 2 = − 5 into
4
7
the third row equation.
Substitute = 3, = 1 into the
−2−3 −5 + 1 3 = 1 first row equation.
4
77 4 3 3 2
7 7 7
1+ 1 3 = 1 − (1) − (3) =
4
4 = 2 + 3 + 9

3 = 1−1 ÷ 1 777
4
4

3 = 3 = 2

Therefore, Therefore,
= 2, = 1, = 3
1 = 2 , 2 = − 5 , 3 = 3
7 4
36

MODE BUTTON

7 − 3 − 3 = 2
− + = −1
− + = 1

Press
3x

43

= 2, = 1, = 3

37

Solve each of the following system by using
Doolittle and Crout method:

x + 2y = 5 x+y+z=2
3x + 2y + z = 10 x + 2y + 3z = 1
2x + 4y + z = 13 2x − y − 3z = 6

= 1, = 2 , = 3 = 2, = 1 , = −1

2x + 3y + 4z = 7 2p + 9q − 7r = −1
x + 2y + 3z = 2 −8 − 39 + 28r = 2
x + 5y + 7z = 1 −3p − 12q + 9r = 1

= 4, = 5 , = −4 = 0, = 2/3 , = 1

44

−a + 3b − 2c = −5 2x − y + 3z = −25
3a − 3b + c = 3 5x + 4y − 3z = −1
−2a + b = −1 3x − 2y − z = −17

= 2, = 3 , = 6 = −5, = 3 , = −4

38

Polynomial Equation defined by
= 0 + 1 1 + 2 2 + ⋯ +

= 0
is the coefficient, is the
variable and is the exponent.
If ≠ 0, has degree n.

45

The Newton-Raphson method is
a method for approximating the
roots of polynomial equations.

( )
+1 = − ′( )

39

STEP 1  Let = 0.
Differentiate to find ′( )

STEP 2  Substitute and ′ into the
formula of Newton Raphson Method

( )
+1 = − ′( )

STEP 3  Choose a suitable initial root, 0.

 Substitute 0 into formula to

determine 1.
 Repeat the process until find the

repeated root.

40

Determine the root of the function ( ) = 3 + 4 2 + 7
correct to three decimal places using Newton Raphson
method with the initial guess of 0 = −4.

Let = 0, differentiate f(x) to find ′( )
( ) = 3 + 4 2 + 7 = 0
Let ( ) = 3 + 4 2 + 7
′ = 3 2 + 8

Substitute f(x) and f’(x) into the formula
( )

+1 = − ′( )
3 + 4 2 + 7

+1 = − 3 2 + 8

Given that 0 = −4, 47

1 = −4 − −4 3+4 −4 2+7 = −4.43
3 −4 2+8 −4

2 = −4.438 − −4.438 3+4 −4.438 2+7 = −4.369
3 −4.438 2+8 −4.438

3 = −4.369 − −4.369 3+4 −−4.369 2+7 = −4.367
3 −4.369 2+8 −4.369

4 = −4.367 − −4.367 3+4 −4.367 2+7 = −4.367
3 −4.367 2+8 −4.367

∴ Since the root of x is repeated, then the root − 4.367

41

0 = −4 3 + 4 2 + 7

+1 = − 3 2 + 8

48

−4.438 (3 . )

−4.369 (3 . )
−4.367(3 . )
−4.367(3 . )

42

Show that = − − + , has a root in the interval [2,3].
Then, solve the equation by using Newton Raphson Method with
the initial guess of = . and give the answer corrects to 4
decimal places.

= − − + When = 3,
When = 2, = (3)4−2 3 3 − 3 + 1 = 25
= (2)4−2 2 3 − 2 + 1 = −1

∴ Since f(2)<0 and f(3)>0, there is a root in [2,3].

Let = 0, find ′( ) Substitute f(x) and f’(x) into the formula

= 4 − 2 3 − + 1 = 0 4 − 2 3 − + 1
Let = 4 − 2 3 − + 1 +1 = − 4 3 − 6 2 − 1
′ = 4 3 − 6 2 − 1

Given that 0 = 2.5,

49

1 = 2.5 − 2.5 4−2 2.5 3− 2.5 +1 = 2.2370
4 2.5 3−6 2.5 2−1

2 = 2.2370 − 2.2370 4−2 2.2370 3− 2.2370 +1 = 2.1340
4 2.2370 3−6 2.2370 2−1

3 = 2.1340 − 2.1340 4−2 2.1340 3− 2.1340 +1 = 2.1181
4 2.1340 3−6 2.1340 2−1

4 = 2.1181 − 2.1181 4−2 2.1181 3− 2.1181 +1 = 2.1177
4 2.1181 3−6 2.1181 2−1

5 = 2.1177 − 2.1177 4−2 2.1177 3− 2.1177 +1 = 2.1177
4 2.1177 3−6 2.1177 2−1

∴ Since the root of x is repeated, then the root 2.1177

43

0 = 2.5 4 − 2 3 − + 1

+1 = − 4 3 − 6 2 − 1

50

2.2370(4 . )
2.1340(4 . )
2.1181(4 . )
2.1177(4 . )
2.1177(4 . )

44