MATRIX: ENGINEERING MATHEMATICS

Page 1

Authors

CHE KHAIRIL YUSRI BIN YUSOF

POLITEKNIK KUCHING SARAWAK

MINISTRY OF HIGHER EDUCATION

KM22, JALAN MATANG, 93050 KUCHING, SARAWAK.

Phone No. : (082) 845596/7/8

Fax No. : (082) 845023

E-mail : poliku.info@poliku.edu.my

Website : http://www.poliku.edu.my/

Copyright © 2023 Politeknik Kuching Sarawak

e ISBN 978-967-2953-92-0

All rights reserved. No parts of this publication may be copied, stored in form or by any means, electronic, mechanical, photocopying and recording or otherwise or by any means for reproduced without the prior permission of Politeknik Kuching Sarawak.

Published by:

Politeknik Kuching Sarawak

Ministry Of Higher Education

Cataloguing-in-Publication Data

Perpustakaan Negara Malaysia

A catalogue record for this book is available from the National Library of Malaysia

eISBN 978-967-2953-92-0

PREFACE

Alhamdulillah, praise be to Allah, for the completion of the book titled "Matrix." It is indeed a great achievement to have finished this book This book will serve as valuable additional notes and exercises for Polytechnic students and contribute to their teaching and learning experience. I would like to express my utmost appreciation and gratitude to all the parties and individuals who have played a part in contributing to this book. May Allah bless all those who have contributed and make this work beneficial for the students who will utilize it. May it serve as a valuable resource, helping them in their studies and furthering their understanding of the subject matter.

Che Khairil Yusri Bin Yusof

Jabatan Matematik Sains dan Komputer

Politeknik Kuching Sarawak

09 Dzulhijjah 1444 Hijri

26 Jun 2023

ABSTRAK MATRIX Oleh

The Engineering Mathematics 1 course is offered to polytechnic students in semester 1. The feedback received from students during the teaching and learning found that students specifically requested trigonometry topics. In response, the author took the initiative to produce a book that focuses on the topics studied by students. This strategy is seen as more effective as a simple reference for students. Therefore, the author hopes that this book will be one of the effective tools in increasing students' interest in mathematics. This approach is expected to provide them with an effective tool to prepare for the end-of-semester exam.

KEY WORDS : Matrix, Engineering Mathematics dan Polytechnic

Contents 1. Construct Matrix ..................................................................................................................................... 1 1.1 Identify the characters of matrices 1 a. Elements of Matrices ........................................................................................................................ 2 b. Order of matrices 4 1.2 State the transportation of a matrix ………………………………………………………………………………………………5 1.2.1 Symmetrical Matric ...................................................................................................................... 6 1.2.2 Upper Triangular Matrix 7 1.2.3 Lower Triangular Matrix ............................................................................................................... 7 2. Demonstrate the operation Matrices 9 2.1 Calculate the operation Matrices 9 a. Addition ............................................................................................................................................ 9 b. Subtraction 10 c. Multiplication ................................................................................................................................. 11 d. Inverse ............................................................................................................................................. 14 3. Demonstrate Simultaneous Equations 21 3.1 Solve simultaneous equations using inverse matrix ......................................................................... 21 3.2 Solve simultaneous equations using Cramer’s Rule 27 ANSWERS .................................................................................................................................................... 35 REFERENCE…………………………..……………………………………………………………………………………………………………….38

1. Demonstrate the operation Matrices

A matrix is a rectangular arrangement set of numbers into rows and column. Matrix are widely used in various fields, including engineering, physics, statistics , computer science, and many other scientific discipline. There are operations that can be performed in matrices such as, addition, multiplication, transportation and etc.

1.1

In matrices, the numbers entries are known as elements. Horizontal number of matrices called as rows and vertical entries are called as columns. Matrices can be presented using brackets or parentheses.

In Figure 1 shows matrix A has three rows and three columns. Since Matrix A has three rows and three columns, we write the dimensions as 2×3, pronounced as “ three by three”

1
Identify the characters of matrices
���� = �719 465 037�
Figure 1: Arrays of number
(323) �453 223� �17 76 59� �1 2 0�
Row Column
Figure 2: The numbers arranged in a round brackets

a. Elements of Matrices

The elements in matrices can be in numbers, variables or any mathematical expressions. Each element can be identified from its rows and column. ���� = �����11 ����12 ����13 ����21 ����22 ����23 ����31 ����32 ����33

Figure 3: Notation of matrices

A matrix is represented by ‘A’ and the elements are represented by ‘a’ and two subscripts representing the position of the element. In this case, ‘������������ ’, where i the number of rows and j is the number of columns.

For example, from Figure 3, ����32 means the elements in 3rd row and 2nd column.

Example 1.1

1. Given ���� = �2 5 4 0 2 1� Find ����21 and ����12 Solution:

2
�����11 ����12 ����21 ����22 ����31 ����33 � �2 5 4 0 2 1� ����21 = 4 ����12 = 5
3 2. Given ���� = �2 3 1 5 4 8 6 4 7� Find ����32 and ����23 �����11 ����12 ����13 ����21 ����22 ����23 ����31 ����32 ����33 � �2 3 1 5 4 8 6 4 7� ����32 = 4 ����23 = 8 1. Given the matrix ���� = �0 8 7 2 6 7 2 3 6�, state the element of: a. ����23 b. ����12 c. ����32 2. Given the matrix ���� = � 5 2 3 2 1 0 1 9�, state the element of: a. ����14 b. ����11 c. ����23 3. Given ���� = �2 8 3 5 4 0 � , find the element of: a. ����11 b. ����31 c. ����22 Exercise 1.1

b. Order of matrices

The size of the matrix, known as the order of the matrix. It is determined by the number of rows and columns. If there are ‘j’ rows and ‘k’ column, the order of matrix is represented as ���� × ���� and is read as j by k. For example,

3 columns

B =�213 397�

2 rows

The order of matrix of Figure 4 is represented as 2 × 3 matrix, pronounced as ‘two by three’. In contrast, matrix B has two rows and three columns.

Example 1.2

1. Identify the order of these following matrices.

a. �130 211�

b. �120 181 563�

Solution:

a. 2 ×3

b. 3×3

Exercise 1.2

1. Determine the order for each matrix below.

a. [ 101 2]

b. �20 7 4�

4
Figure 4: 2 × 3 matrix

1.2 State the transportation of a matrix

The transpose of matrix is obtained by interchanging the rows and column. Means that the rows will become columns and column will become rows. If A is a 2 × 4 matrix, when we transposed it will become 4 × 2 matrix. The transpose of matrix A is known as �������� .

Solution:

Note that A is a 2 × 3 matrix, so �������� will be a 3 × 2 matrix. By the definition, the first column of �������� is the first column of A. While the second column of �������� is the second row of A. Therefore,

5
���� = �5 7 3 8 2 1� ; �������� = �5 7 3 8 2 1�
���� = �1 2 3 8 7 6�
1. Find the transpose of:
�������� = �1 2 3 8 7 6�
Example 1.3

Find the transpose of the following matrices.

Symmetrical matrix is always a square matrix. A square matrix is a matrix that has an equal number of rows and columns. So, if a matrix is symmetric, it means that its elements are symmetric with respect to the main diagonal (the diagonal from the top left to the bottom right).

To better understand the concept of the diagonal of a matrix, let's consider matrix B and its transpose ���� ���� . The diagonal elements of a matrix are the elements that lie on the main diagonal, which extends from the top left to the bottom right of the matrix. When we take the transpose of matrix B, denoted as ���� ���� , the diagonal elements remain the same. This means that if we represent the original matrix B and its transpose ���� ���� , the diagonal elements will be identical in both representations.

To illustrate this, let's denote the diagonal elements of matrix B as follows, where the diagonal elements are in bold:

As you can see, the diagonal elements 1, 2 and 3 remain unchanged in both the original matrix B and its transpose ���� ���� .

6
1.2.1 Symmetrical Matric
���� = 7 2 3 3 0 4 2 4 6 ���� = �3 −1 2 0 5 5 9 6 � ���� = (4 3 2 1)
���� = ����� 7 8 3 ���� 9 4 5 ����� ���� ���� = ����� 3 4 7 ���� 5 8 9 �����
Exercise
1.3

In summary, the diagonal of a matrix refers to the elements that lie on the main diagonal, running from the top left to the bottom right of the matrix. Taking the transpose of a matrix does not alter its diagonal elements.

1.2.2 Upper Triangular Matrix

An upper triangular matrix is a square matrix in which all the entries below the main diagonal (including the diagonal itself) are zero. In other words, all the nonzero entries of an upper triangular matrix lie on or above the diagonal.

Here's an example of an upper triangular matrix:

In this matrix, all the entries below the main diagonal (the numbers 0) are zero, while the entries on or above the main diagonal can be non-zero.

1.2.3 Lower Triangular Matrix

A lower triangular matrix is a square matrix in which all the entries above the main diagonal are zero. The main diagonal of a matrix consists of the elements where the row index is equal to the column index. In a lower triangular matrix, all the elements above the main diagonal are zero, while the elements on and below the main diagonal can be any value. Here's an example of a lower triangular matrix:

In this example, all the entries above the main diagonal (highlighted by zeros) are zero, while the entries on and below the main diagonal can have any value.

7
���� = �213 054 007�
���� = �200 350 149�

Identify the diagonal of each matrix, and state whether each matrix is diagonal, upper triangular, or none of the above.

Solution:

• The matrix A is upper triangular because the only nonzero entries lie on or above the diagonal.

• The matrix B is diagonal. By the definition, matrix B is both upper and lower triangular.

• The matric C is upper triangular.

Determine the type of the following matrices:

8
Example 1.4
���� = �1 3 2 0 5 4 0 0 6� ���� = �2 0 0 0 6 0 0 0 2� ���� = �2 3 1 0 6 7 0 0 8 0 0 0�
a. ���� = �2 4 3 1 � b. ���� = �2 0 0 0 1 0 0 0 3� c. ���� = � 1 0 0 2 6 0 2 3 7� Exercise 1.3

2. Demonstrate the operation Matrices

Matrix operations are fundamental mathematical operations that involve manipulating matrices. Some of the commonly used matrix operations such as addition, subtraction and multiplication.

2.1 Calculate the operation Matrices

Addition and subtraction of matrices can only be performed on matrices of the same dimensions. In other words, matrices must have the same number of rows and the same number of columns in order to be added or subtracted.

a. Addition

When performing addition or subtraction, the corresponding elements in the matrices are added from each other.

For example, given two matrices A and B of the same dimensions:

The addition of A and B would be:

So, the answer would be like this;

9
���� = �1 2 3 4� ���� = �5 6 7 8�
���� + ���� = �1 + 5 2 + 6 3 + 7 4 + 8�
���� + ���� = � 6 8 10 12�

b. Subtraction

Similarly with addition, the subtraction of the matrices should have the same number of rows and column. In subtracting two matrices, we subtract the respective element in each row and column.

For example, given two matrices A and B of the same dimensions:

The subtraction of A and B would be:

So, the answer would be like this;

the following:

10
���� = �7 3 5 8� ���� = �9 2 1 6�
���� ���� = �7 9 3 2 5 1 8 6�
���� ���� = � 2 1 4 2� 1. Given matrices ���� = �0 2 6 3� and ���� = �6 1 2 0�. Determine
a. ���� + ���� b. ���� ���� Exercise 2.1

c. Multiplication

In order for two matrices to be multiplied, the number of columns in the first matrix must be equal to the number of rows in the second matrix.

• Scalar Multiplication

Scalar multiplication of a matrix involves multiplying each element of the matrix by a scalar (a single number). Let's say we have a matrix A and a scalar k. The scalar multiplication of matrix A by scalar k, denoted as kA, is performed by multiplying each element of A by k.

For example, let’s consider the following 4×4 matrix A:

If we multiply this matrix by a scalar ���� =2, we would get:

The answer would be like this:

• Multiplication of Two Matrices

If A is an ���� × ���� matrix, then kA will also be an ���� × ���� matrix with the same dimensions.

Let's denote the matrices as follows:

11
���� = �12 34�
2���� = �2×12×2 2×32×4�
2���� = �24 68�
Matrix ����: ���� × ���� Matrix ����: ���� × ����

The dimensions of the matrices are given by the number of rows (m) and the number of columns (n) in Matrix A, and the number of rows (n) and the number of columns (p) in Matrix B.The multiplication of matrices A and B is defined as follows:

To calculate each element of Matrix C, you perform a dot product between the corresponding row of Matrix A and the corresponding column of Matrix B. The result is the element in the corresponding row and column of Matrix C. For example, to work out for the 1st row and 1st column;

To get the product, we can use the “Dot Product” where we multiply and sum up the matching members. In this case;

Here is another example for 1st row and 2nd column:

12
������������������������ ���� = ������������������������ ���� × ������������������������ ����
���� = �321 763� × �63 37 24� = �26 �
(321) (632) = (3×6) + (2×3) + (1×2)
= 26
���� = �321 763� × �63 37 24� = �2627� (321) . (374) = (3×3) + (2×7) + (1×4) “Dot Product” “Dot Product”

We do the same thing for 2nd row and 1st column:

For 2nd row and 2nd column:

The final answer would be like this:

13
2.2 = 27
Exercise
(763) . (632) = (7×6) + (6×3) + (3×2) = 66
(763) . (374) = (7×3) + (6×7) + (3×4) = 75
���� = �321 763� × �63 37 24� = �2627 6675�
a. �23 7 1� + � 64 03� b. �7 5 43 �−�94 2 7� c. 2 �21 13 02� � 43 12 20� d. �21 14��32 12�
1. Calculate the following:

If the number has a reciprocal, in matrix we have inverse.

get reciprocal of a number, 1 need to be divided by the number.

The reciprocal of a number 1 5 can also be identified as 5−1

Inverse of matrix:

For matrix we write ����−1 instead of 1 ���� because matrix cannot be divided. But there are other similarities between reciprocal and inverse.

Reciprocal:

If we multiply a number by its reciprocal, we will get 1. For example:

5 × 1 5 = ����

Inverse matrix:

If we multiply a matrix by its inverse, we will get the Identity matrix , I

For example:

���� × ����−1 = ����

14 d. Inverse
To
5
A ����− Inverse
Reciprocal NOTES!

The identity Matrix (has the number of rows and column) can it be 2×2 or 3×3, 4×4 in size.

The formula to find the inverse of 2×2 matrix: in other words, the positions of a and d was swapped and put negatives sign in front of b and c

A matrix can invert based on determinant conditions:

• If the determinant of the matrix is nonzero.

• If the determinant of the matrix is equal to 0. For example:

15
���� = ��������� �������� � ����−1 = 1 |����| � ����−���� −�������� � To find determinant of 2×2 matrix, ���� = ��������� �������� � |����| = ���� ���� ���� ���� = |�������� �������� | ����−1 = 1 |�������� �������� | � ����−���� −�������� �
���� = �21 47� ����−1 = 1 (2×7) (1×4) � 7 1 42 � = 1 10 � 7 1 42 � = � 0.7 0.1 0.40.2 � �������� �������� is called the determinant inverse NOTES! inverse inverse

Let’s check either this is the correct answer. Remember that ���� × ����−1 = ����. Lets check the answer by multiply the matrix by its inverse.

As we know that there is no concept of division. But we can multiply by an inverse to achieve the same thing.

To obtained determinant for ���� × ���� matrix, two different methods can be use are cofactor expansion and cross multiplication. Let’s see the example how to solve using these two methods.

16
�21 47� × � 0.7 0.1 0.40.2 � = �(2×0.7) + (1× 0.4)(2× 0.1) + (1×0.2) (4×0.7) + (7× 0.4)(4× 0.1) + (7×0.2)� = �1.4 0.4 0.2+0.2 2.8 2.8 0.4+1.4� = �10 01�
���� = � ������������ ������������ ����ℎ ���� � • Cofactor expansion ∴ |���� | = � ������������ ������������ ����ℎ ���� � =

The inverse formula of a ���� × ���� matrix is:

���� 1 = 1 |����|

������������ (����)

Where:

• |����| is the determinant of matrix A

• ������������ (����) is the adjoint of matrix A

Let’s see the example how to solve ���� × ���� matrix.

We apply the inverse formula of ���� × ���� matrix.

Step 1:

First, find the determinant of the matrix.

17
���� = �1 2 3 0 2 1 3 4 0�
���� 1 = 1 |����| ������������ (����)
|����| = 1 �2 1 4 0� 2 �0 1 3 0� 3 �0 1 3 0� = 1[(2 × 0) (1 × 4)] 2[(0 × 0) (1 × 3)] 3[(0 × 0) (1 × 3) = 1( 4) 2( 3) 3( 3) = 4 + 6 + 9 = 11

• To find the cofactor, the matrix obtained need to multiply with a sign chart

18
Find the minor. If ���� = �1 2 3 0 2 1 3 4 0�, minor ���� = �����11 ����12 ����12 ����21 ����22 ����23 ����31 ����32 ����33 � where, ����11 = �2 1 4 0� = (0) (4)(1) = 4 ����12 = �0 1 3 0� = (0) (3)(1) = 3 ����13 = �0 2 3 4� = (0) (2)(3) = 6 ����21 = �2 3 4 0� = (0) (3)(4) = 12 ����22 = �1 3 3 0� = (0) (3)(3) = 9 ����23 = �1 2 3 4� = (1)(4) (2)(3) = 2 ����31 = �2 3 2 1� = (2)(1) (3)(2) = 5 ����32 = �1 3 0 1� = (1)(1) (0) = 1 ����33 = �1 2 0 2� = (1)(2) (2)(0) = 2 Therefore, minor ���� = � 4 3 6 12 9 2 5 1 2 �
Step 2:
Step 3: Find the cofactor.
�������� = � (+) 4 ( ) 3 (+) 6 ( ) 12 (+) 9 ( ) 2 (+) 5 ( )1 (+)2 � �������� = � 4 3 6 12 9 2 5 1 2 � �+ + + + +� Sign chart
19 Exercise 2.3
4: Find the adjoint. ������������(����) = ���� ���� ������������(����) = � 4 12 5 3 9 1 6 2 2 �
5: Find the inverse, A-1. ���� 1 = 1 |����| × ������������(����) = 1 11 � 4 12 5 3 9 1 6 2 2 � = ⎝ ⎜ ⎜ ⎛ 4 11 12 11 5 11 3 11 9 11 1 11 6 11 2 11 2 11 ⎠ ⎟ ⎟ ⎞ 1. Given a matrix ���� = �3 2 1 5�, find |���� |. 2. Find the determinant of matrix ���� = �4 5 3 7� 3. Find the cofactor of the matrix ���� = �4 3 1 5�. 4. Inverse the following 2 × 2 matrix.
Step
Step

I. ���� = �25 13

II. ���� = �41 52

5. Inverse the following 3×3 matrix.

I. ���� = �15 2 621 8 −40

II.

20
���� = � 4 21 503 126

3. Demonstrate Simultaneous Equations

There are two methods that can be use to solve the simultaneous equations; using inverse matrix or Cramer’s rule. To solve the simultaneous equations using either these two methods, the linear equations must be in this form:

Let’s consider a system of equations with two variables:

Let's define matrix A as the coefficient matrix, matrix X as the variable matrix, and matrix B as the constant matrix. The system of equations can be written as:

To solve for X, multiply both sides of the equation by the inverse of A:

Then calculate the inverse of A.

21
3.1 Solve simultaneous equations using inverse matrix
����11 ���� + ����12 ���� + ����13 ���� = ����1 ����21 ���� + ����22 ���� + ����23 ���� = ����2 ����31 ���� + ����32 ���� + ����33 ���� = ����3
2���� +3���� =8 Equation 4���� 2���� =2 Equation
�������� = ���� , then ���� = ����−1 ���� Where, ���� = �23 4 2����������� = �8 2�
����−1 . �������� = ����−1 ����
1 2

To find the inverse of A, calculate its determinant:

Then, swap the diagonal elements and change the sign to obtain the adjoint matrix.

Multiply the adjoint matrix by the determinant to get the inverse matrix of A.

Then, solve for X. Multiply both sides of the equations with ����−1 .

22 ����−1 = ��������� �������� �
det ���� =
=
(2× 2) (3×4)
14
������������ ���� = � 23 42�
����−1 = 1 det ���� ∙������������(����) ����−1 = 1 14 � 23 42� = �1 7 3 14 2 7 1 7 �
�1 7 3 14 2 7 1 7 � ���������� = �8 2� ���� = ����−1 ���� = �1 7 3 14 2 7 1 7 � �8 2� = ��1 7 ×8� + � 3 14 ×2� �2 7 ×8� + � 1 7 ×2��

Hence, ���� = 5 7 and ���� = 2. You can check the solution by substituting the value of ���� and ���� into the equations by verify the left-hand side is equal to right-hand side.

Let’s try to solve these three linear equations below using inverse matrix method.

First, change the equation into matrix form:

23 �5 7 2� = ����������
2���� + 3���� ���� = 1 ���� 2���� + 2���� = 3 3���� + ���� 4���� = 5
�2 3 1 1 2 2 3 1 4� ����� ���� ���� � = � 1 3 5 �
Find the determinant. |����| = �2 3 1 1 2 2 3 1 4� |����| = 2 � 2 2 1 4� 3 �1 2 3 4� + ( 1) �1 2 3 1 � = 2[( 2 × 4) (2 × 1)] 3[(1 × 4) (2 × 3)] 1[(1 × 1) ( 2 × 3)] = 2(6) 3( 10) 1(7) = 35
Step 1:
24 Step 2: Find the minor. ���� = �2 3 1 1 2 2 3 1 4� ����11 = � 2 2 1 4� = ( 2 × 4) (2 × 1) = 6 ����12 = �1 2 3 4� = (1 × 4) (2 × 3) = 10 ����13 = �1 2 3 1 � = (1 × 1) ( 2 × 3) = 7 ����21 = �3 1 1 4� = (3 × 4) ( 1 × 1) = 11 ����22 = �2 1 3 4� = (2 × 4) ( 1 × 3) = 5 ����23 = �2 3 3 1� = (2 × 1) (3 × 3) = 7 ����31 = � 3 1 2 2 � = (3 × 2) ( 1 × 2) = 4 ����32 = �2 1 1 2 � = (2 × 2) ( 1 × 1) = 5 ����33 = �2 3 1 2� = (2 × 2) (3 × 1) = 7 Therefore, minor ���� = � 6 10 7 11 5 7 4 5 7� Step 3: Find the cofactor. �������� = � (+)6 ( ) 10 (+)7 ( ) 11 (+) 5 ( ) 7 (+)4 (−)5 (+) − 7� �������� = � 6 10 7 11 5 7 4 5 7�
25 Step 4: Find the adjoint. ������������(����) = ���� ���� ������������(����) = � 6 11 4 10 5 5 7 7 7� Step 5: Find the inverse ���� 1 . ���� 1 = 1 |����| × ������������(����) = 1 35 � 6 11 4 10 5 5 7 7 7� = ⎝ ⎜ ⎜ ⎛ 6 35 11 35 4 35 2 7 1 7 1 7 1 5 1 5 1 5 ⎠ ⎟ ⎟ ⎞ Step 6: Find ���� = ���� 1 ���� ����� ���� ���� � = ⎝ ⎜ ⎜ ⎛ 6 35 11 35 4 35 2 7 1 7 1 7 1 5 1 5 1 5 ⎠ ⎟ ⎟ ⎞ � 1 3 5 � ����� ���� ���� � = ⎝ ⎜ ⎜ ⎛ 6 35 (1) 11 35 ( 3) 4 35 (5) 2 7 (1) 1 7 ( 3) 1 7 (5) 1 5 (1) 1 5 ( 3) 1 5 (5)⎠ ⎟ ⎟ ⎞

1. Solve the system using inverse matrix method.

2. Solve the system using inverse matrix method.

26 ����� ���� ���� � = ⎝ ⎜ ⎜ ⎛19 35 11 14 9 5 ⎠ ⎟ ⎟ ⎞ Therefore, ���� = 19 35 , ���� = 11 14 , ���� = 9 5 .
���� 3���� = 1 4���� + 3���� = 11
2���� + 3���� ���� = 2 ���� + 2���� + ���� = 3 ���� ���� + 3���� = 1 Exercise 3.1

3 2 Solve simultaneous equations using Cramer’s Rule

Cramer's rule is a method used to solve systems of linear equations by determining the value of each variable individually. It involves calculating determinants. Here's how you can use Cramer's rule to solve a system of linear equations:

Solve 2 × 2 linear equations using Cramer’s Rule

Consider a system of linear equations:

Step 1: Change into matrix form.

Step 2: Find the determinant, |����|

Step 3:

Substitute column �����1

with

, then find x-matrix: |

27
����1 ���� + ����1 ���� = ����1 ����2 ���� + ����2 ���� = ����2
�����1 ����1 ����2 ����2 � ���������� = �����1 ����2 �
���� = �����1 ����1 ����2 ����2 �
where
�������� | |�������� | = �����1 ����1 ����2 ����2 �
����2
�����1 ����2 �

Substitute column �����1 ����2 � with �����1 ����2 �, then find x-matrix:

Solve the equations for ���� and ����

28
� ��������� � = �����1 ����1 ����2 ����2 �
���������
���� = |�������� | |����| = �����1 ����1 ����2 ����2 � �����1 ����1 ����2 ����2 � ���� = ��������� � |����| = �����1 ���� ����2 ����2 � �����1 ����1 ����2 ����2 �
Step 4: Step 5:

Let’s try to solve this ���� × ���� using Cramer’s Rule.

29
12���� + 3���� = 15 2���� 3���� = 13
�12 3 2 3� ���������� = �15 13� where ���� = �12 3 2 3� Step 2: Find the determinant, |����|. |����| = �12 × ( 3)� (3 × 2) = −42
3:
column �����1 ����2 � with �����1 ����2 �,
x-matrix: |�������� | |�������� | = �15 3 13 3�
�����1 ����2 �
�����1 ����2 �,
x-matrix: ��������� �.
1:
Change into matrix form.
Step
Substitute
then find
Substitute column
with
then find
Step
Step 4:

So, the answer is (2, 3)

30 ��������� � = �1215 2 13� Solve the equations for ���� and ���� ���� = |�������� | |����| = �15 3 13 3� | 30| = �15 × ( 3)�− (3× 13) | 42| = 84 42 =2 ���� = ��������� � |����| = �1215 2 13� | 42| = (12 × 13) (15 ×2) | 42| = 126 42 = 3
. Step 5:

Solve 3 × 3 linear equations using Cramer’s Rule

Consider the system of equations:

Step 1:

Change into matrix form.

Step 2:

Find the determinant, |

Step

Step 4:

with

, then find x-matrix:

31
����11 ���� + ����12 ���� + ����13 ���� = ����1 ����21 ���� + ����22 ���� + ����23 ���� = ����2 ����31 ���� + ����32 ���� + ����33 ���� = ����3
�����11 ����12 ����13 ����21 ����22 ����23 ����31 ����32 ����33 � ����� ���� ���� � = �����1 ����2 ����3 � where ���� = �����11 ����12 ����13 ����21 ����22 ����23 ����31 ����32 ����33 �
����|
�����11 ����21 ����23 �
�����1 ����2 ����3 �
x-matrix: |�������� |
3: Substitute column
with
, then find
�����12 ����22 ����32 �
�����1 ����2 ����3 �
Substitute column
���������

Substitute column

with

, then find x-matrix: |

Solve the equations for ���� , ���� and

Let’s try to solve this ���� × ���� using Cramer’s Rule

32
�����13 ����23 ����33 �
�����1 ����2 ����3 �
�������� |
���� . ���� = |�������� | |����| = �����1 ����12 ����13 ����2 ����22 ����23 ����3 ����32 ����33 � �����11 ����12 ����13 ����21 ����22 ����23 ����31 ����32 ����33 � ���� = ��������� � |����| = �����11 ����1 ����13 ����21 ����2 ����23 ����31 ����3 ����33 � �����11 ����12 ����13 ����21 ����22 ����23 ����31 ����32 ����33 � ���� = |�������� | |����| = �����11 ����12 ����1 ����21 ����22 ����2 ����31 ����32 ����3 � �����11 ����12 ����13 ����21 ����22 ����23 ����31 ����32 ����33 �
���� + ����−���� =6 3���� 2���� + ���� = 5 ���� +3����− 2���� = 14
into matrix form. �11 1 3 21 13 2� ����� ���� ���� � = � 6 5 14 � where ���� = �11 1 3 21 13 2�
5:
6:
1:
Change
Step
Step
Step
33
Find the determinant, |����|. ���� = �1 1 1 3 2 1 1 3 2� |����| = 1 � 2 1 3 2� 1 �3 1 1 2� + 1 �3 2 1 3 � = 1[( 2 × 2) (1 × 3)] 1[(3 × 2) (1 × 1)] + ( 1)[(3 × 3) ( 2 × 1)] = 1(1) 1( 7) 1(11) = 3 �������� = � 6 1 1 −5 −2 1 14 3 2� |�������� | = 6 � 2 1 3 2� − 1 � 5 1 14 2� − 1 � 5 2 14 3 � = 6[( 2 × 2) (1 × 3)] 1[( 5 × 2) (1 × 14)] 1[( 5 × 3) ( 2 × 14)] = 6(1) 1( 4) 1(13) = 3 �������� = �1 6 1 3 5 1 1 14 2� ��������� � = 1 � 5 1 14 2� 6 �3 1 1 2� 1 �3 5 1 14 � = 1[( 5 × 2) (1 × 14)] 6[(3 × 2) (1 × 1)] 1[(3 × 14) ( 5 × 1)] = 1( 4) 6( 7) 1(47) = 9 �������� = �1 1 6 3 2 5 1 3 14 � |�������� | = 1 � 2 5 3 14 � 1 �3 5 1 14 � + 6 �3 2 1 3 �
Step 2:

=1[( 2× 14) ( 5×3)] 1[(3× 14) ( 5×1)] +6[(3×3) ( 2×1)]

=1( 13) 1(47) +6(11)

=6

Solve the equations for ���� , ���� and ����

So, the answers is (1,3 , -3).

Step 3: Exercise 3.2

1. Solve the system of equations using Cramer’s Rule.

I. 2���� −���� =5

II. 12���� +3���� = 15

���� 3���� = 13

III. 2���� + ���� + ���� =0 ���� −����−���� =0

���� +4���� +6= ���� =0

IV. 2���� + ���� + ���� =1

���� −���� +4���� =0 ���� +2����− 2���� =3

34
���� = |�������� | |����| = 3 3 =1 ���� = ��������� � |����| = 9 3 =3 ���� = |�������� | |����| = 6 3 = 2
���� + ���� =4
2

a. A= not triangular matrix

b. B= upper and lower triangular

c. ���� = lower triangular

35
1.1
���� = �087 267 236�
����23 =7
����12 =8
����32=3
���� = � 5 2 321 0 1 9�
����14 =1
����11 =5
����23 = −1
���� = �2 8 35 40 �
����11 =2
����31 =4
����22 =5 Exercise 1.2 ���� = � 732 204 346� ���� = � 32 10 59 56� ���� = �4 3 2 1�
ANSWERS Exercise
1.
a.
b.
c.
2.
d.
e.
f.
3.
d.
e.
f.
Exercise 1.3
36 Exercise 2.1 1. ���� + ���� = �63 83� 2.����−���� = � 6 1 4 3� Exercise 2.2 1. Calculate the following: a. �23 7 1� + � 64 03� = � 47 72� b. �7 5 43 �−�94 2 7� = � 2 9 2 10 � c. 2 �21 13 02� � 43 12 20� = � 0 −1 34 24 � d. �21 14��32 12� = �76 7 10� Exercise 2.3 1. -17 2. 13 3.� 5 1 34 � 4. i) ����−1 = � 3 5 12 � ii) ����−1 = � 2 3 −1 3 −5 3 4 3 �
37 5. i) ���� −1 = ⎝ ⎜ ⎛ 1 15 2 15 3 20 −2 15 4 15 13 60 2 3 11 15 −7 15 ⎠ ⎟ ⎞ ii) ���� −1 = ⎝ ⎜ ⎛ 3 26 7 26 −3 26 33 52 25 52 −7 52 5 26 −3 26 5 26 ⎠ ⎟ ⎞ Exercise 3.1 1. ���� = 2 3 , ���� = 7 13 2. ���� =1, ���� =1, ���� =1 Exercise 3.2 1. i) ���� =3, ���� =1 ii) ���� =2, ���� = 3 iii)���� =3, ���� =0, ���� =0 iv) ���� = 3, ���� =5, ���� =2

REFERENCE

1) https://www.algebrapracticeproblems.com/inverse-of-amatrix/#:~:text=The%20inverse%20of%20a%20matrix%20is%20a%20matrix,1%20%C2%B7A%3DI%20Where%20I%20is%20the%20identity%20matrix

2) https://www.mathsisfun.com/algebra/matrix-inverse.html

3) https://www.algebrapracticeproblems.com/upper-lower-triangular-matrix/

4) https://math.libretexts.org/Bookshelves/Linear_Algebra/Book%3A_Linear_Algebr a_(Schilling_Nachtergaele_and_Lankham)/07%3A_Eigenvalues_and_Eigenvect ors/7.05%3A_Upper_Triangular_Matrices

5) https://www.cuemath.com/algebra/triangular-matrix/

38

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