AM015 LECTURE & TUTORIAL NOTES

AM015/4. Matrices and Systems of Linear Equations

CHAPTER 4: MATRICES AND SYSTEMS OF LINEAR EQUATIONS

LECTURE 1 OF 5

At the end of lesson, student should be able to:

a) Identify the different types of matrices

*Row, column, zero, diagonal, upper triangular, lower triangular and identity matrices
b) Perform operations on matrices

*Addition, subtraction, scalar multiplication and multiplication of two matrices and
up to 3 x 3 matrix
c) Find the transpose of a matrix
*Include properties

4.1 MATRICES

Definition

 A matrix is a rectangular array of numbers enclosed between brackets.
 The general form of a matrix with m rows and n columns is

 a11 a12 a13  a1n 
 
 a 21 a 22 a 23  a 2n 

 a 31 a 32 a 33  a 3n  m rows
      
 
a m1 a m2 a m3  a mn 

n columns

 The order or dimension of a matrix of m rows and n columns is m x n.
 The individual numbers that make up a matrix are called its entries or

elements, aij and they are specified by their row and column position.

 The matrix for which the entry is in ith row and jth column is denoted by [ aij ].

Example 1

Let  5 6 1
A 2
 2 3  7

(a) What is the order of matrix A?

(b) If A= [ aij ], identify a21 and a13 .

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AM015/4. Matrices and Systems of Linear Equations

Types of Matrices

1) Row Matrix is a (1 x n) matrix (one row).

Example 2

A  1 2 3 4 5 B  1 0 7 8 4 3 5

2) Column Matrix is a (m x 1) matrix (one column)

Example 3

2
3
A  0 B  5
4
7

3) Square Matrix is a (n x n) matrix which has the same number of rows as columns

Example 4

A  1 3 2 x 2 matrix. 1 3 2
1 8 B  3 1 2 3 x 3 matrix

2 3 1

4) Zero Matrix is a (m x n) matrix which every entry is zero, and denoted by O.
Example 5

0 0 0 0 0 O  0 0
O  0 0 0 O  0 0 0 0
0
0 0 0 0

5) Diagonal Matrix

 a11 a12 a13 a1m 
 
 a21 a22 a23 a2m 

Let A =  a31 a32 a33 a3m 
 
 
am1 am2 am3 amm 

 The diagonal entries of A are a11 , a22 , a33 ,…, amm

 A square matrix which non-diagonal entries are all zero is called a
diagonal matrix.

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AM015/4. Matrices and Systems of Linear Equations

Example 6

A  2 0 1 0 0 a 0 0
0 3 B  0 2 0 C  0 0 0

0 0 3 0 0 b

6) Identity Matrix is a diagonal matrix in which all its diagonal entries are 1 and
denoted by I.

Example 7

A  1 0  I22 1 0 0
0 1 B  0 1 0  I33

0 0 1

7) Lower Triangular Matrix is a square matrix and aij  0 for i  j

a11 0 0
a21 a22 
0 

a31 a32 a33 

Example 8

a 0 0 1 0 0
A = b f 0 B = 3 2 0

c d e 3 2 3

8) Upper Triangular Matrix is a square matrix and aij  0 for i  j .

a11 a12 a13 
 
 0 a22 a23 

 0 0 a33

Example 9

1 2 3 a b c 
A = 0 2 4 B = 0 d 
e 
0 0 3
0 0 f 

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AM015/4. Matrices and Systems of Linear Equations
Operations on Matrices
1) Addition and Subtraction of Matrices

For m x n matrices, A = [ aij ] and B = [ bij ],
A + B = C = cij mxn , where cij  aij  bij .
A – B = D = dij mxn , where dij  aij  bij.

PROPERTIES

A+B=B+A Commutative

(A + B) + C = A +(B + C) Associative

A + (-A) = (-A) + A = O O - zero matrix

Note: The addition or subtraction of two matrices with different orders is not
defined. We say the two matrices are incompatible.

Example 10

Simplify the given quantity for A  1 2 , B  4 3 and C  1 .
3 4  5 6 2

(a) A + B (b) A – B (c) A + C

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AM015/4. Matrices and Systems of Linear Equations
2) Scalar Multiplication

If c is a scalar and A  aij  then cA  bij  where bij  caij .

PROPERTIES

(α + β)A = αA + βA α, β constant

 α (A + B) = αA + αB -

 α (βA) = (αβ)A -

Example 11

2 4 1
  2
Given A   8 5  , find  A.
7 
6

3) Multiplication of Matrices

The product of two matrices A and B is defined only when the number of columns in A
is equal to the number of rows in B.

If the order of A is m n and the order of B is n  p , then AB has order m  p.

Amn Bn p  ABm p

A row and a column must have the same number of entries in order to be multiplied.

b1 
b2 
 R  a1 a2 a3 ... an and 
b..3. 
C  

bn 

 RC  a1b1  a2b2  a3b3  ...  anbn

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AM015/4. Matrices and Systems of Linear Equations

PROPERTIES

A (BC) = (AB)C Associative
A (B + C) = AB + AC Distributive

Example 12

1 2 3 and B = 2 1
Let A = 2 0 5 3 4 . Calculate AB.
 2 1

Example 13

2 3 2  3 0 2
Find 1 1 1 4 
0 1 

4 1 0 1 1 0 

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AM015/4. Matrices and Systems of Linear Equations

Transpose of a Matrix
The transpose of a matrix A, written as AT, is the matrix obtained by interchanging the
rows and columns of A. That is, the ith column of AT is the ith row of A for all i’s.

If Amn  aij  then AT nm  aji 

A  aa1211 a12 a13  a11 a21 a31 
a22 a23  AT  a12 
 then a22 a32 

a31 a32 a33 33 a13 a23 a33 33

Properties of the transpose matrix
(A ± B)T = AT ± BT
(AT)T = A
(AB)T = BTAT
(kA)T = kAT

Example 14

Find the transpose of the following matrices

a) Let 2 1 3 3
A  1 b) Let B  2 5 4

3 1 3 5

Example 15

Let A  1 2 B  3 4 and C  1 4 .
3 4 , 2 1 3 2

Show that (a) (A + B)T = AT + BT
(b) (BC)T = CTBT

Exercise

Let A  1 2 and B  2  1 . Show that AB ≠ BA.
 4 3 
 3 2 

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AM015/4. Matrices and Systems of Linear Equations

LECTURE 2 OF 5

At the end of lesson, student should be able to:
a) Find the minors and cofactors of a matrix. *For 2 x 2 matrices and 3 x 3 matrices
b) Find the determinant of a matrix. *For 2 x 2 and 3 x 3 matrices. *Use basic

properties of determinant

4.2 DETERMINANT OF MATRICES
Determinant of 2 x 2 Matrices

Given A = a b
 
 c d 

Then determinant A = ab = ad – bc

cd

Example 1

2 5 3 2
Given A = 3 8 and B = 5 2 , find A , B , AB

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AM015/4. Matrices and Systems of Linear Equations

Minor and Cofactor
Let A be n x n matrix,
The minor Mij of the element aij is the determinant of the matrix obtained by deleting the
ith row and jth column of A.

1 2 1
Consider the matrix A  3 4 
2 

1 4 3 

 Minor
M11 is the determinant of the matrix obtained by deleting the first row and first
column from A.

1 2 1

M11  3 4 2 
14 3

Similarly

1 2 1

M 32  3 4 2 
14 3

Therefore,

a11 a12 a13 
A  a21 
If a22 a23  , then

a31 a32 a33 

M11  a22 a23 and M 32  a11 a13
a32 a33 a21 a23

 Cofactor M11 M12 M13 
The cofactor Cij of the element aij is C  M21 M 22 
M 32 M 
Cij = ( - 1 )i+j Mij   M 31

23

M33 

C11 = C32 =

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AM015/4. Matrices and Systems of Linear Equations

Example 2

 2 4 2
Given matrix A  2 0 
4  . Find the minor of matrix A. Hence, find the cofactor.

 4 3 3

Determinant of 3 x 3 matrix

The determinant of 3 x 3 matrix can be evaluated as the sum of the products of the
elements of any one row (column) with their respective cofactors.

Expansion of the cofactor

A  aij Cij ; i 1, 2,..., n and j 1, 2,...,n

Example:

By expanding along the first row By expanding along first column

Elements in 1st row: a11, a12 , a13 Elements in 1st column: a11 , a21 , a31

A  a11C11  a12C12  a13C13 A  a11C11  a21C21  a31C31

or or

A  a11M11  a12M12  a13M13 A  a11M11  a21M 21  a31M31

Note: Choose row or column that has the most zero.

Example 3

 4 10  2
  3  5 , find the determinant A by choosing
Given A   8

 6  3 3 

(i) 1st row
(ii) 2nd column

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AM015/4. Matrices and Systems of Linear Equations

Example 4

 2 2 0 
Find the determinant of A   3 1 
3 

 2  4 1

Properties of determinant

1. If a square matrix B is obtained from a square matrix A by multiplying each element

of any row or column of A by some real number k, then B  k A

Example 5

Given the |A| = 5. Hence find |B|, |C| and |D| by using the determinant properties.

1 2 3 5 10 15  5 10 15  5 10 15
A  2 3 5 B  2
3 5  C  10 15 25 D  10 15 25

3 4 2 3 4 2   3 4 2  15 20 10

Note: kA  kn A , where A is a square matrix (n x n) and k is a constant

2. If a square matrix B is obtained from a square matrix A by interchanging any two

rows (columns), then B   A .

Example 6

Given A  2 3 Then, A  2(4)  3(1)
 1 4 .  11

Given B  1 4 Row 1 and 2 are interchanged
 3
 2

Then, B 



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AM015/4. Matrices and Systems of Linear Equations

3. If A is a square matrix, then A  AT

Example 7

Given A  2 3 A  11
 1 4 and

AT  2 1 Then, AT 


3 
4 



4. If A and B are square matrices, then AB  A B .

Example 8

Given A  2 3 B  1 4 . The determinant A and B are 11
 1 4 and 3 2

and -14 respectively.

AB 

AB =



Exercise :

4 1 0
1. Given matrix B  2 3 1 . Find the minor of matrix A. Hence, find the

 2 1 3

cofactor.

 2 5 1 
2. Find the determinant of A   3 0 
1  .

 2 5  4

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AM015/4. Matrices and Systems of Linear Equations

LECTURE 3 OF 5

At the end of lesson, student should be able to:

a) Compute the inverse of a non-singular matrix using Adjoint Matrix

b) Compute the inverse of a non-singular matrix using AB=kI
*For 2 x 2 and 3 x 3 matrices. **Emphasize that A-1≠1/A

4.3 INVERSE MATRICES

If A and B are two square matrices such that AB = BA = I, where I is the identity matrix,
then B is the inverse of A and A is the inverse of B. The inverse of A is denoted by A-1.

There are 2 methods to obtain inverse of matrices:
1) Adjoint Method: A-1 = 1 adj A
A

2) Use the property AB = kI

1) Finding Inverse by Using Adjoint Method

The inverse of a matrix A is denoted by

A1  1 adj A , given that A  0

A

*** Remember! A1  1
A

If A  0 ,

~ A is a non-singular matrix
~ Inverse matrix exists

If A  0 ,

~ A is a singular matrix
~ Inverse matrix does not exist

Adjoint Matrix

Let C  cij  be the cofactor matrix of A.

Adjoint of matrix A (adj A) is defined as the transpose of the cofactor matrix that is

adj A = CT  cij T  cji 

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AM015/4. Matrices and Systems of Linear Equations
Example 1

1 2 3
Given A  3 2 4 . Find the adjoint of A.

1 1 3

i. Inverse of a 2 x 2 matrix

Let A  a b , then A1 is given by
 c 
d 

A1  ad 1 bc d b
 c a 

Example 2

Find the inverse matrix for A  3 1
5 4

Example 3

Show that A 2  4 has no inverse.
 1 2 

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AM015/4. Matrices and Systems of Linear Equations

ii. Inverse of a 3 x 3 matrix
Example 4

 1 3 2
 2 2 . Find
Given B   0

2 1 0

a) The determinant of matrix B

b) Find the cofactor of matrix B

c) Hence, find the inverse of matrix B by using adjoint method

2) Use the property of AB = kI

Properties of Inverse Matrix
( A-1 )-1 = A

( AT )-1 = ( A-1)T
(( AB )-1 = B-1A-1

A1 = 1
A

Note: If AB = I where A and B are square matrices, then B is called the inverse

matrix of A and is written as A1 . Thus AA-1 = A-1A = I.

If

AB = kI

A-1AB = kA-1I

IB = kA-1

A-1 = 1

B
k

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AM015/4. Matrices and Systems of Linear Equations

Example 5

1 2 3  1 1 1
Given A  2 4 and B  10 
3 4 2  . It is known that AB = kI, where k is a

1 5 7  7 3 1

constant and I is a 33 matrix. Find k and hence deduce A-1.

Example 6

 1 1 2
 2 2 . Find A2 – 6A + 11I, with I as an identity matrix 3 x 3. Show that
Given A =  0

1 1 3

A (A2 – 6A + 11I) = 6I, hence deduce A-1.

Exercise :

3 1 2 
Find the inverse matrix of K  1 3 
4  .

 2 1 1

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AM015/4. Matrices and Systems of Linear Equations

LECTURE 4 OF 5

At the end of lesson, student should be able to:
a) Write a system of linear equations in the form of AX=B (for 2 x 2 and 3 x 3

matrices)

**Apply to some practical problems

b) Solve the unique solution to AX=B using:

i. Inverse Matrix ii. Cramer’s Rule

4.4 SYSTEM OF LINEAR EQUATIONS WITH THREE VARIABLES

Consider the system of linear equations with three unknown x, y and z.

a11x  a12 y  a13z  b1
a21x  a22 y  a23z  b2
a31x  a32 y  a33z  b3

All the linear systems above can be written as a single matrix equation as below

a11 a12 a13  b1  x
With A = a21  b2   y
a22 a23  , B =  and X = 

a31 a32 a33  b3   z

Thus, the system of linear equation can be written as AX=B

a11 a12 a13  x b1 
a21   y b2 
a22 a23    

a31 a32 a33  z  b3 

A X=B

Solving System of Linear Equations

System of linear equations can be solved by

1) Using the Inverse Matrix
 adjoint method
 property of AB = kI

2) Cramer’s Rule

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AM015/4. Matrices and Systems of Linear Equations

1) Using the Inverse Matrix to solve AX = B

If A is a n x n square matrix that has an inverse A-1, X is a variable matrix and B is a
known matrix, both with n rows, then the solution of matrix equations
AX = B is given by

X = A-1 B

Proof:

AX=B ( 3 x 3 square matrix)
A-1 ( A X ) = A-1 B
( A-1A ) X = A-1B

I X = A-1 B
X = A-1 B

a) Adjoint method

Example 1

Solve the following system of equations by using adjoint method.

4x 3y  2
x2y 1

Example 2
Solve the following system of equations by using adjoint method.

3x  y  2z  11
3x  2y  2z  10
x z 5

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AM015/4. Matrices and Systems of Linear Equations
Example 3

 1 2 3
Given A  1 0 4 .

 0 2 2
a. i) The determinant of matrix A

ii) Find the cofactor of matrix A
iii) Hence, find A-1 by using adjoint method
b. A company produces there grades of mangoes: X, Y and Z. The total profit for
1kg grade X, 2kg grade Y and 3kg grade Z mangoes is RM20. The profit for 4kg
grade Z is equal to the profit for 1kg grade X mangoes. The total profit for 2kg
grade Y and 2kg grade Z mangoes is RM10.
i) Obtain a system of linear equation.
ii) Write down the system in (i) as a matrix equation.
iii) Find the profit for each grades of mangoes.

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AM015/4. Matrices and Systems of Linear Equations

LECTURE 5 OF 5

At the end of lesson, student should be able to:

a) Solve the unique solution to AX=B using Inverse Matrix
b) Solve the unique solution to AX=B using Cramer’s Rule

b) Property of AB=kI

Example 1

 1 5 1 1 2 1
 1 3 B = 3 1 
Given A =  3 0 

 9 3 7 0 3 1 

Find AB and A-1. Hence, solve the following linear equations.

x5y  z 7
3x y3z 5
9x3y7z 1

2) Cramer’s Rule

Step 1 : Write the system of equation in the matrix form AX  B

Step 2 : Find the determinant of matrix A

Step 3 : Replacing the column of A with n x 1 matrix B

a11 a12 a13  x b1 
a21 a22 a23  y
 = b2 

a31 a32 a33  z b3 

A X= B

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AM015/4. Matrices and Systems of Linear Equations
Step 4: Then the solution is given by

b1 a12 a13
b2 a22 a23
x  b3 a23 a33

A
a11 b1 a13
a21 b2 a23
y  a31 b3 a33

A
a11 a12 b1
a21 a22 b2
z  a31 a23 b3

A

Example 2
Solve the following system using Cramer’s Rule

x  y  2z  3
x  y  3z  11
2x  3y  z  9

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AM015/4. Matrices and Systems of Linear Equations

Example 3

Ali, Bob and Ravi bought tickets for three separate performances. The table below shows
the number of tickets bought by each of them.

Ali Concert Orchestral Opera
Bob 2 1
Ravi 1 1 1
2 2 1
1

(a) If the total cost for Ali was RM 122, for Bob RM 87 and for Ravi RM 146, represent
this information in the form of three equations.

(b) Find the cost per ticket for each of the performances.
(c) Determine how much it would cost Hassan to purchase 2 concerts, 1 orchestral

and 3 opera tickets.

Exercise:

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

x3y  z  12
x yz0
2x y  z  8

2. A farmer needs to plant 400 fruit seedlings consisting of durian, rambutan and mango
seedlings. The number of durian seedlings must be the same as the total number of
rambutan and mango seedlings. The table below shows the number and price of
each seedlings.

Type of seedlings No of seedlings Cost per seedling
Durian x RM 1.20
y RM 1.80
Rambutan z RM1.00
Mango

The farmer spends RM536 to plant the 400 fruit seedlings.

a) Based on the above information, write a system of linear equations.
b) Write a matrix equation based on part (a).
c) Hence, determine the number of each type of seedlings planted by the farmer

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AM015/4 Matrices Sesi 2021/2022

TUTORIAL CHAPTER 4 : MATRICES AND SYSTEM OF LINEAR EQUATIONS

4.1 Matrices

Tutorial 1 of 4

1. Given that A  1 1 , B  1 3 0 , C  1 5 3 4 0
0 2  3 6 2 4 3 and D  2 4 3 evaluate
5 2  1 3 2

the following; (b) D2 CB

(a) BC  2A

2. Given that A  2 3 , B  2 4 and C  3 1
5 1  1 6 5 4 . Find

(a) ATB (b) (BC)T (c) (A+B)T

2 1 3
3. If A  4 8 6 , find the minor of A.

0 7 5

4. Write the cofactor matrix of the following;

 2 4 2  4 10 2
 3 5
(a) A  2 5 4  (b) B   8

 4 1 3 6 3 3 

ANSWER

1   9 2 31 1 2  2 (a) 1 38 (b) 14 33 (c) 0 6
a) 11 3 b) 22 27 12  7 14 25 7 5
6 
10 19 17 

3  2 20 28 4 19 10  22   6 6  6 
16  (a) 14 2 18 b)  24 0  48
10 14    6  4 2   44 4  68

18 0 12

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AM015/4 Matrices Sesi 2021/2022

4.2 Determinant of Matrices.

Tutorial 2 of 4

 2 4 2
1. Given that matrix A  2 5 
4  , find the determinant using cofactor expansion by

 4 1 3

a) first row b) third column

2. Given that A  3, B  10 , find

(a) A2 (b) AB (c) AT

011 232

3. Given that determinant 2 3 2  8 . Hence, find 0 1 1 .

0 1 3 0 1 3

21 1 63 3 21 2

4. If 0 5 2  21. Hence, deduce the values of a) 0 5 2 b) 0 5 4

1 3 4 1 3 4 1 3 8

1 (a) -34 (b) -34 ANSWER (b) -30 (c) 3
38 2 (a) 9 (b) 42
4 (a) 63

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AM015/4 Matrices Sesi 2021/2022

4.3 Inverse of a matrix.

Tutorial 3 of 4

1. Find the inverse matrix for matrix 2 1 0 by using the adjoint matrix.
A  4 1 5
1 2 
1

1 1 1  6 4 2
Given that matrix A  1 2
2. 0 and B   3 5 1  . Prove that AB  kI , where I is an
 
2 3 7 3 1 1 

identity matrix and k is a constant. Find k and hence, find B1 .

1 2 1

3. Given that matrix A  1 1 0 , find the matrix A2 . If A3  A2 6A  I , where I is an

3 1 1
identity matrix, find the inverse matrix of A.

1 7 2 5 ANSWER 1 1 1
 27 27 27  2

13  6 6 6
 27  4 27 10 27  
k=6 B1   1 0 1 
 6 1 3 
 1  1 2   7 
 9 9 9  1 2 6
3

3 1 1 1 
1  2 
1 

 4 5  3

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AM015/4 Matrices Sesi 2021/2022

4.4 System of linear equations with three variables.
Tutorial 4 of 4

2 1 1  1 3 1
1. If P  1 0 1 and Q  1 5 1 , find PQ and deduce the matrix P1 . Express the
3 1 4 1 1 1 

2x  y  z  3

following system in its matrix form. x  z  1 . Hence solve the linear system.

3x  y  4z  0

2. Solve the following systems of equation system by using,

a) The Inverse matrix b) The Cramer’s Rule

x2y z 5 2x  3y  4z  1

2x  3y  2z  1 2x  2y  3z  2

x 3y  2z  2 x  2y  2z  3

3. Three weight lifters, Halim, Amin and Ahmad entered a weight-lifting contest in which they
could make up their total weight by adding three different sized weights. Halim used 4 of

weight X, 3 of weight Y and 5 of weight Z for a total lift of 295 kg . Amin used 2 of weight X,

5 of weight Y and 3 of weight Z for a total lift of 295 kg . Ahmad used 3 of weight X, 4 of weight

Y and 5 of weight Z for a total lift of 310 kg.
(a) Use the above information to determine the three separate weights by using
Cramer’s Rule method.
(b) If Amin replaced one weight Z by a weight Y, would this change the result of the
competition?

3 2 2  1 4 2  MN. Hence find M-1.
4. (a) If M  4 1 2 and N  8 5
2  , find

2 1 3  2 1 5

(b) In a supermarket, there were three promotion packages A, B and C which offer shirts,

long trousers and neckties. The number of each item and the promotion price for each

package is given in the following table:

Package No. of No. of No. of Promotional price

Shirts trousers neckties (RM)

A32 2 256

B41 2 218

C21 3 173

By using x, y and z respectively to represent the price of a shirt, a pair of trousers and a

necktie, write down a matrix equation to represent the information above. By using (a)

solve the equation to determine the price of each item.

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AM015/4 Matrices Sesi 2021/2022

 1 2 1
 2 . Find M 2  2M 6I , where I is the 33 identity
5. (a) M is a matrix given by M   1 0

1 3 1

 matrix. Show that M M 2  2M  6I  9I . Hence, deduce the inverse matrix M 1 .

(b) A factory produces three types of medals for a sports championship, namely gold, silver
and bronze. The total profit obtained from the sale of 1 gold, 2 silver and 1 bronze medal
is RM 20. The profit obtained from the sale of 1 gold and 2 bronze medals is RM 13. The
profit obtained from the sale of 3 silver and 1 bronze medal is RM 11 more than the profit
obtained from the sale of 1 gold medal. By using x, y and z respectively to represent the
profit obtained from the sale of 1 gold, 1 silver and 1 bronze medal respectively, obtain a
matrix equation to represent the above information given. By using information from (a)
determine the profit obtained from the sale of each type of medal.

6. Use the Cramer’s Rule method to solve the following system of linear equations.

2x +3 y – z = 1
3x + 5y + 2z = 8
x – 2y – 3z = – 1

Answers

1 2 0 0 2 (a) x  1 , y   23 , z  52
3 21 21
PQ  0 2 0 x  3, y  1, z  2
0 0 2 (b) x = 3 , y = -5 and z = -2

3 a) weight X = 25, weight Y = 40, 4 (a)  1 4  2 
 9 9 9
weight Z = 15
(b)Yes, Amin lifts 320 kg M 1   8  5  2 9 
 9 9 

 2  1 5
9 9 9

(b) x = RM 30, y = RM 68,z = RM 15

5 2 1  4  6 x =3
9 9 
 3  y = 1

1 2 1
(a)  9 
 3 5 9  z= 2
 1 9 2 
9
3

(b) Profit obtained from the sale of gold

medal = RM 7, silver medal = RM 5 and

bronze medal = RM 3

5

AM015/4 Matrices Sesi 2021/2022

EXTRA EXERCISE

5 2 3  a 2 36
1. If P  1 3,Q  
1 4   b 2 24  and PQ = 4I , where I is the 33

3 1 2 26 2 c 

identity matrix, determine the values of a, b and c.

 1 2 1 
 1 find
2. A and B are square matrices such that BA  B1 . If B   1 0

1 1 2 

A1 and hence, find A.

2 3 4  9 27 3 
1. Given C  5 0 1 and D   7  26 
2 18  . Find CD. Hence, determine [7m]

8 9 3  45 6  15

C-1.

2. (a) If Y  7 2 [5m]
3 1 , show that Y satisfies the equation Y 2  8Y 1 0 where I

is an identity matrix 2x2.Hence, find Y-1.

1 2 0 2 4
  1 3 2 , determine matrix R such
(b) Given P   5 4  and Q 

 3  2

 2  2 5 [6m]
that (PQ)T  R   1 4 1 .

 0 3 4

(c) Table below shows the floor area of the rooms in Amir’s house.

Room Floor Area (square metre)

Living room 25

Bedroom 1 15

Bedroom 2 10

Bedroom 3 10

Kitchen 10

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AM015/4 Matrices Sesi 2021/2022

Amir intends to replace his home floor to ceramic tiles, parquet or mosaic. If [3m]
he uses ceramic tiles for the living room, parquet for all the bedrooms and
mosaic for the kitchen, the total cost is RM 2625. If he uses ceramic tiles for
the living room and bedroom 1, and mosaic for the other bedrooms and
kitchen, the total cost is RM 2450. However, if bedrooms 2 and 3 use parquet
and the other rooms and kitchen use ceramic tiles, the total cost is RM 3200.

Let the prices per square metre of ceramic tiles, parquet and mosaic be x, y [9m]
and z respectively.

(a) Obtain a system of linear equations to represent the given information.
Hence, state the equations in the form of AX=B.

(b) Based on (a), find the inverse of matrix A. Hence, determine the price (RM)
per square metre for each ceramic tiles, parquet and mosaic.

3 1. Given = [22 34] and = [12 −11] where −1 −1 = [10 13]. Find B. [5 ]

2. Given = [−34 4 ]. Show that 2 = −7 and 3 = −7 . Hence, find [8 ]
−3

4 5.

3. 3. Aini spent RM 11 to buy 4 pens, 2 rulers and 1 eraser. Nora spent RM 9 for 2

pens, 3 rulers and 2 erasers. While, Dila spent RM 13 for 3 pens, 4 rulers and 3

erasers.

4. (a) (i) Write the information given above as a system of linear equations in the [2 ]
form of matrix equation, AX = B.

5. (ii) Find the cofactor matrix of A and hence determine the determinant of matrix A.

6. (iii) Determine the adjoint matrix of A and hence, find the inverse of A. [4 ]
[4 m]
7. (b) What is the price of each pen, ruler and eraser ? [2 m]
[3 m]
8. (c) How much will each person spend if the price of each pen, ruler and eraser is
increased by 20% ?

9.

7

AM015/4 Matrices Sesi 2021/2022

 1 1 1 2  3 0 0 1
4  0 , 1  0 0 and  
1. Given A=  0 1 S=  , V= 1 U =  p 

1 0 0 P 0 0 2 1

(a)Compute W = STV. Hence find the value of p such that WU = p2 . [7m]
[6m]
(b) Compute (VA)T . Hence, show that (VA)T = ATVT

2. A system of linear equations is given as follows,

2x + 3y + 4z = 11

4x + 3y + z = 10

x + 2y + 4z = 8

(a) Write the above system of linear equations in the form of matrix equation
AX = B where A, X and B are the coefficient matrix, the variable matrix
and the constant matrix respectively. Hence determine

(i) the determinant of A, |A|,

(ii) the matrix of cofactors of A, and

(iii) the adjoint matrix A, adj(A). [9m]
[6m]
(b) Find the inverse matrix, A-1. Hence solve the above system of linear
equations.

8

AM015/4 Matrices Sesi 2021/2022

5 1. The table below shows the weight (in kilogram, kg) of three types of fruits
namely mango, pineapple and watermelon which were supplied by a
supplier to three restaurants K, L and M on a daily basis.

Restaurant Fruits (kg)

Mango Pineapple Watermelon

K555

L 10 15 10
M 10 15 15 [4m]

The payment received by a supplier from K, L and M restaurants are RM66, [10m]
RM147 and RM153 respectively. If x, y and z are the prices for each kilogram of
the fruits,

a. State a system of linear equations for the above information given
above. Hence, write in the form of matrix equation, AX = B.

b. Based on 4(a) , find the inverse matrix of A. Hence, find the price per
kilogram for each of the fruits.

6  1 4 5
1. (a) Given matrix A = 4 0 2
 8 0 3

Find the cofactor and adjoint for the matrix. Hence find A1

(b) Find the values of w,x,y and z in the following matrix equations.

3 7  3 x 2  4 2y
6 1 5 z 5w 7 

9

AM015/4 Matrices Sesi 2021/2022

ANSWERS (EXTRA EXERCISE)

1. a  22,b  14,c  44

1  1 11 1
 8 8 
2 1 5 
 3 1
2. A1   2 A   1 3 1
8 8 2
2 4 2 
 1 5 12
 8 8

  3 9 1
2 1.  47 47 
 47 

CD 141I C1  1471  26 6
141 47 
 
  15 2  5 

 47 47 47

2. Shown, Y 1  1  2 3. x = RM 50, y = RM 35, z = RM 15
 3 7 

3 1. = [195 1182]
2. 4 = 2. 2 = (−7 )(−7 ) = 49

5 = 4. = 49 . [−34 −43] = [−114976 −119467]

3. a) (i) 4 + 2 + = 11
2 + 3 + 2 = 9
3 + 4 + 3 = 13

4 2 1 11
[2 3 2] [ ] = [ 9 ]
3 4 3 13

1 0 −1
(ii) cof A = [−2 9 −10]

1 −6 8

| | = 3

10

AM015/4 Matrices Sesi 2021/2022

(iii) −1 = 1
| | .

−2 1 −2 1
1 9 13
1 −10 −6] = [ 0 3 3
= 3 [0 8 −1
−1 3 −2]

−10 8

33 3

b) =

= −1
2

= [1]
1

Price of pen is RM 2, ruler is RM 1 and eraser RM 1.

c) Increased by 20%.....price of pen = RM 2.40, price of ruler = RM 1.20,
price of eraser = RM 1.20
Aini will spend RM 13.20, Nora will spend RM 10.80 and Dila will spend
RM 15.60.

4 1. (a) W= 6 1 2 p, p = 2 , p= -3

 3 0 2
(b) (VA)T= 3 1 
0  , Shown.

 3 0 0 

2 3 4  x  11
2. 4 1   10
3  y  =

1 2 4  z   8 

10 15 5   10 4 9
(ii) cofactors of A = 4 1 (iii) Adj (A) = 15 
(i) |A| = -5 4 4 14 

9 14 6  5 1 6

2 4 9
5 
 4 5  x = 2 , y = 13 , z = 3
 5 14  555
(b) A-1 =  3 X=A-1B
5
 
 1 1 6 

 5 5

11

AM015/4 Matrices Sesi 2021/2022

5 (a) 5 + 5 + 5 = 66
10 + 15 + 10 = 147
10 + 15 + 15 = 153

 5 5 5   x   66 
10 10   147
15  y  =

10 15 15  z  153

A . X= B

3 0  1
 5 
 5

A-1 =  2 1 0 
5 5 
 
0 1 1 

 5 5 

x  RM9.00 , y  RM3.00, z  RM1.20

6  0 28 0 
(a) Cofactor of A = 12 37 32
 8 22 16
 0 12 8 
Adj(A)  28 37 22
 0 32 16 

 0 12 8
 112 
 37 112 
 28 112 22 
A-1 = 112
112 
 
0 32 16 

 112 112 

(b) y1 z2 w  9
2 5
x  1
3

12

AM015 – Functions And Graphs

CHAPTER 5: FUNCTIONS AND GRAPHS

LECTURE 1 OF 5
At the end of the lecture, you should be able to

i. define a relation and a function.
ii. understand the concept of one-to-one and onto.
iii. identify a function from the graph by using vertical line test.
iv. sketch the graphs of constant, linear, quadratic and cubic functions.
v. state the domain and range of the constant, linear, quadratic and cubic functions.

5.1 FUNCTIONS
A) Relations

A relation is a correspondence between a first set (x), called the domain and a second set
(y), called the range, such that each member of the domain corresponds to at least one
member of the range.

AB

2. .6

3. .9

5 . . 10

domain range

B) Function

A function is defined as a relation in whom every element in the domain has a unique image

in the range. In other words, a function is

i) One-to-one relation
ii) Many-to-one relation

AB AB

a1 a .1
b2
c3 b
.2

c .3
.4

One-to-one relation and onto One-to-one relation and not onto

Page 1 of 28

AB AM015 – Functions And Graphs
AB
a.
b. .2 a. .1
c. b. .2
c. .3

Many-to-one relation and onto Many-to-one relation and not onto

Mapping is another name for function. A mapping or function f from a set A to a set B is usually
written as f : A  B.
If an element x, object of set A is mapped into an element y in set B we say that y is an image
of x. The image of x is thus represented by f(x) and we write y = f(x).

AB

f(x)

X. .Y

C) Identify Relation and Function
To determine whether a relation is a function, we can use:
i) arrow diagram as shown before.
ii) vertical lines test.

Vertical Line Test
Vertical lines are drawn parallel to the y – axis. If the vertical line cuts the graph at only one point,
then the graph is a function.

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AM015 – Functions And Graphs
Example 5.1
Consider the graphs below and state whether the graphs represent functions or not.

f(x) f(x)
y = x3 x = y2

x 0
0
x

The graph is a function because The graph is not a function
the vertical line cut the graph at only because the vertical line cuts the
one point. graph at more than one point.

D) Domain and Range
The domain of the function is the set of all first components of the ordered pairs of the
function. The range of a function is the set of all second components of the ordered pairs of
the function.
AB

a. .d
b. .e
c.
range
domain

There are two methods to determine the domain and range for a function.
 graphical approach
 algebraic approach (p/s: NOT COVERED IN LECTURE, BUT WILL DO IN TUTORIAL)

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AM015 – Functions And Graphs

Graphical Approach

i) Constant Function
f(x)

c f(x)=c

x

0 Range : {c}

Domain :  ,

Example 5.2
Sketch the graph of f(x) = 3. Hence, find the domain and range.

ii) Linear Function f(x)

f(x)
f(x) = ax + b, (a > 0)

b b
0x f(x) = ax + b, (a < 0)

x
0

Domain :  ,
Range :  ,

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AM015 – Functions And Graphs

Example 5.3

Sketch the graph of the following functions. Hence, find the domain and range.
a) f(x) = 5 – 2x

b) f(x) = 2x, – 2 < x  3

iii) Quadratic Function f(x)

f(x) x
f(x) = ax2, a > 0 0

0 x
f(x) = ax2, a < 0
Domain :  ,
Domain :  ,
Range : [0, )
Range : (, 0]

Page 5 of 28

f(x) AM015 – Functions And Graphs
f(x) = ax2 + bx + c, a > 0
f(x)

c
xx

00

c

x1   b
2a

y1  ax12  bx1  c

Domain :  ,  Domain :  ,  f(x) = ax2 + bx + c, a < 0
Range : [y1, )
Range : (, y1 ]

Example 5.4

Sketch the graph of the following functions. Hence, find its domain and range.
(a) f(x) = x2 – x – 2

(b) f(x) = –2x2 + 3x + 2,  3  x  4.

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iv) Cubic Function AM015 – Functions And Graphs

f(x) f(x)
f(x) = ax3, a > 0 f(x) = ax3 + d, a < 0

0x d
x

0

Domain :  , Domain :  ,
Range :  , Range :  ,

f(x) f(x) f(x) = ax3 + bx2 + cx + d, a > 0
f(x) = ax3 + bx2 + cx + d, a < 0
d
d
x
x 0
0

Domain :  , Domain :  ,
Range :  , Range :  ,

Example 5.5

Sketch the graph of the following functions. Hence, find its domain and range.
(a) f(x) = x3 + 3

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AM015 – Functions And Graphs

(b) f(x) = –2x3
(c) f(x) = x2 (2 – x )
(d) f(x) = x3 – x2 – 6x + 2

Exercise
Sketch the graph of the following functions. Hence, state the domain and range.

a) f (x)   7
3

b) f (x)  2x  5
c) f (x)  x2  2x 1

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AM015 – Functions And Graphs
LECTURE 2 OF 5
At the end of the lecture, you should be able to

i. sketch the graphs of surd, absolute value and piecewise functions.
ii. state the domain and range of the surd, absolute value and piecewise functions.

v) Surd Function f(x)

f(x)

f(x) = √

0x 0x

Domain : [0, ) Domain : [0, ) f(x) = −√
Range : [0, ) Range : (, 0]

f(x) f(x)

f(x) = √ +

a x
0

-a 0 x

Domain : [a, ) f(x) = −√ −
Range : [0, )
Domain : [a, )
f(x) Range : (, 0]

f(x) = √− − f(x) x

0a

-a 0 x f(x) = −√ −

Domain : (,  a] Domain : (, a]
Range : [0, ) Range : (, 0]

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AM015 – Functions And Graphs
Example 5.6
Sketch the graph of the following functions. Hence, find the domain and range.

(a) f(x) = x  3

(b) f(x) =  x  5

(c) f(x) = 3  x

vii) Absolute Value Function f(x)

f(x) f(x) = |x| 0x

x f(x) = –|x|

0 Domain :  ,

Domain :  , Range : (, 0]

Range : [0, )

Page 10 of 28

AM015 – Functions And Graphs

f(x) f(x)

f(x) = |x+a|

x f(x) = |x| + a

a
0x

–a 0

Domain :  , Domain :  ,
Range : [0, )
Range : [a, )

Example 5.7
Sketch the graph of the following functions. Hence, find the domain and range.
(a) f(x) = |x+3|

(b) f(x) = |x| – 2

viii) Piecewise Function

(a) f ( x)   x 2, x0
 x 5, x0


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AM015 – Functions And Graphs

2x  4,  4  x  1
(b) f (x)  3  4x, -1 x 1

 x, x 1

Exercise
Sketch the graph of the following functions. Hence, state the domain and range.
a) f (x)  2  x 1
b) f (x)  x 1

LECTURE 3 OF 5
At the end of the lecture, you should be able to

i. represent a composite function by an arrow diagram.
ii. find composite function.
iii. find one of the functions when the composite and the other functions are given.

Definition of Composite Functions:
It is also possible to take the output values from one function and use them as the input values
for another function. The functions which are composed in this way are called composite
functions or function of a function.

Consider two functions f(x) and g(x).
We define f  g(x) = f [g(x)] meaning that the output values of the function g are used as the
input values for the function f.

Page 12 of 28

This can be represented in an arrow diagram: AM015 – Functions And Graphs

f og C
f[g(x]
AB

gf
x g(x)

Similarly, we define g  f(x) = g [f(x)] meaning that the output values of the
function f are used as the input values for the function g.

This can be represented in an arrow diagram. C
g[f(x)]
g of

AB

fg
x f(x)

Note :

i. ( f  g)(x)  f(x). g(x)

ii. (g  f)(x)  g(x). f(x)
iii. ( f  f)(x)  f(x)2
iv. f 2 (x)  [f(x)]2
v. f 2 (x) = ff(x)

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AM015 – Functions And Graphs

Example 5.8
If f(x) = 3x + 1 and g(x) = 2 – x, compose the

(a) (f  g)(x)

(b) (g  f)(x)

Example 5.9
If f(x) = 2x – 1 and g(x) = x3, find the values of

(a) gf(3)

(b) fg(3)
(c) f 2(3)

Example 5.10 (x  1).

The functions f and g are defined by f (x)  2  x and g(x)  3
x 1

(a) Show that f 2(x) = x.
(b) Find an expression for g2(x)

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AM015 – Functions And Graphs

Example 5.11
Given that f(x) = 2x + 3 and f0g(x) = 10x – 9, find g(x).

Example 5.12
Given the function f (x)  x2  3x 1, x   3 . Find the function g(x) if f [g(x)]  2x 3.

2

Example 5.13
If g(x) = 2x + 3 and fg(x) = 10x – 9, find f(x).

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AM015 – Functions And Graphs
Example 5.14
If fg(x) = 4x2 – 2x + 1 and g(x) = 2x + 1, find gf(x). Subsequently, find the values of x that
satisfy fg(x) = gf(x).

Example 5.15

Find the function of g(x) , given f (x)  x2 1, x  0 and ( f  g)(x)  x2  2x  2, x  1.

Exercise

Given fg(x)  x2 1. Find
a) g(x) if f (x) 1 2x
b) f (x) if g(x)  x  3

Page 16 of 28