AM015 20212022

7/7/2022

NOTE:

cofactor of a12 m11 m12 m13 
minor matrix : M  m21 
c12 = (-1)1+2 m12 m22 m23 

m31 m32 m33 

cij = (-1)i+j mij EXAMPLE 2
Cofactor matrix,
1 2 0 

  1 m11  1 12 m12  1 13 m13  Let A=  3 1 1  ,
11  1 22 m22   
  1 32 m32
 1 21  123  2 0 3
C  1 31 m21  133 m23 
 m31 m33 
find all the minors and cofactors of A.
 
Hence, form the matrix minor and cofactor A.

m11 m12 m13 
C  m21 m22 
m32 m23 
m31
m33 

SOLUTION 1 2 0 1 2 0

11 
m11 = 0 3
3A  1 1 A   3 1 1 
m12 = 3 1    
23
31 2 0 3

m13 = 2 0 
 
 2 0 3 
20
M21 = 0 3 =

10
m22 = 2 3

12
m23 = 2 0

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 1 2 0
3
A   2 1 1  minor A 
 
cofactor A 
 0 3 

m31 = 2 0
11

10

m32 =

31
12

m33 =

31

Conclusion LEARNING OUTCOMES

Minor and Cofactor At the end of this topic, students should
be able to :
The minor, mij , is the determinant of
(n -1) x (n -1) matrix obtained by (b) determine the determinant of a matrix by
using expansion of the cofactors.
deleting the i th row and j th column of A

The cofactor, c ij , of the element aij is

cij = (-1)i+j mij

Determinant of 3x3 matrix EXPANSION OF THE COFACTORS

In matrices, determinants are the special A method for evaluating determinants.
numbers calculated from the square matrix. Expansion by cofactors involves following
The determinant of a 3 x 3 matrix is any row or column of a determinant and
calculated for a matrix having 3 rows and 3 multiplying each element of the row or
columns. column by its cofactor. The sum of
these products equals the value of the
Determinant of A is denoted determinant.
as det A or |A| .

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a11 a12 a13
A a21 
If A aij  33 a22 a23 

By expanding along the first row. a31 a32 a33 By expanding along the first column.

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

A  a11c11  a12c12  a13c13 A  a11c11  a21c21  a31c31

cij = (-1)i+j mij cij = (-1)i+j mij

A  a11m11  a12m12  a13m13 A a11m11 a21m21 a31m31

EXAMPLE 1: SOLUTION

3 1 4  3 1 4

Let A  1  A  aij cij A  1 2 7 
 
2 7 , find A by expanding along
5 1 10
5 1 10
a21c21 a22c22 a23c23
a) second row
b) first column ( i = 2, j = 1,2,3 )
a) Expansion along 2nd row;

3

A  a2 jc2 j
j 1

cij = (-1)i+j mij 3 1 4 A  aij cij 3 1 4 

A  1 2 7  b) Expansion along 1st column, A  1 2 7 
 

5 1 10 5 1 10
A  a21c21  a22c22  a23c23

(j=1 , i = 1,2,3 )

3

A  a i1c i1

i 1

 a11c11  a21c21  a31c31

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cij = (-1)i+j mij 3 1 4 

A  1 2 7 


5 1 10 HINT

A  a11m11  a21(m21)  a31m31 Choose the row or column that
has the most zeroes.

When aij = 0 , aij cij = 0

A  aij cij

EXAMPLE 2

A  a21(m21)  a22m22  a23(m23 )

2 1 3

Find the determinant of A  0 0 1 


0 5 4

by using expansion of the cofactors.

2 1 3

A  0 0 1 


0 5 4

Properties of Determinant: Properties of Determinant

( i ) k A  k n A Where k is scalar and n (i) kA  k n A

1. is an order of the EXAMPLE 2 −2 and 2 4 −4
4 1 =0 8 2
matrices If 1 0
=0 0 −4 0 0
(ii) AB  A B −2

(2ii.i ) A T  A are a 3x3 matrix, verify that
|B| = 2 |A|.
(iv) The value of the determinant changes sign
3w. hen two rows or columns are intercharged.

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Solution Properties of Determinant

1 2 −2 (ii) AB  A B
A= 0 4 1 =
EXAMPLE | for the
−2 0 0
2 4 −4 Find | |,  | | and | 20 1
matrices    = 0 −1 −2
B= 0 8 2 =
−4 0 0 1 −2 2 3 1 −2
= 0 3 2    
Solution
101

Properties of Determinant

(iii) A T  A

EXAMPLE

Show that A T  A for matrix
3 1 −2

=2 0 0
−4 −1 5

Solution 31 −2 Properties of Determinant
=2 0 0
3 5
A= 2 −4 −1
(iv) The value of the determinant changes sign
−4 1 −2
00 when two rows or columns are intercharged.
−1 5
EXAMPLE

Given that the matrix A and matrix B as below:

3 1 −2 200

= 2 0 0 B= 3 1 −2

−4 −1 5 −4 −1 5

Show that | | = − .

15

Solution 7/7/2022

3 1 −2 TOPIC 3.0 MATRICES AND
A= 2 0 0 SYSTEMS OF LINEAR
EQUATIONS
−4 −1 5
3.3 Inverse of Matrix
Interchanged 1st row and 2nd row,

200
B = 3 1 −2

−4 −1 5

LEARNING OUTCOMES Adjoint Matrices

At the end of this topic, students should Let C  cij be the cofactor matrix of A.
be able to :
Adjoint of matrix A (adj A) is defined as the
compute the inverse of a non-singular matrix transpose of the cofactor matrix that is
up to 3x3 using
(a)adjoint matrix; and Adj A  CT  cij T  c ji 
(b)the property of AB=KI.

REMEMBER!! Example 1
Cofactor, cij  (1)i j mi j
1 2 3
 m11  m12 m13  Given A  3 2 4
  m22 
C    m21  m32  m23  1 1 3

Find the adjoint of A.

  m31 m33 

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Solution 1 2 3  m11 m12  m13 The cofactor matrix of A is
A  3 2 4 C  m21 m22 m23
a) m32  m33
1 1 3  m31

c11  2 4 c12   3 4 c13  32
1 3 11
= 13
==

2 3 c22  1 3 c23  1 2
c21   1 3 1 3 1 1

= ==

c31  2 3 c32   1 3 c33  1 2
4 3 4 3 2
2

= ==

Example 2 Extra example

T

1 2 2

Given P  2 10 5 . Find adj P.
1 3 3

Solution 1 2 2  m11  m12  m13  Solution 1 2 2  m11  m12  m13 

P  2 10 5 C   m21  m22  m23  P  2 10 5 C   m21  m22  m23 
1 3 3  m31  m32  1 3 3  m31  m32 

 m33   m33 

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There are 2 methods to obtain
the inverse of a matrix.

a) Adjoint Method A1  1 adj A .
A

b) Use the property AB = kI.

FINDING INVERSE BY USING Notes A 0
ADJOINT METHOD

A1  1 adj A ~ A is a non-singular matrix
A ~ Inverse matrix exists

A1  1 A 0
A
~ A is a singular matrix
~ Inverse matrix does not exist

Example 3

 1 3 2

Find the inverse of B   0 2 2 .


2 1 0

Solution

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1

= | |  If AB  BA  I ,

 B  A1 and A  B 1

Hence, if AB  I
then A1  1 B


and also B 1  1 A


Notes FINDING INVERSE BY USING THE PROPERTY OF

If A B  I where A and B are square AB  kI
matrices, then B is called the inverse
matrix of A and is written as A-1. Example 4
Thus
1 2 3  1 1 1
A AB-1 = AB-1 A = I
Given A  2 3 4 and B  10 4 2 .
Identity  7 3 1
matrix 1 5 7

It is known that A B  kI , k is a

constant,

I is a 3x3 identity matrix.

Find k and hence deduce A-1.

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Solution AB  2 I

 1 2 3   1 1 1 Hence,
 
AB   2 3 4  10 4 2 

 1 5 7   7 3 1

Example 5 3 1 1 Solution A 2  mA  nI  0

GivenA  1 3 1 ,

1 1 3

Find the values m and n such that

A2  mA  nI  0

where I is the 3x3 identity matrix and 0 is the
zero matrix.

Use this relation to obtain A-1 and

show that A3  39 A  70I .

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CONCLUSION TOPIC 3.0 MATRICES AND
SYSTEMS OF LINEAR
adj ACT  cij T  cji  EQUATIONS

A 1  1 adj A 3.4 System of Linear
A Equations with Three Variables

LEARNING OUTCOMES SYSTEM OF LINEAR EQUATIONS WITH THREE VARIABLES

At the end of this topic, students should Consider the system of linear equations with three
be able to : unknown x1, x2 and x3.

(a) express a system of linear equations in a11 x1  a12 x2  a13 x3  b1
the form of AX=B.
a 21 x1  a 22 x 2  a 23 x3  b2
(a) solve the unique solution to AX=B using
inverse matrix. a31 x1  a32 x2  a33 x3  b3

The linear equations above can be express as a

single matrix equation as:

a11 a12 a13   x1  b1 
a21    b2 
a22 a 23   x2  = 

a31 a32 a33   x3  b3 

 a11 a12 a13   x1  b1  Example 1
A = a21 a22 
a23  , X =  x2  , B = b2  Express the system of linear equations
   in the form of AX =B.
a31 a32 a33 
 x3  b3  5x  8y  5z  36
4x  6 y  6z  30
Thus, the system of linear equations 5x  9 y  7z  40
can be express as
Solution

AX = B

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Example 2 Solution

Two companies P and Q decided to award prizes to their a)
employee for three work ethical values, namely
punctuality (x), creativity (y) and efficiency (z). b)
Company P decided to award a total of RM3850 for the
three values to 6, 2 and 3 employees respectively; while
company Q decided to award RM3200 for the three
values to 4, 1 and 5 employees respectively. The total
amount for all the three prizes is RM1000.

a) Form a system of linear equations.
b) Express the system of linear equations in the form of

AX=B.

(A) Using the Inverse Matrix method in solving Example 3
System of Linear Equations
3 1 2

If the number of equations in a system equals Find the inverse of A   3 2 2  by using
to the number of variables and the coefficient  
matrix has an inverse, then the system will 1 0 1 
always have a unique solution that can be adjoint method.
found by using the inverse of the coefficient
matrix. Hence, solve the system of linear equations

by using the inverse matrix method.

3x1  x2  2x3  11

3x1  2x2  2x3  10

AX = B  X = A-1B x1  x3 5

Solution :

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=

=

Example 4

3 1 2  2 1 2

Given A   3 2 2  and A1 1 1 0 
  2 1 3 
1 0 1 

Solve the system of linear equations by
using the inverse matrix method.

9x1  3x2  6x3  15
9x1  6x2  6x3  27
3
3x1  3x3

Solution :

9 3 6   x1  15 
9     
6 6   x 2    27 

3 0 3  x3   3 

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3.0 MATRICES AND SYSTEM OF LIEAR EQUATIONS AM015_2022/2023

3.1 Matrices

3 0 2 1 3 x 0 x  y
 2 1 0   0 , R  0 1 
1. Given the matrices P  , Q   1 2 2 3 2  and
 0
0 

7 3 18
 
S   3 2 2  . If QP  2R  S , determine the values of and .

 0 8 0 

3 04 13

2. Given = [ 5 1 6] and = [0 0 ]. Find

−1 2 7 3 −4

(a) (b)

(c) ( ) (d) Is ( ) = ?

500 −2 0 0

3. The matrices A and B are given by = [1 8 0] and = [−1 −5 0 ]. Find the

135 −1 −3 −2

values of and if = + , where is the 3 × 3 identity matrix.

3.2 Determinant of Matrices

1 0 −1
4. If = [2 + 3 5 ] , | | = 38 and > 0, find

3 −2 + 2

(a) the value of > 0.

(b) | |.

2 1 6 
5. Given the matrix W  3 5 3 , find W . Hence, find the value of V if

6 2 8 

3 5 3
V  4 
2 12  .

6 2 8 

1

3.0 MATRICES AND SYSTEM OF LIEAR EQUATIONS AM015_2022/2023

6. Find the inverse of the matrix = [42 13]. Hence, find the matrix such that [24 13] =
[1100 76].
111

7. Given matrix = [1 0 1], find
0 −1 2

(a) minor ,
(b) cofactor matrix of ,
(c) adjoint ,
(d) determinant of .
(e) Hence , find inverse of .

121
8. Given that matrix = [1 −1 0], find the matrix 2. If 3 − 2 − 6 = , where is

311
an identity matrix, find the inverse matrix of .

1 0 1
Given the matrix , A  3 x  2 
9. 4  . If A  3, find the values of

1 1 x  3

(i) x

(ii) AT

(iii) A 2

3.3 Inverse of a Matrices 0
−5]. Find −1 by using adjoint method.
21 2
10. Given = [4 −1

1 −1

2 −1 1
11. Find the inverse matrix of = [1 0 1] by using adjoint method.

3 −1 4

2

3.0 MATRICES AND SYSTEM OF LIEAR EQUATIONS AM015_2022/2023

3 6 3  1 1 4 
 03   
12. Given that matrices P   3 3 and Q   1 2 5  . Find PQT . Hence, find P1
9 3 1 1 3

.

13. Matrices A and B are given as

 1 2 3  4 1 4 
  ,  
A   1 0 4 B   1 1 3.5  .
0 2 2 1 1 1

Find AB and hence find A1 .

3.4 System of Linear Equations with Three Variables

21 0
14. Given = [4 −1 −5]. Find −1 by using adjoint method.

1 −1 2
Hence, solve the following system of linear equations.

4x  2y  2
(a) 8x  2y 10z  10

2x  2y  4z  12

2x1  4x2  x3  7
(b) x1  x2  x3  2

5x2  2x3  7

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3.0 MATRICES AND SYSTEM OF LIEAR EQUATIONS AM015_2022/2023

7 2 5 
 
2 1 0  27 27 27 
Given = [4 −1 −5] and 4
15. −1 2 A1   13  10  . Solve the following system of
1  27 27 27 
 1 
 1 2 

 9 9 9 

linear equations:

2 + = 7

4 − − 5 = 2

− + 2 = −7

1 3  1 
 2 2 
2 −1 1  2 5 
Find the inverse matrix of = [1 0 1] and 2
16. −1 4 P1   1 1  1  .
3 2 2 
 
 1 1 

 2 2 2 

Express the following system of linear equations in the form of PX=Q.

2 − + = 3

+ = 1

3 − + 4 = 0

Hence, solve the system of linear equations.

1 2 0  2 2 2
   .
17. Given that M   3 2 1  and N   1 1 1 Show that MN=4I, where I is a 3x3
2 4 1 8 0 4

identity matrix. Hence,
(a) determine M 1 .

(b) solve the following system of linear equations:

4

3.0 MATRICES AND SYSTEM OF LIEAR EQUATIONS AM015_2022/2023

x2y 5
3x  2y  z  10
2x  4y  z  13

18. An osteoporosis patient was advised by a doctor to take enough magnesium, vitamin D and
calcium to improve bone density. In a week, the patient has to take 8 units magnesium, 11
units vitamin D and 17 units calcium. The following are three types of capsule that contains
the three essential nutrients for the bone:

Capsule of type P: 2 units magnesium, 1 unit vitamin D and 1 unit calcium.
Capsule of type Q: 1 unit magnesium, 2 units vitamin D and 3 units calcium.
Capsule of type R: 4 units magnesium, 6 units vitamin D and 10 units calcium.

Let x, y and z represent the number of capsule of types P, Q and R respectively that the
patient has to take in a week.

(a) Obtain a system of linear equation to represent the given information. Hence,

x
 y
express the system of linear equations in the form of AX  B, where X   .

 z 

(b) Find the inverse of matrix A from part (a) by using the adjoint method.

Hence, find the values of x, y and z.

(c) The cost for each capsule of type P, Q and R are RM10, RM15 and RM17

respectively. How much will the expenses be for 4 weeks if the patient follows
the doctor’s advice?

ANSWERS

1. = 5, = 3

15 −7 [−157 23 −2301]
−9
2. (a) [23 −9 ] (b)

20 −31

(c) [−157 23 −2301] (d) Yes. ( ) = .
−9

3. = −1, = 3

4. (a) = 1 (b) | | = 38

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3.0 MATRICES AND SYSTEM OF LIEAR EQUATIONS AM015_2022/2023

5. – 94, 188

6. 3 − 1 = ( 10 11

−1 = ( 2 2); 2)
−2 1 −10 −5

11 = 1 12 = 2 13 = −1 1 −2 −1

7. (a) 21 = 3 22 = 2 23 = −1 (b) Cofactor = [−3 2 1 ]

31 = 1 32 = 0 33 = −1 1 0 −1

1 −3 1

(c) Adjoint = [−2 2 0 ] (d) | | = −2

−1 1 −1

−1 3 −1
22 2
(e) −1 = [ 1 −1 0 ]
1 −1 1

2 22

612 −1 −1 1
8. 2 = [0 3 1] ; −1 = [−1 −2 1 ]

764 4 5 −3

9. (i) x = 0 , x = 4 ii) | | = −3 iii) | 2| = 9

7 2 5 
 
 27 27 27 

10. A1   13 4  10 
 27 27 27 
 
 1 1 2 

 9 9 9 

1 3  1 
 2 2 
 2 

11. P1   1 5  1 
2 2 2 
 
 1 1 1 

 2 2 2 

  1 1 1
 3 3 
 3 
2
12. PQT  3I, P1    1 3 1
 3 5 
 3 
 3
4 1
3

6

3.0 MATRICES AND SYSTEM OF LIEAR EQUATIONS AM015_2022/2023

4 1  4 
 5 5 
 5 

13. B  5I , A1    1 1 7
 5 5 10 
 
 1 1  1 
5 5 5

7 2 5 
 
 27 27 27 

14. A1   13 4  10 
 27 27 27 
 1 1 2 
 9 9 9 

(a) x 1, y  1, z  2 (b) x1  2, x2 1, x3  1

15. x  2 , y  17 , z  1
33

 2 1 1 x   3
     
16.  1 0 1   y    1  , x  3, y 1, z  2
3 1 4 z 0

 1 1 1
 
 2 2 2 

17. M 1   1 1 1  , x  1, y  2, z  3
 4 4 4 
 
 2 0 1 

2x  y  4z  8 2 1 4  x  8 
18. (a) x  2y  6z 11  160    1171
 1 2 y  
x  3y 10z 17 1 3 z

1 1 1 
 2 2 
(b) A1   2 4  x 1, y  2, z 1
 1 2 

 1 5 3 
 
4 4 4 

(c) RM 228

7

TOPIC 4 At the end of the lesson, students should be
LINEAR PROGRAMMING able to:
(a) represent the system of linear inequalities
4.1 SYSTEM OF LINEAR INEQUALITIES
graphically.
(b) identify the feasible region.
(c) determine the system of linear inequalities

defined by the feasible region.

SYSTEMS OF INEQUALITIES

Inequalities involve one of the four signs (notation) By using the following notations we can transform the
above statements into mathematical expressions.
SIGNS MEANINGS EXAMPLES ON
NUMBER < “less than”
< less than > “greater than”
> LINE ≤ “less or equal to” / “at most” / “not exceed” /
≤ greater than
≥ 5 “not more than”
less than or ≥ “greater or equal to” / “at least” / “not less than”
equal to 5 = “equal to” / “exactly”

greater than or 5
equal to
5

Example BOUNDARIES OF INEQUALITIES

Write inequalities for the statements below: SIGNS BOUNDARIES DIAGRAMS

Solution < Broken/dashed line
> Broken/dashed line
(a) y is less than x ≤ Full/solid line
(b) y is greater than twice the value of x ≥ Full/solid line
(c) the sum of x and y is less than 7
(d) the maximum value of y is x
(e) 2 y minus 3x is less than 4.

BOUNDARIES OF INEQUALITIES GRAPHING LINEAR INEQUALITIES
PROCEDURE:
y y
y > x-3 y ≥ x-3 Step 1:
Sketch the graph of the linear equation, Ax + By = C by
3 x x using solid or dashed line.
-3 3 x Step 2:
y (a) Choose a test point that is not on the boundary line.
y < x-3 -3
y (b) Substitute the coordinates into the inequality.
3 Step 3:
-3 y ≤ x-3 (a) Determine the half-plane of the inequalities.
(b) Shade the region.
x3
-3

GRAPHING LINEAR INEQUALITIES (b) Note :

Example (b) 2x ≤ 3 (c) x ≤ 2y 1. Draw the line 2x = 3.
(a) y > -5 2. Determine the region which will satisfies 2x ≤ 3
Solution and shade.

(a) Note : y
1. Draw the dash line y = -5.
2. Determine the region which will satisfies y > -5 and shade. x

y

x

(c) Note : DEFINITION

1. Draw the line x = 2y. Feasible Region:
2. Determine the region which will satisfies x ≤ 2y
and shade. The graph of all ordered pairs of real
numbers (x,y) that simultaneously satisfy all
y the inequalities in the system is called
feasible region.
x

Example Solution

Solve the following system graphically. Graph the line x + y = 9 and shade the region that satisfies
x+y ≥9 the linear inequality x + y ≥ 9.
2x – y ≥ 0 Graph the line 2x – y = 0 and shade the region that satisfies
the linear inequality 2x – y ≥ 0.

y

x

Example Solution

Sketch and shade the feasible region R, Graph the line x + y = 5 and shade the region that satisfies the linear inequality x + y ≥ 5.
which satisfies the following inequalities: Graph the line y = x and shade the region that satisfies the linear inequality y ≤ x.
Graph the line x = 5 and shade the region that satisfies the linear inequality x ≤ 5.
x+y ≥5
y≤x y
x<5
x

Example Solution

Determine the systems of inequalities which the system which satisfies the feasible region
satisfies the feasible region in the diagrams are
below:

y
y=x

2

R

2x + 3y = 6
x

3

Example Solution

y First Line
4
3 Example y = 3x

R y
8
x
5 R

Note : x
1. Find the equation of the two lines. 2
2. Repeat the steps as in above note.
Note :
Second Line 1. Find the equation of the two lines.
2. Repeat the steps as in above note.
Solution
Third Line
First Line

Second Line

Thus, the inequalities that defined the shaded
region are

TOPIC 4 At the end of the lesson, students should be
able to:
LINEAR PROGRAMMING (a) formulate the system of linear inequalities

4.2 SOLVING PROBLEM INVOLVING in economics and business problems.
SYSTEM OF LINEAR INEQUALITIES (b) draw and shade the feasible region.

(c) determine and draw the objective PROBLEM INTERPRETATION AND MODEL
function of the problem. CONSTRUCTION

(d) compute the optimal value of an DEFINITION 1:
objective function by using vertices of Problem interpretation is a process of converting
the feasible region. daily problem into a form that can be solved using
mathematical approach method.

STEPS :
1. Determine the variables.
2. Form related inequalities or equations for

mathematical models.
3. Determine the objective function of the problem.

PROBLEM INTERPRETATION AND MODEL Example:
CONSTRUCTION
An express postal company wishes to send at least
DEFINITION 2: 900 parcels to a town. 4 lorries and 8 vans are
available for transportation. Each lorry can be used
Mathematical model is a process of translating to transport 150 parcels while a van can
information into equations or inequalities using accommodate 90 parcels. The transportation cost of
mathematical approach. each lorry and each van is RM100 and RM80
respectively. The maximum total cost of
transportation is RM 800.
(a) Determine the variables of this problem.
(b) Form the equations or inequalities system from

this problem.

Solution : The number of parcel that can be carried by x lorry and y
van must at least 900.
(a) Variables of the problem are
x = the number of lorries As the number of lorries and vans cannot exceed 4 and 8
y = the number of vans respectively, then
The total cost of transportation of x lorries and y vans does
(b) All the information given are summarized as follow: not exceed RM800.

Type of Number of Transportation Total Number of As we are looking for x and y that minimize the cost :

vehicles Parcel Cost Vehicles

Therefore the equation and inequalities system that exists Interpretation
in this problem are
For this example, the cost value of the company is
Subject to obtained from the function.

C = 100x + 80y
This function is known as the objective function.

All the equations and inequalities which formed
the condition and limitation in order to get the
solution is known as the constraints.

CORNER POINT METHOD CORNER POINT METHOD

A corner point of a feasible region is a point 1. Graphically identify the region of feasible solutions.
in the feasible region that is the intersection 2. Determine the coordinates of each corner point on the region
of two boundary lines. Therefore, the corner-
point method for solving linear programming of feasible solutions.
problems is as follows: 3. Substitute the coordinate of the corner points into the

objective function to determine the corresponding value.
4. An optimal solution occurs in a maximization problem at the

corner point yielding the highest value of the objective function
and in a minimization problem at the corner point yielding the
lowest value of the objective function.

Example: Solution :

Minimize z = 3x + 6y Step 1:
Sketch the graph and shade the feasible region.

Subject to

4x + y ≥ 20
x + y ≤ 20
x + y ≥ 10
x,y ≥ 0

Step 2: Step 3:
Determine the coordinate of each corner point by
Draw the objective function using graph paper.

Corner point (x,y) z = 3x+6y

Notes:
Substitute the coordinates of the corner point into the
objective function z.

Step 4: Example :

A farmer has a choice of two types of cattle feed,
type A and type B. A kilogram of type A cost RM11
and provides 20 units of protein and 30 units of
carbohydrate. A kilogram of type B cost RM15, and
provides 20 units of protein and 45 units of
carbohydrate. To keep the herd healthy, they must
receive at least 200 units of protein and 360 units of
carbohydrate per day. What is the cheapest way of
feeding the herd?

Solution : The objective function :
The system of linear inequalities are;
The information are summarize in the table below:

Type of Unit of Unit of Cost per kg
Food
Protein Carbohydrate (RM)

Let :
x is the unit of cattle food type A
y is the unit of cattle food type B.

The feasible region is shown in the graph below;

Corner point (x,y) C = 11x+15y

AM015 4.0 LINEAR PROGRAMMING

TOPIC 4.0: LINEAR PROGRAMMING
4.1 System of Linear Inequalities

1. Determine the region (R), which satisfies the inequalities:

a) 2y  x  5 b) x  y  7 c) y  x 1
x  1 3x  y  5 2y  5x
x0

d) 2x  y  1 e) 5y  8x 10
x2y  2 2x  5
y0 2x  55y
2y 5

Ans: -Refer to lecturer-

2. For each of the following, find the inequality that define the shaded region, R:

(a)

R

(b) Ans: y  3x
Ans: y  x  7
R

AM015 4.0 LINEAR PROGRAMMING

3. Find the system of linear inequalities defined by the following feasible region, R:

(a)

R Ans:
2y  4 x
(b) 4y  x4

R x0

(c) Ans:
80 y3
(30,40) y x2
2y  3x  6
R
Ans:
| 4x  3y  240
50 60 2x  y  100

x0
y0

AM015 (4, 6) 4.0 LINEAR PROGRAMMING

(d) R Ans:
10 x  y  2
8 |
2 4 10 x  y  10
2x  y  8

y 1

4.2 Solving Problem Involving System of Linear Inequalities

1. a) Maximize: f (x, y)  2x 3y Ans:
The maximum value of f (x, y) is 18
subjected to when x  0 and y  6

x y 6 Ans:
8 y  2x The minimum value of f (x, y) is 5
when x 1 and y  3
x0
y0

b) Minimize: f (x, y)  2x  y

subjected to

6  3x  y
4 y x

x0
y0

2. A dealer sells two types of calculators, namely Cashio and Conan. He can sell up
to a maximum of 500 units of calculator with the condition that he must sell at least
100 unit of Cashio calculator and 180 units of Conan calculator. He is given a
commission of RM2 for every units of Cashio calculator sold and RM4 for every
units of Conann calculator sold. Given he sold x unit of Cashio calculator and y unit
of Conan calculator and the commission he obtained is at least RM1000.
(a) Write down 4 inequalities beside x  0 and y  0 for the above constraints.

AM015 4.0 LINEAR PROGRAMMING

(b) Determine the objective function.
(c) Graph the feasible region and label it R
(d) Compute the maximum commission obtainable by the dealer.

Ans: The maximum commission obtainable is RM1800

3. A company has 200 employees. The company provides not more than 4 standard
buses and a few mini buses to ferry the employees to and fro for work. Each
standard bus can accommodate 40 passengers, whereas each mini bus can
accommodate 20 passengers. The daily operational cost for a standard bus and a
mini bus are RM50 and RM30 respectively. The company has only 9 drivers. Let x
and y denote the number of of standard buses and the number of mini bueses
respectively.

(a) Write down 3 inequalities besides x  0 and y  0 for the above constraints
(b) Determine the objective function.
(c) Graph the feasible region and label it R.
(d) Determine the minimum daily operational cost and the corresponding

number of each type of bus required.
Ans: The minimum daily operational cost is RM260.
4 standard buses and 2 mini buses.

4. A patient who is suffering from iron and vitamin B difiency are advised by

nutritionist to take at least 2400 mg of iron, 2100 mg of vitamin B-1 (Thiamine),

and 1500 mg of vitain B-2 (Riboflavin) over a period of time. Two vitamin pilss

are recommended, brand P and brand , which cost 6 cents and 8 cents per pills

respectively. Each brand pill contains iron, vitamin B-1 and vitamin B-2 as shown

in the following table:

PRODUCT IRON VITAMIN B-1 VITAMIN B-2

Brand P 40 mg 10 mg 5 mg

Brand Q 10 mg 15 mg 15 mg

AM015 4.0 LINEAR PROGRAMMING

The patient wishes to determine the possible combination of pills that he should
purchase in order to meet the minimum iron and vitamin requirements at the lowest
cost. Let x and y denote the number of pill Brand P and the number of pill Brand
Q respectively.
(a) Formulate the above problem as a linear programming model.
(b) Graph the feasible regionand label it R.
(c) Determine the number of each pills should be purchased to meet the

minimum iron and vitamin requirements at the lowest cost.
Ans: 30 pills Brand P and 120 pills Brand Q

5. A farmer has 20 hectares of land for growing paddy and corn. The farmer has to
decide how much of each to grow. The cost per hectare for paddy is RM 30 and for
corn is RM 20. The farmer has budgeted RM 480. Paddy requires 1 man-day for
hectare and corn requires 2 man-days per hectare. There are 36 man-days available.
The profit on paddy is RM 100 per hectare and on corn is RM 120 per hectare. Find
the number of hectares of each crop the farmer should sow to maximize the profits
and also the maximum profit.
Ans: The farmer should sow 4 hectares of
paddy and 16 hectares of corn. The maximum
profit is RM2320

6. A manufacturer produces two product, X and Y each day. The production contain
3 types of material. The supply that company receives each day for material A,
material B and material C are 6g, 4g and 6g respectively. Product X requires 2g of
material A, 1g of material B and 1g of material C each day. Product Y requires 1g
of material A, 1g of material B and 2g of material C each day. The profit obtained
from each unit of product X and product Y are RM34 and RM19 respectively.
Assume that x the number of product X and y the number of product Y produced
each day. Find the number of product X and Y that will maximize the profit.

Ans: 2 unit of product X and 2 unit of product Y

AM015 4.0 LINEAR PROGRAMMING

7. Abadi has 200 litres of high grade soya sauce and 360 litres of low grade soya
sauce. He wishes to produce 2 new grades of soya sauce, namely X and Y. X is
produced by mixing high grade soya sauce and low grade soya sauce in the ratio
1:3, whereas Y is produced by mixing high grade soya sauce and low grade soya
sauce in the ratio 2:3. Abadi makes a profit of RM3 for every litre of X sold and
RM4 for every litre of Y sold. The soya sauce are packed in 1-litre bolttles. Find
the highest profit obtained and the number of bottles X and Y that must be sold to
make highest profit.
Ans: The highest profit is RM2080 when 160 bottles of X and 400
bottles of Y are sold.

Subtopic: FUNCTIONS AND GRAPHS
5.1 Functions 5.1 Functions
5.2 Composite Functions
5.3 Inverse Functions L
5.4 Polynomials

L

LEARNING OUTCOMES DEFINITION OF A FUNCTION

(a) DEFINE A FUNCTION A function is defined as a relation in which
every element in the domain has a unique
(b) IDENTIFY A FUNCTION FROM THE image element in the range .
GRAPH BY
In other words, a function is
USING VERTICAL LINE TEST one to one relation or many to one relation

(c) SKETCH THE GRAPH OF A FUNCTION 1A 1A
FOR 2B 2B
3C 3C
CONSTANT, LINEAR, QUADRATIC, SURD,
ABSOLUTE VALUE, PIECEWISE, L L
EXPONENTIAL
AND LOGARITHMIC FUNCTION.

ONTO FUNCTION ONE TO ONE FUNCTION

A function which each element of the range is A function which every element of the range
mapped to at least one element of the domain corresponds to exactly one element of the

domain

1A 1A 1A 1A
2 2B 2B
3C 2B 3C
3B 3C

ALL THE RANGES THEY CANNOT
HAVE AT LEAST ONE SHARE SAME THING!

FRIEND FROM THE
DOMAIN

L L

1

How to identify function and one to one VERTICAL LINE TEST
function?
STEP
VERTICAL LINE TEST
DRAW A VERTICAL LINE PARALLEL TO Y AXIS
L
CHECK:
IF THE VERTICAL LINE CUTS THE GRAPH AT ONLY ONE
POINT, THEN THE GRAPH IS A FUNCTION

EXAMPLE 1 f(x)

f(x)

X
X

L

SKETCH: BASIC GRAPHS. LET’S IMPROVISE! THE BASIC GRAPHS

f x  x f x  x f x  x2 f x  x2 LINEAR QUADRATIC

f x  x 3 f x  x2 3
y
y

3 x 3
-3 x

RECIPROCAL SQUARE ROOT

f x  1 f x  x f x  x y f x  1 f x  x3
y
x TL x 3

x 3
-3 -3

x
L

DOMAIN AND RANGE

Can you see the differences?
All the graphs be affected by their intercepts, value of x
and y which show the importance of domain and range

Let f(x) be a function then ,
a) horizontal axis (x – axis) represents the
domain
b) vertical axis (y – axis) represents the
range

TL L

2

Example 1 Notes :

XY Domain : Domain and range can be determined by
X= using
-1 0 • graph
-2 1 Codomain :
1 Y= or
2 4 • algebraic approach
3 9 Range :
L
16

L

Find the domain and range of a function (b) Linear function
by using graph f(x) = x - 1

(a) Constant : f(x) = 2

y y Domain of f =

Domain of f =  -1 1 Range of f =

2 x

0 x Range of f = 2 L

L

(c) Quadratic function If a > 0, then (d) Cubic function
If a < 0, then f(x) = x3 – 5x2 + 6x
f(x) = x2 – 3x -10
 (x  3)2  49  x x2 x3
24
f(x)
(by completing the square)
Domain of f =
f(x) x
-49/4 Range of f =
Domain of f = 
x
Range of f = [- 49/4 , )
0 23
L
L

3

(e) Square root function (f) Absolute value function

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

f(x) y Domain of f =
f(x)
Domain of f = [0, )
Range of f = [0, ) Range of f = [ 0 , )

(x) 0x

L L

(g) Rational function HINTS

f x  3x  2 1  k0
k
2x 1
k k0
y Domain of f = \  1
  1 k0
2  k

2 y = 3/2 Range of f = \ 3
-2/3 
x 2

X = -1/2

L L

Example 2 Graphical Approach
(a)
Sketch the functions and find the
domain and range for each of the f(x)
function given.
0
(a) f(x) = -2x2 + 4x – 5
(b) f(x) = |x + 2|

L L

4

Algebraic Approach

Graphical Approach L L
(b)
Algebraic Approach L
f(x) (b)
L
x L
5
Example 3 SOLUTION

Sketch the functions and find the
domain and range for each of the
function given.

a) f (x)  x  2 b) f (x)  1
x2

L

Example 4

Find the domain and range for each of the
function given below.

(a) f (x)  3  x  2

(b) f (x)  1 L
x 1

L

SOLUTION

LL

LL

6

Example 5 SOLUTION

Find the domain and range for L

2x  2 x  2
 2 x0
f (x)   2
x6
2x  2

L

Conclusion

X Y Domain : FUNCTIONS AND GRAPHS
X ={ -1,-2,1,2,3} 5.1 Functions

-1 0 L

-2 1 Codomain :

1 4 Y = {0,1,4,9,16}

29

3 16 Range : {1,4,9}

Image : 1,4,9

L

LEARNING OUTCOMES Exponential Function

(c) SKETCH THE GRAPH OF A FUNCTION The exponential function is expressed by
FOR f(x) = ax , where x is a real number, a > 0
and a  1
CONSTANT, LINEAR, QUADRATIC, SURD,
ABSOLUTE VALUE, PIECEWISE, a) f (x)  ax, x, a 1
EXPONENTIAL
AND LOGARITHMIC FUNCTION. b) f (x)  ax, x, 0  a  1

(d) STATE THE DOMAIN AND RANGE

7

a) f (x)  a x , x , a  1 3.When x  , f (x)  
4.When x  , f ( x)  0
Basic properties Df  

1. f (x)  0 R f  0, 

y 5.Range of f (x) is (0,) y

2.When x  0, f ( x)  1 f (x)  ax ,a  1 6 . f ( x ) is 1  1 function f (x)  ax ,a  1

Intercept the y-axis at (0,1) x 1
1 x
0

b) f (x)  a x , x  , 0  a  1 3.When x  , f ( x)  0
y 4.When x  , f (x)  
Basic properties
5.Range of f ( x) is (0, )
y 1. f ( x)  0
f (x)  a x ,0  a  1 2.When x  0, f ( x)  1 1 6. f ( x) is 1 1 function

1 Intercept the y-axis at (0,1) f (x)  ax,0  a  1 x Df  ,
x
Rf  0, 

Exponential Functions of the form
f(x) = eax+b

REMARK

y x=0 is a reflection axis Any positive number can be used as the

base for an exponential function.

The most important exponential function

has a base e, where e is a number such

that

1 f (x)  ax ,a  1 e 1 1 m as m  
m 
f ( x )  ax ,0  a 1 x

8

m  1  1 m Example 1
 m 
1
10 2
100
1000 2.59374 Sketch the graph and find the
10000 domain and the range.
100000 2.70481
1000000
10000000 2.71692

2.71815  1  x
 
2.71827 f ( x)  2

2.71828

2.71828 f (x)  2x

e  2.71828 L T

SOLUTION Example 2
Sketch the graph of the following
and find the domain and the range.

a. f ( x)  2x 1

b. f ( x )  2x 1

LT

SOLUTION SOLUTION

LL

9

Example 2 SOLUTION

Sketch the graph of the following
and find the domain and the range.

a. f ( x )  ex

b. f ( x )  e2x

1x
c. f ( x )  e2

T L

Therefore how will the graph of the following Example 3
Sketch the graph of the following
     functions look like ? 1 x and find the domain and the range.

f x  e2x, f x  ex, f x e 2 a. f ( x )  ex1
b. f ( x )  2ex
f ( x )  e2x x
T
y f(x )  e 1x

f ( x )  e2

1
x

0T

SOLUTION SOLUTION

LL

10

Logarithmic Function

Logarithmic function is written as
f(x) = logax,where x > 0 and a is a
positive constant other than 1

a) f ( x)  loga x, x   , a  1

b) f ( x)  loga x, x   , 0  a  1

TL

a) f ( x)  loga x, x   , a  1

Basic properties y Domain of f  0, y

Range of f   1x

1. x  0 , undefined function 1x f ( x)  loga x,
2.When x  a, f ( x)  1 f (x)  loga x, a1
3.When x  1, f ( x)  0 a1
4.When x  0 , f ( x)  , for 0  x  1
L 5.When x  , f ( x)  , for x  1 L

b) f ( x)  loga x, x   , 0  a  1 D f  0, 

Rf   y

y f ( x)  log a x,
0a1
f ( x)  loga x, x
0a1 1

Basic properties 1 x

1 . x  0 ,undefined function

2.Whenx  a, f (x)  1 4.When x 0 , f ( x ) , for 0  x 1

3.When x  1, f ( x )  0 L 5.When x  , f ( x)  , for x  1 L

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Example 1 SOLUTION
Sketch the graph of

f ( x )  log 1 x and f ( x)  log2 x
2

state the domain and the range.

T L

Example 2: SOLUTION L

Sketch all the following graphs L
on the same axes, then state the
domain and the range 12

(a) f (x)  ln x

(b) f (x)  ln  1 x 
 2 

(c) f (x)  ln(2x) T

Example 3: SOLUTION

Sketch the graph and state the domain and
the range of the following functions

(a) f (x)  ln (x 1)
(b) f (x)  ln (x 1)
(c) f (x)  ln (1 x)

T

SOLUTION SOLUTION

LL

FUNCTIONS AND GRAPHS OBJECTIVES
5.2 Composite Functions
(a) Represent a composite function by an
L arrow diagram

(b) Determine the composite of two functions
(c) Determine one of the functions when the

composite and the other functions
are given

L

Composite functions f  g and g  f This can be represented in an arrow diagram:
ABC
DEFINITION

xg g(x) f f [g(x)]

Composite fg
Function

L L

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Example 1 SOLUTION
If f(x) = 5 and g(x) = 10, find
the following as a function of x

(a) (f  g)(x)

(b) (g  f)(x)

L L

Example 2 SOLUTION L
If f(x) = 3x + 1 and g(x) = 2 - x, find
the following as a function of x L
(a) (f  g)(x)
(b) (g  f)(x) 14
(c) [f  ( g  f )](x)

L

SOLUTION SOLUTION

L

SOLUTION Example 2
SOLUTION
SOLUTION If f x   1  x, g x   x 2  3, h x   x  7,

k(x)  25  x

Find e f  h

a f  g

b  g  f f  h f

c g  h g  f k

d  h  g h k  f

LL

SOLUTION

L L

SOLUTION

LL

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