Focus SPM KSSM 2021 Tingkatan 5 - Maths DLP

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CC035231
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Form FOCUS

MATHEMATICS KSSM 5 SPM


MATHEMATICS SPM


REVISION FOCUS SPM KSSM Form 5 – a complete and
REVISION
REVISI
precise series of reference books with special
üKeywords üSPM Tips features to enhance students’ learning as a whole.
üConcept Maps üRemember! This series covers the latest Kurikulum Standard
Sekolah Menengah (KSSM) and integrates Form
Sijil Pelajaran Malaysia (SPM) requirements.
REINFORCEMENT
REINFORCEMENT
A great resource for every student indeed! KSSM 5
& ASSESSMENT
& ASSESSMENT
üTry This! üSPM Model Paper MATHEMATICS
üSPM Practices üComplete Answers REVISION

REINFORCEMENT
EXTRA FEA TURES
EXTRA FEATURES
ASSESSMENT
üDaily Applications üCloned SPM Questions
üCalculator Corner üSPM Highlights EXTRA
üHOTS Questions üQR Codes



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CC035231 Textbook
ISBN: 978-967-2965-63-3 KSSM NEW SPM ASSESSMENT
Mathematics Form 5
Ng Seng How • Ooi Soo Huat FORMAT 2021
PELANGI Samantha Neo • Yong Kuan Yeoh




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Format: 190mm X 260mm TP Focus F5 Maths BI pgi_1 imp
















MATHEMATICS

SPM





Form
KSSM 5



Ng Seng How
Ooi Soo Huat
Samantha Neo
Yong Kuan Yeoh














© Penerbitan Pelangi Sdn. Bhd. 2021
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ISBN: 978-967-2965-63-3
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TP_Focus_F5.indd 4 2/2/21 4:26 PM

CONTENTS









Mathematical Formulae iv


Chapter
1 Variation 1


1.1 Direct Variation 2
1.2 Inverse Variation 7
1.3 Combined Variation 11
SPM Practice 1 14


Chapter
2 Matrices 17


2.1 Matrices 18
2.2 Basic Operation on Matrices 20
SPM Practice 2 32


Chapter
3 Consumer Mathematics: Insurance 35


3.1 Risk and Insurance Coverage 36
SPM Practice 3 45


Chapter
4 Consumer Mathematics: Taxation 48


4.1 Taxation 49
SPM Practice 4 58


Chapter
5 Congruency, Enlargement and Combined Transformations 60


5.1 Congruency 61
5.2 Enlargement 65
5.3 Combined Transformation 73
5.4 Tessellation 80
SPM Practice 5 83



ii





00 Cont Focus SPM Math F5_2021.indd 2 18/02/2021 2:01 PM

Chapter
6 Ratios and Graphs of Trigonometric Functions 86


6.1 The Value of Sine, Cosine and Tangent for Angle q, 0° < q < 360° 87
6.2 The Graphs of Sine, Cosine and Tangent Functions 96
SPM Practice 6 103



Chapter
7 Measures of Dispersion for Grouped Data 109


7.1 Dispersion 110
7.2 Measures of Dispersion 120
SPM Practice 7 127


Chapter
8 Mathematical Modeling 131


8.1 Mathematical Modeling 132
SPM Practice 8 142






SPM Model Paper 144



Answers 161





























iii





00 Cont Focus SPM Math F5_2021.indd 3 18/02/2021 2:01 PM

Chapter Learning Area: Relationship and Algebra


2 Matrices












My seat is at the 2 row
nd
from the front and there
are 3 chairs on my right.

Oh, at a 24 .























• Column matrix – Matriks lajur
• Determinant – Penentu
• Element – Unsur
• Identity matrix – Matriks identiti
• Inverse matrix – Matriks songsang
• Matrix – Matriks
• Order – Peringkat
• Rectangular matrix – Matriks segi empat tepat Concept
• Row matrix – Matriks baris map
• Scalar multiplication – Pendaraban skalar
• Square matrix – Matriks segi empat sama
• Zero matrix – Matriks sifar
Matrix is a set of numbers arranged in a rectangular or a square array. Matrix is always used in
the fields of business, engineering and information technology. For example, car design engineers
use matrix to represent parts of the car in computer-aided simulation. Animations and graphics
in cartoons can also be created by using matrix. Managers in a company can also manage their




inventory by using matrix.





17





02 Focus Math F5_E2021.indd 17 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

(b) The number of male and female participants of a
2.1 Matrices Mathematics Quiz are 16 and 12 respectively.
(c) The table below shows the number of accidents
A Representing information from real in three cities, X, Y and Z.
situations in the form of matrices Accidents
City
1. Matrix is a set of numbers arranged in rows and Car Motorcycle
columns to form a rectangular or a square array. X 10 26
For example,
Y 11 8 Chapter 2
columns 15 14
Z
Chapter 2
1 2 9
rows 3 4 7 8 4 Solution
4
4 3
Minion Pro 10pt = 1 60 40 15 60 48 55
2 2. Matrix is usually represented by capital letter (a) 3 48 38 15 or 40 38 23
and enclosed by bracket [ ] or ( ). 55 23 10 15 15 10
Arial 8.5 pt = 1 16
2 3. Row matrix is a matrix that has only one row. (b) [16 12] or 3 4
For example, A = [5 –2 11] 12

Kaedah Alternatif 4. Column matrix is a matrix that has only one (c) 3 10 26 4 3 10 11 15 4
11
8 or
column. 15 14 26 8 14
3
For example, B = 3 4 Try Question 1 in Try This! 2.1
35
5. Square matrix is a matrix that has the same
number of rows and columns. B Determining the order of a matrix,
hence identifying certain elements in
For example, C = 3 1 0 4 a matrix
9 8
1. The order of a matrix is determined by the
6. Rectangular matrix is a matrix that has different number of rows and columns of the matrix.
number of rows and columns.

For example, D = 3 16 8 –5 4 2. Matrix with m rows and n columns has an
order of m × n and is read as “matrix m by n”.
–3 9 21
For example, matrix A = 3 1 2 9 4 has 2 rows
4 7 8
1 and 3 columns. Therefore, the order of matrix A
Represent the following information in the form of is 2 × 3.
matrices. 3. Each number in a matrix is called an element
th
(a) The table below shows the sales of stationery in a of the matrix. The element located at the i row
school bookshop for March to May. and j column of matrix A can be represented
th
as a ij .
Exercise Drawing
Pen For example, for matrix
book paper 1 2 9
March 60 40 15 A = 3 4 7 8 4 ,
April 48 38 15 the element at the first row and 3 column, that
rd
is a 13 = 9.
May 55 23 10
4. a 13 is read as “a one three”.



18





02 Focus Math F5_E2021.indd 18 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

SPM Tips 4

• Row matrix has an order of 1 × n. Determine whether each of the following pairs of
• Column matrix has an order of m × 1. matrices is equal.
• Square matrix has an order of m × m. 6 11 6 11
• Rectangular matrix has an order of m × n. (a) P = 3 2 9 4 and Q = 3 2 9 4
4
(b) R = 3 4 and S = [4 1]
1
2 7 5 0 7 5 0
(c) T = 3 4 and U = 2 84 Chapter 2
Determine the order of the following matrices. 3 0.4 8 3 3 5
3
3 4 (d) V = 3 –2 4 4 and W = 3 –2 1 4
Chapter 2
4 15
1 15
(a) –1
8 Solution
(b) 3 21 7 4 (a) P and Q are equal.
(b) R and S are not equal because they do not have
–5 2
the same order.
Solution (c) T and U are equal.
(a) 3 × 1 3 rows and 1 column (d) V and W are not equal because their
corresponding elements are not equal, which are
(b) 2 × 2 2 rows and 2 columns v 21 ≠ w 21 and v 12 ≠ w 12 .

5
3
It is given that matrix E = 3 1 2 + y 4 and matrix
z – 4
–3
3 –4 6 4 F = 3 1 6 4 . Determine the values of x, y and z if
Given that matrix P = 12 0 , determine x 3z
9 –3 E = F.
(a) the element at the first row and 2 column. Solution
nd
(b) p 22 . E = F
(c) p 31 . 1 2 + y 1 6
=
Solution 3 –3 z – 4 4 3 x 3z 4 Element at the 2
nd
(a) 6 x = –3 row and first column
(b) 0 2 + y = 6 Element at the first
(c) 9 y = 6 – 2 row and 2 column
nd
= 4
Try Questions 2 – 3 in Try This! 2.1
z – 4 = 3z Element at the 2 row
nd
nd
–4 = 2z and 2 column
z = –2
C Determining whether two matrices
are equal Try Questions 4 – 6 in Try This! 2.1
1. Two matrices are equal if they are of the same Try This! 2.1
order and their corresponding elements are 1. Represent the following information in the form of
equal. For example, matrices.
if 3 a b 4 3 p q s 4 , then a = p, b = q, c = r (a) The table below shows the number of Form 4
=
and Form 5 members of two clubs in a school.
c d
r
and d = s. Science Club History Club
Form 4 35 40
Form 5 48 37

19





02 Focus Math F5_E2021.indd 19 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

(b) Ali’s year-end exam scores for English, 2. The corresponding elements can be added or
Mathematics and Science are 76, 83 and 80 subtracted to obtain a single matrix of the same
respectively. order. For example,
(c) The table below shows the selling prices of rice a b p q a ± p b ± q
±
=
by brand. 3 c d 4 3 r s 4 3 c ± r d ± s 4
Selling price
Rice
5 kg 10 kg 6
Brand A RM26 RM45 Determine whether the following pairs of matrices
Brand B RM18 RM32 can be added and subtracted. Chapter 2
–7
Brand C RM13 RM20 (a) [1 2], 3 4
5
2. Determine the order of the following matrices. 0 4 8 9
Chapter 2
,
(a) [–3 11] (b) 3 1 –3 4 3 11 6 4
(b) 3 3 –7 2 4 Solution
19
8
6
(c) 3 1 4 4 (a) Cannot. It is because both matrices are of
different orders.
2
5
(b) Can. It is because both matrices are of the same
3 5 0 1 4 order.
3. Given that matrix A = 13 8 –4 , determine
–1 2 9 7
(a) the element at the first row and 3 column.
rd
(b) a 22 . Express each of the following as a single matrix.
(c) a 31 . (a) [7 5] + [2 –11]
–
4. Determine whether each of the following pairs of (b) 3 2 3 4 3 1 9 4
matrices is equal. 1 –2 6 –5
(a) 3 10 –2 1 4 3 10 2 1 4 Solution
,
5
8
7
5
7
8
(a) [7 5] + [2 –11] = 37 + 2 5 + (–11)4
3 4 = [9 –6]
3
(b) [3 4 9], 4
3 – 9
3
9 (b) 3 2 1 –2 4 3 1 9 4 3 2 – 1 (–2) – (–5) 4
=
–
1 – 6
6 –5
0.5 3 4
7
(c) 3 4 , 1 7 1 –6
2 = 3 –5 3 4
5. It is given that matrix P = 3 8 x 4 and matrix 8
–7 9
8
Q = 3 2x + y –2 4 . Determine the values of x and y 12 7 2 3
9
if P = Q. Given that matrices M = 3 –3 5 4 , N = 3 9 –4 4 and
P = 3 1 6 4 , calculate
–10 8
1
3 2 4 x + 3 4 2 0.25 2
6. It is given that 0 2y 4z – x = 3 0 7 5 4 . (a) M + N
Determine the values of x, y and z. (b) N – P
Solution
2.2 Basic Operation on Matrices (a) M + N = 3 12 7 4 3 2 3 4
+
9 –4
–3 5
= 3 12 + 2 7 + 3 4
A Adding and subtracting matrices –3 + 9 5 + (–4)
1. Addition and subtraction can only be performed = 3 14 10 4
1
6
on matrices that have the same order.
20
02 Focus Math F5_E2021.indd 20 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices


(b) N – P = 3 2 3 4 3 1 6 4 Solution:
–
9 –4
8
–10
[98 92] – [45 50] + [50 45]
= 3 2 – 1 3 – 6 4 = [103 87]
The final stocks of flour and sugar are 103 and 87
9 – (–10)
–4 – 8
respectively.
= 3 1 –3 4 Try this HOTS Question
19 –12
The table below shows the scores of two teams in
9 a competition.
2
16
8
=
Given that 3 4 + S – 3 4 3 4 , find matrix S. First Second Chapter 2
3
4
–5
Team
round round Penalty
Solution A 15 17 2
Chapter 2
8
16
2
3 4 + S – 3 4 3 4 B 15 19 3
=
–5
4
3
16
8
2
+
S = 3 4 3 4 3 4 It is given that the penalty will be deducted from
–
–5
4
3
the total score of both rounds, and the team with
8 – 16 + 2
= 3 3 – (–5) + 4 4 a total score above 30 will be eligible for final
round. Which team will proceed to final round?
–6
= 3 4 Find your answer using matrix method.
12
Answer: Team B
10
–
=
Given that 3 3 x 4 3 1 2x 4 3 2 –1 4 , find
y 8
7 0
4y + 1 8
the value of x and of y. B Multiplying a matrix by a number
Solution 1. Multiplying a matrix by a number is known as
3 3 x 4 3 1 2x 4 3 2 –1 4 scalar multiplication.
=
–
7 0
4y + 1 8
y 8
Compare the corresponding elements: 2. The process of multiplying matrix A by a number
x – 2x = –1 n is a repeated addition.
–x = –1 nA = A + A + A + … + A
x = 1 n times
4y + 1 – y = 7 3. To multiply a matrix by a number, we multiply
3y = 6 each element in the matrix by the number.
y = 2 a b na nb
n 3 4 3 4
=
Try Questions 1 – 4 in Try This! 2.2 c d nc nd
Example of HOTS 4. Addition of matrices obeys the:
HOTS Question
(a) Commutative Law
The table below shows the stock records of flour and A + B = B + A
sugar sold at Ismail’s shop.
(b) Associative Law
Number of packets (A + B) + C = A + (B + C)
Flour Sugar 5. Addition and subtraction of matrices obey the
Stock 98 92 Distributive Law:
Sold 45 50 n(A + B) = nA + nB
Received 50 45 n(A – B) = nA – nB
Calculate the final stock for flour and sugar.
21
02 Focus Math F5_E2021.indd 21 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

3
–2
6. Zero matrix is a matrix with all its elements (b) 5(P – Q) = 5 13 4 3 42
–
5
2
are zero, for example 3 0 0 4 and is represented –2 – 3
0
0
by O. = 5 3 5 – 2 4
7. The results of addition and subtraction of a = 5 3 4
–5
matrix with a zero matrix are the matrix itself. 3
Minion Pro 10pt = 1 A + O = A and A – O = A = 3 5 × (–5) 4
2 5 × 3
–25
= 1 0 0 2 Chapter 2 11 = 3 4 Chapter 2
0 –0
15
6 3
Arial 8.5 pt = 1 Given that T = 3 1 –9 4 , find Alternative Method
2 (a) 2T (b) 1 T
= 1 0 0 2 Solution 3 (a) Addition of matrices obeys the Distributive Law,
which is n(P + Q) = nP + nQ.
0
0
(a) 2T = 2 3 1 –9 4 Hence, 5(P + Q) = 5P + 5Q
3
–2
Kaedah Alternatif 6 3 = 5 3 4 3 4
+ 5
5
= 3 2 × 1 2 × (–9) 4 5 × (–2) 2 5 × 3
2 × 3
2 × 6
+
= 3 2 –18 4 = 3 5 × 5 4 3 5 × 2 4
12
6
= 3 –10 + 15 4
25 + 10
3
(b) 1 T = 1 1 –9 4
5
3 3 6 3 = 3 4
35
1
× 1 1 × (–9) (b) Subtraction of matrices obeys the Distributive
= 3 3 × 3 4
1
3 × 6 1 Law, which is n(P – Q) = nP – nQ.
3 3 Hence, 5(P – Q) = 5P – 5Q
3
–2
3 1 –3 4 = 5 3 4 3 4
– 5
3
= 2 1 5 2
–
= 3 5 × (–2) 4 3 5 × 3 4
5 × 5
5 × 2
12 –10 – 15
= 3 4
25 – 10
–2
3
Given that P = 3 4 and Q = 3 4 , calculate –25
5
2
15
(a) 5(P + Q) = 3 4
(b) 5(P – Q)
13
Solution Express each of the following as a single matrix.
–2
3
(a) 5(P + Q) = 5 13 4 3 42 (a) 3[–7 5] + 0.5[20 8]
+
2
5
= 5 3 –2 + 3 4 (b) 5 3 –1 2 4 3 4 –6 4
– 2
–2 3
8
–5
5 + 2
1
= 5 3 4 Solution
7
(a) 3[–7 5] + 0.5[20 8]
= 3 5 × 1 4 = [–21 15] + [10 4]
5 × 7
= [–21 + 10 15 + 4]
5
= 3 4 = [–11 19]
35
22
02 Focus Math F5_E2021.indd 22 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices


– 2
(b) 5 3 –1 2 4 3 4 –6 4 SPM Highlights
–2 3
–5
8
– 2
+ 3
–8 7
= 3 –5 10 4 3 8 –12 4 3 11 7 4 3 4 2 4 3 3 –1 4 =
10 –9
5
4
–
–10 16
–10 15
A 3 18 8 4 C 3 17 15 4
6
–20 –2
3
= 3 –5 – 8 10 – (–12) 4 B 3 12 8 4 D 3 17 15 4
15 – 16
–10 – (–10)
6
= 3 –13 22 4 Solution 3 –22 2
0
–1
3 11 7 4 3 4 2 4 3 3 –1 4 Chapter 2
– 2
+ 3
4
–8 7
5
10 –9
14 11 7 12 6 6 –2
–
+
= 3 10 –9 4 3 –24 21 4 3 8 10 4
Chapter 2
1
It is given that M − [9 –1] + 3[2 7] = 4[5 8]. 17 15
2 = 3 4
Find matrix M. –22 2
Answer: D
Solution
1 M − [9 –1] + 3[2 7] = 4[5 8] Example of HOTS
HOTS Question
2
1 M = 4[5 8] + [9 –1] – 3[2 7] Mr Yusof has two vegetable stalls, A and B, in a
2 market. The table below shows the sales of cabbage
= [20 32] + [9 –1] – [6 21] and carrot at his stalls.
= [20 + 9 – 6 32 – 1 – 21] Stall
M = 2[23 10] Vegetable A B
= [46 20]
Cabbage 60 92
Carrot 56 75
15 It is given that the sales of cabbage and carrot at
stall A need to be halved. Calculate the total sales of
3x
x
–20
=
Given that 3 4 3 4 3 4 , find the value of x cabbage and the total sales of carrot by using matrix
+ 2
method.
y
–5
9
and of y. Solution:
1 60 92 122
3 4 3 4 3
Solution 2 56 + 75 = 103 4
x
–20
3x
3 4 3 4 3 4 The total sales of cabbage and carrot are 122 and
+ 2
=
103 respectively.
–5
y
9
Try this HOTS Question
–20
3 3x + 2x 4 3 4 The table below shows Ali’s and Hamid’s expenses
=
9
–5 + 2y
in January.
Compare the corresponding elements: Expense Ali Hamid
3x + 2x = –20 Food
5x = –20 RM200 RM150
x = –4 Entertainment RM50 RM70
It is given that Ali’s expense is not change while
–5 + 2y = 9 Hamid’s expense is doubled in February. Calculate
2y = 14 the difference between Ali’s and Hamid’s expenses
y = 7 in February by using matrix method.
Try Questions 5 – 10 in Try This! 2.2 Answer: 3 100 4
90
23
02 Focus Math F5_E2021.indd 23 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

–1
C Multiplying two matrices (b) 3 10 3 43 4
6
–4 7
1. To multiply two matrices, A and B, the number Order: 2 × 2 2 × 1
of columns in matrix A must be the same as the
number of rows in matrix B, which is Same
A × B = AB Hence, the pair of matrices can be multiplied
Order: m × n n × p m × p and the order of the product is 2 × 1.
5
3 4
Number of = Number of (c) [12 4 7] 0 Chapter 2
columns in A rows in B 9

The order of AB is m × p Order: 1 × 3 3 × 1
Chapter 2
2. In multiplication of two matrices, each element Same
in the rows of the first matrix is multiplied by Hence, the pair of matrices can be multiplied
corresponding element in the columns of the and the order of the product is 1 × 1.
Minion Pro 10pt = 1 second matrix. Then, sum up the results of –2
2 multiplication of these elements. (d) 3 4 [6 1]
3
p
(a) [a b] 3 4 = [ap + bq] Order: 2 × 1 1 × 2
1 q
Arial 8.5 pt =
2 1 × 2 2 × 1 1 × 1 Same
Hence, the pair of matrices can be multiplied
(b) 3 a b 43 p q s 4 3 ap + br aq + bs 4 and the order of the product is 2 × 2.
=
c d
cp + dr cq + ds
r
2 × 2 2 × 2 2 × 2 17
Find the product of each of the following.
a b
3 4 q 3 4 5
ap + bq
p
(c) c d 3 4 = cp + dp (a) 3 4 [1 2]
e f ep + fq –4
43
3 × 2 2 × 1 3 × 1 8 –1 6 0
1
(b) 3 2 3 –2 4
16 9 4
Determine whether the following pairs of matrices (c) [3 –4] 3 4
12
can be multiplied. If yes, state the order of the Solution
product. 5
7
–1
–4
(a) 3 43 9 11 4 (b) 3 10 3 43 4 (a) 3 4 [1 2]
6
–4 7
0 8
2
= 3 5 × 1 5 × 2 4
–4 × 1 –4 × 2
3 4 –2
5
(c) [12 4 7] 0 (d) 3 4 [6 1] = 3 5 10 4
9 3 –4 –8
Solution 8 –1 6 43 0
1
7
(a) 3 43 9 11 4 (b) 3 2 3 –2 4
4
0 8
2
Order: 2 × 1 2 × 2 = 3 (8 × 0) + (–1 × 1) + (6 × 4) 4
(2 × 0) + (3 × 1) + (–2 × 4)
Not the same = 3 0 – 1 + 24 4
0 + 3 – 8
Hence, the pair of matrices cannot be multiplied. 23
= 3 4
–5
24

02 Focus Math F5_E2021.indd 24 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

9
(c) [3 –4] 3 4 Solution 1 3
12
= [(3 × 9) + (–4 × 12)] [4x 9] 3 2 –x 4 = [x 2y]
= [27 – 48] [4x + 18 12x + (–9x)] = [x 2y]
= [–21] Compare the corresponding elements:

4x + 18 = x 12x − 9x = 2y
18 3x = −18 3x = 2y
x = −6 2y = –18
8
It is given that P = 3 3 9 4 and Q = 3 5 1 –3 4 . Find y = –9 Chapter 2
–2 –4
(a) PQ Try Questions 11 – 14 in Try This! 2.2
(b) QP
Chapter 2
(c) Q Example of HOTS
2
HOTS Question
Solution Madam Dahniah’s catering company receives orders
(a) PQ = 3 3 9 43 5 8 4 from two customers, A and B. The table below shows
the number of orders of both customers.
–2 –4 1 –3
= 3 15 + 9 24 + (–27) 4 Customer A Food Drink
–16 + 12
–10 + (–4)
45
60
= 3 24 –3 4 Customer B 82 50
–14 –4
It is given that the preparation costs of the food and
drink are RM30 and RM8 for each order respectively.
(b) QP = 3 5 8 43 3 9 4 Calculate the total preparation cost for both
1 –3 –2 –4
customers’ orders.
= 3 15 + (–16) 45 + (–32) 4 Solution: 1800 + 360
9 + 12
3 + 6
60 45 30
=
8
2460 + 400
= 3 –1 13 4 3 82 50 43 4 3 3 2160 4 4
9 21
=
2860
(c) Q = QQ Total preparation cost = 2160 + 2860
2
= RM5020
= 3 5 8 43 5 8 4 Try this HOTS Question
1 –3 1 –3
= 3 25 + 8 40 + (–24) 4 Roshan’s shop sells black cartridge and colour
5 + (–3)
8 + 9
cartridge for printer. The table below shows the
= 3 33 16 4 sales of the cartridges for three months.
2 17
Colour cartridge
Black cartridge
January 10 6
REMEMBER! February 12 4
March 8 5
(a) When multiplying matrices P and Q, PQ is not
necessarily same as QP. It is given that the profits earned from selling each
5 2 8 2 black cartridge and colour cartridge are RM10 and
(b) Q ≠ 3 2 24 RM8 respectively. Calculate the total profits, in
2
1 (–3)
RM, earned in January, February and March in the
form of matrix.
148
3 4
19 Answer: 152
Solve the equation [4x 9] 3 1 3 4 = [x 2y]. 120
2 –x
25





02 Focus Math F5_E2021.indd 25 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

D Explaining the characteristics of 22
identity matrix
It is given that matrix P = 3 12 4 4 and matrix
1. Identity matrix is a square matrix with its –1 2 –7 6
diagonal elements from top left corner to bottom Q = 3 3 –5 4 . Express each of the following as a
right corner are 1 and all other elements are 0. single matrix.

1 0
4
and 0
0 .
For example, 3 0 1 3 1 0 1 0 4 (a) PI + IQ
(b) (P – Q)I
0
1
0
Solution Chapter 2
2. The product of matrix A with identity matrix I is (a) PI + IQ
matrix A itself, which is
43
43
Chapter 2
+
AI = IA = A = 3 12 4 1 0 4 3 1 0 –1 2 4
–7 6 0 1
0 1 3 –5
+
20 = 3 12 4 4 3 –1 2 4
–7 6
3 –5
Write identity matrices for the following orders. 11 6
(a) 4 × 4 = 3 –4 1 4
(b) 1 × 1
(b) (P – Q)I
Solution
–
= 13 12 4 4 3 –1 2 423 1 0 4
1 0 0 0
0 1
3 –5
–7 6
3 0 0 1 04
0 1 0 0
(a) = 3 13 2 43 1 0 4
0 0 0 1 –10 11 0 1
= 3 13 2 4
(b) [1] –10 11
Try Questions 15 – 17 in Try This! 2.2
21
Determine whether the following matrices are identity
matrices for 3 3 5 4 . E Explaining the meaning of inverse
matrix and hence determining the
4 1
inverse matrix for a 2 × 2 matrix
(a) 3 1 0 4 (b) 3 0 1 4 1. When two matrices of order 2 × 2, A and B
0 1
1 0
Solution are multiplied, if AB = BA = I, then B is the
inverse matrix of A and A is the inverse matrix
43
(a) 3 3 5 1 0 4 3 3 + 0 0 + 5 4 of B.
=
4 1 0 1
4 + 0 0 + 1
−1
= 3 3 5 4 2. Inverse matrix of A is written as A .
Then, AA = A A = I.
−1
−1
4 1
Hence, 3 1 0 4 is an identity matrix. 3. When matrix A = 3 a b 4 , the inverse matrix of
0 1
c d
A, A can be derived by the formula below:
−1
43
(b) 3 3 5 0 1 4 3 0 + 5 3 + 0 4 A = 1 3 d –b 4
=
0 + 1 4 + 0
4 1 1 0
−1
= 3 5 3 4 where ad – bc ≠ 0. ad – bc –c a
1 4
Hence, 3 0 1 4 is not an identity matrix. 4. ad – bc is known as determinant of matrix A,
1 0
|A|, where ad – bc ≠ 0.
26
02 Focus Math F5_E2021.indd 26 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

5. If ad – bc = 0, then the inverse matrix of A, A 24
–1
does not exist.
Find the inverse matrix for each of the following
matrices.
REMEMBER! 2 6
(a) E = 3 4
1 3 10
A ≠
–1
A
(b) F = 3 –8 –3 4
12 4
23 Solution Chapter 2
10 –6
1
3

Determine whether the following matrices are the (a) E = (2 × 10) – (6 × 3) –3 2 4
–1
Chapter 2
inverse matrices of 3 2 5 4 . = 1 10 –6 4
3
1 3
(a) 3 3 –5 4 (b) 3 –2 1 4 2 –3 2
–3
5 –3
5
–1 2

3
= 3 – 14
Solution 2
43
4
1
3
(a) 3 2 5 3 –5 4 (b) F = [(–8) × 4] – [(–3) × 12] –12 –8 4
3

–1
1 3 –1 2
= 3 6 + (–5) –10 + 10 4 = 1 4 3 4
3
3 + (–3)
–5 + 6
= 3 1 0 4 4 –12 –8
3
4
0 1

1
= 3 4
43
3 3 –5 2 5 4 –3 –2
1 3
–1 2
25
= 3 6 + (–5) 15 + (–15) 4 Find the value of m if P does not have an inverse
–2 + 2
–5 + 6
= 3 1 0 4 matrix.
0 1
(a) P = 3 m 12 4
Hence, 3 3 –5 4 is the inverse matrix of 3 2 5 4 3 4
–1 2
1 3
10 1 – 2m
because the product is an identity matrix. (b) P = 3 2 1 4
43
(b) 3 2 5 –2 1 4 Solution
(a) P does not have an inverse matrix, therefore
1 3 5 –3
ad – bc = 0
= 3 –4 + 25 2 + (–15) 4 |P| = 0.
1 + (–9)
–2 + 15
m(4) – 12(3) = 0
= 3 21 –13 4 4m – 36 = 0
13 –8
4m = 36
m = 9
≠ 3 1 0 4 (b) P does not have an inverse matrix, therefore
0 1
Hence, 3 –2 1 4 is not the inverse matrix of |P| = 0. ad – bc = 0
5 –3
3 2 5 4 because the product is not an identity 10(1) – (1 – 2m)(2) = 0
1 3
10 – 2 + 4m = 0

matrix. 4m = –8
m = –2
27
02 Focus Math F5_E2021.indd 27 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

26 2 Write the equations 3 a b 43 4 3 4 where
m
x
=
n
y
cd
Given that 3 –2 1 4 Q = 3 1 0 4 , find matrix Q. in the form of a, b, c, d, m and n are
–4 3
0 1
matrix,
Solution AX = B. constants, while x and y
are variables.
Given that the product of 3 –2 1 4 and Q is an identity
–4 3
matrix, therefore Q is the inverse matrix of 3 –2 1 4 . 3 Solve by multiplying
–4 3
Q = 1 3 3 –1 4 inverse matrix: x 1 d –b m Chapter 2
AX = B
3
[(–2) × 3] – [1 × (–4)] 4 –2 A AX = A B 3 4 = ad – bc –c a 43 4
n
y
–1
–1
3
Chapter 2
= – 1 3 –1 4 IX = A B
–1
2 4 –2 X = A B
–1
3
3 – 1 4
2
= –2 2 28
1
Write the following simultaneous linear equations in
Minion Pro 10pt = 1 27 the form of matrix.
2 4 5 2x = 4 – 3y
= 1 0 0 2 It is given that G = 3 –2 –3 4 and the inverse matrix of 3x + 4y = 5
0
0
3
G is 1 –3 n 4 . Find the value of m and of n. Solution
Arial 8.5 pt = 1 m 2 4 2x + 3y = 4
2 Solution 3x + 4y = 5
= 1 0 0 2 The inverse matrix of G Matrix form:
0
0
x
4
=
= 1 3 –3 –5 4 3 2 3 43 4 3 4
[4 × (–3)] – [5 × (–2)] 2 4 3 4 y 5
Kaedah Alternatif 1 –3 –5
= – 3 4 29
2 2 4
Compare with the inverse matrix given: Solve the following equation. 2
3 2
x
=
3
3
4
y
– 1 –3 –5 4 = 1 –3 n 4 3 8 6 43 4 3 4
2 2 4 m 2 4
Solution
Hence, m = –2 and n = –5 3 2
The inverse matrix of 3 8 6 4
Try Questions 18 – 22 in Try This! 2.2 1 6 –2
= 3 4
18 – 16 –8 3
3
F Using the matrix method to solve = 1 6 –2 4
simultaneous linear equations 2 –8 3
2
x
1. Simultaneous linear equations can be solved 3 4 = 1 6 –2 43 4
3
using matrix method in accordance to the steps y 2 –8 3 4
below: 1 12 + (–8)
= 3 4
1 Write simultaneous 2 –16 + 12
3 4
linear equations in ax + by = m = 1 4
the form of cx + dy = n 2 –4
2
ax + by = c. = 3 4
–2
Hence, x = 2 and y = –2.
28
02 Focus Math F5_E2021.indd 28 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

30 In matrix form,
43 4 3 4
Solve the following simultaneous linear equations 3 26 30 x y = 125
5
6
25
using matrix method.
x
43 4
x + 2y = 4 3 4 = 1 3 5 –30 125
3x + 4y = 6 y 130 – 180 –6 26 25
1 625 + (–750)
3
Solution = – 50 –750 + 650 4
x
4
1 –125
3
3 1 2 43 4 3 4 = – 50 –100 4 Chapter 2
=
6
3 4
y
2.5
1
4 –2
4
x
3
3 4 = 4 – 6 –3 1 43 4 = 3 4
2
Chapter 2
y
6
Hence, x = 2.50 and y = 2.00.
3
= – 1 16 + (–12) 4
2 –12 + 6 Try Questions 26 – 27 in Try This! 2.2
3 4
= – 1 4 SPM Highlights
2 –6
–2
= 3 4 Haris, Jimmy and Lim sold cakes and burgers on
school canteen day. Haris gained a profit of RM61.20
3
Hence, x = –2 and y = 3. from selling 12 cakes and 36 burgers. Jimmy sold 25
cakes while Lim sold 15 burgers. The profit difference
between selling 25 cakes and 15 burgers is RM19.50.
Calculate the profit, RMx, of selling a cake, and the
REMEMBER! profit, RMy, of selling a burger.
–2
Do not leave the answer in matrix form, 3 4 .
3
Write the answer as x = –2 and y = 3. Solution
12x + 36y = 61.20 ⇒ x + 3y = 5.10
25x – 15y = 19.50 ⇒ 5x – 3y = 3.90
Try Questions 23 – 25 in Try This! 2.2 1 3 x 5.10
3 5 –3 43 4 3 3.90 4
=
y
–3 –3 5.10
G Solving problems involving matrices 3 4 = 1(–3) – 3(5) –5 1 43 3.90 4
x
3
1
y
1 –15.30 – 11.70
3
31 = – 18 –25.50 + 3.90 4
Mr Salleh has stalls in market A and market B. The = – 1 –27.00 4
3
table below shows the number of packets of nasi lemak 18 –21.60
and mi goreng sold in both markets in a day. = 3 1.50 4
1.20
Market Nasi lemak Mi goreng Hence, x = 1.50, y = 1.20.
A 52 60
B 48 40
Try This! 2.2
It is given that Mr Salleh earned RM250 in market
A and RM200 in Market B on that day. Calculate the 1. Determine whether the following pairs of matrices
profit RMx of selling a packet of nasi lemak, and the can be added and subtracted.
profit, RMy, of selling a packet of mi goreng. (a) 3 0 –3 1 4 3 5 6 –1 4
,
8
9
7
2 10 4
Solution (b) 3 –2 5 4 3 4
3
,
52x + 60y = 250 ⇒ 26x + 30y = 125 1 6 8
48x + 40y = 200 ⇒ 6x + 5y = 25 (c) [11 4], [7 2]
29
02 Focus Math F5_E2021.indd 29 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

2. Express each of the following as a single matrix. (c) [7 2] 3 1 –1 4
(a) 3 2 3 4 3 –4 5 4 (d) 3 9 10 3 12
+
1 –9
6 –7
4
43 4
4 –5
1
–
(b) 3 0 3 10 4 3 5 6 1 4 12. Find the product for each of the following.
9
–8 –2
–3 –4 9
3. Given that [–4 9] + T – [5 –2] = [1 –6], find 6 –3 1
matrix T. (a) [3 –4] 3 2 5 –1 4
2
4. Find the values of x and y for each of the following. (b) 3 –7 6 43 4
10
5
–1
8
=
(a) 3 5 2x 4 3 2 5x 4 3 3 9 4 4 2 –3 7 –1 Chapter 2
–
3y 7
8
3

y
1
x
4
(b) 3 3x – 2 4 3 4 3 4 (c) 3 3 4 43 –3 5 4
+
=
7
1 + y
2y
5
Chapter 2
(d) 3 4 [1 3 –4]
5. Express each of the following as a single matrix. –2
1
(a) 3[–5 7] 13. It is given that S = [–4 3], T = 3 4 , U = 3 5 1 4 and
–3 2
–2
3
(b) 1 12 –3 4 V = 3 1 –4 4 . Find
4 8 20 3 6
6. It is given that P = 3 2 6 4 , Q = 3 1 –4 4 and (a) ST
(b) VT
–3 5
8 3
R = 3 –7 9 4 . Find (c) SU
(d) TS
10 –2
2
(a) R + Q (e) U
(b) P + (Q + R) 14. Solve the following equations.
(c) Q – R + P
5
6
=
(a) 3 x –1 43 4 3 4
3 2y
4
–9
–6
4
7. Given that M = 3 4 and N = 3 4 , calculate 3 12 21
2
1
(a) 2(M + N) (b) 3 4 [1 – x 7] = 3 2y –35 4
–5
(b) 0.8(M – N)
15. Write the identity matrices for the following orders.
8. Express each of the following as a single matrix. (a) 2 × 2
(b) 3 × 3
(a) 2 3 2 –1 4 3 4 3 4 16. Determine whether the following matrices are
+
3 12
7 –15
1
(b) [13 4] – 5[3 2] + [12 –8] 4 5
2 identity matrices for 3 3 2 4 .
9. Find matrix U for each of the following. 1 0
1 6 –4 4 (a) 3 0 1 4
(a) 3 4 + 3 3 4 – U = 3 4
3 12 2 1 (b) 3 1 1 4
(b) 2U – 3 10 6 4 = 4 3 4 –2 4 0 0
2
1
9
5
10. Find the values of p and q for each of the following. 17. It is given that M = 3 –1 5 4 , N = 3 3 10 4 and
8
4 –2
6
q
6
3
–
(a) 5 3 4 3 4 3 4 K = 3 5 2 4 . Express each of the following as a
=
–7 1
11
–p
4
=
(b) 3 1 p 4 + 3 3 2 p 4 3 7 –8 4 single matrix.
(a) MI – IN
3
14 11
q – 1
2
5
(b) (K + N)I
11. Determine whether the following pairs of matrices (c) (KI)M
can be multiplied. If yes, state the order of the
product. 18. Determine whether the following matrices are
3
8
1
3 4 inverse matrices of 3 –2 –5 4 .
(a) 9 [2 –1]
8 (a) 3 –5 –8 4
0
(b) [5 –3 4] 3 4 2 3
6
(b) 3 –3 2 4
–8 5
30
02 Focus Math F5_E2021.indd 30 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

19. Find the inverse matrix for each of the following. 24. Solve the following equations.
19
x
(a) P = 3 4 7 4 (a) 3 5 2 43 4 3 4
=
4
2
3
y
5
16
–2
x
=
(b) Q = 3 8 –6 4 (b) 3 1 3 43 4 3 4
–7
y
3
8
–5 4
–3
p
=
(c) R = 3 1 3 4 (c) 3 3 9 43 4 3 4
3
–2 –7
q
–2 4
–18
p
=
20. Find the value of x if L does not have an inverse (d) 3 4 –3 43 4 3 4
1
2
q
1
matrix. 25. Solve the following simultaneous linear equations Chapter 2
(a) L = 3 3 –6 4 using matrix method.
x
2
(a) 4x + 3y = –5
x + 2y = 10
Chapter 2
(b) L = 3 12 8 4 (b) 4y = 1 + 3x
2x + 1
2
8 y – 5 = 5x
21. It is given that EF = 3 1 0 1 4 . Find matrix (c) y = 3x – 1
x = 3y – 5
5

0
(d) 6y + 4 = x
(a) E if F = 3 3 9 4 . 2 y + 3x = –8
2
7
26. Ahmad has two part-time jobs in college. His salary
(b) F if E = 3 –2 6 4 . as a tutor is RM20 per hour while his salary as a
–4 11
helper in a bookshop is RM12 per hour. It is given
that he allocates 18 hours each week for the part-
22. Given that H = 3 5 3 4 and the inverse matrix of time jobs and receives salary of RM280. If the time
worked as a tutor is x hours and the time worked as
10 7
1 7 –3 a helper in the bookshop is y hours, find the value of
H is 3 4 , find the value of m and of n. x and of y using matrix method.
m n 5
27. It is given that the perimeter of an equilateral triangle
23. Write the following simultaneous linear equations in is equal to the perimeter of a regular pentagon.
the form of matrix. The length of side of the equilateral triangle is
(a) 6x + 5y = 1 2 cm longer than the length of side of the regular
2 x + y = 3 pentagon. Assuming the length of side of the
(b) 4x = 3y + 5 equilateral triangle is x cm and the length of side of
x – 6y = 4 the regular pentagon is y cm,
(c) 5x = 2y – 1 (a) form two linear equations from the above
y = 3x – 5 information.
(d) 6y = 4 + x (b) calculate the value of x and of y using matrix
method.
7 y + 3x = –8 HOTS
Applying
31





02 Focus Math F5_E2021.indd 31 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

SPM Practice 2
SPM Practice




PAPER 1 15 6
8. Given that 3 4 – 3H = 2 3 4 , find matrix H.
1. The table below shows the number of breads and –7 1
1
3
cakes sold in Adila’s shop in January and February. A 3 4 C 3 4
1
–3
Month Bread Cake 1 –1 Chapter 2
3
1
January 50 65 B 3 4 D 3 4
February 42 45 4 1
9. [5 –3] 3 4 =
Chapter 2
Represent the sales of breads and cakes in January 7 –2
with a row matrix. A [1 7] C [13 17]
A [50 65] C [42 45] B [–1 11] D [2 11]
B 3 50 65 4 D 3 50 42 4 10. 3 1 –7 8 –9 4 =
43
65 45
42 45
–2 5
2
4
2. It is given that matrix G = 3 15 –1 10 4 and g 23 is A 3 36 23 4 C 3 7 –16 4
9
2
–2
3
16 18
3
an element of matrix G. State the value of g 23 . –20 –23 9 –16
A –2 C 3 B 3 4 28 4 D 3 2 7 4
B –1 D 9 8
11. Given that [x –3] 3 4 = [10], find the value of x.
x
=
3. It is given that 3 3x –5 4 3 12 –5 4 . Find the A 4 C 2
y + 1
3
6
3
value of x – y. B 3 D 1
A –2 C 4 12. Given that 3 2 m 43 4 3 –10 4 , find the value of
–1
=
B –1 D 5 8 5 4 4n
m and of n.
8
–5
+
–
4. 3 4 3 4 3 –16 4 = A m = 1, n = 3 C m = –3, n = 1
D m = –2, n = 3
B m = 2, n = 4
9
12
7
A 3 –13 4 C 3 –19 4 13. It is given that matrix P = 3 6 –2 4 , matrix
10
28
3 10
6
–3
B 3 4 D 3 4 Q = 3 1 8 4 and matrix I = 3 1 0 4 . Find the value
–9 4
4
0 1
5
of PI – Q.
3
5. 3 9 3 4 + 1 16 20 4 – 5 3 5 –3 4 = A 3 7 6 4 C 3 5 –10 4
2 4 14
1
–7 8
2
–6 14
12 6
A 3 32 10 4 C 3 8 17 4 B 3 14 –6 4 D 3 5 12 4
0 13
–2 5
6
7
–10 6
B 3 20 17 4 D 3 –8 28 4
–10 5
–2 24
14. Find the inverse matrix of 3 7 3 4 .
5 2
6. Given that 3 3 –2 p 7 4 – 2 3 5 3 4 3 –16 9 4 , A 3 2 –3 4 C 3 –2 3 4
=
4
1 – q 8
2
5
find the value of p and of q. –5 7 5 –7
A p = 4, q = 5 C p = 5, q = 4 B 3 –7 5 4 D 3 7 –3 4
B p = –4, q = 5 D p = 5, q = –4 3 –2 –5 2
15. Sofia bought 3 boxes of crayons and 5 boxes of
1
7. Given that m[3 –2] – [12 n] = [6 –11], find the SPM colour pencils. Suraya bought 4 boxes of crayons
4
value of m + n. 2017 and 6 boxes of colour pencils. It is given that each
box of crayon costs RM8 while each box of colour
A −12 C 12 pencil costs RM12. Which of the following is the
B 4 D 23 correct method to calculate the total payment to be
made by Sofia, p, and Suraya, q?
32
02 Focus Math F5_E2021.indd 32 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

A 3 3 5 43 4 3 4 20. It is given that the inverse matrix of 3 2 1 4 is
8
p
=
q
4 6
12
6 5
SPM
1 5 –1
3
12
p
B 3 3 5 43 4 3 4 2019 m n 2 4 . Calculate the value of m + n.
=
q
8
4 6
A 2 C 4
p
8
C 3 3 4 43 4 3 4 B –4 D –2
=
q
12
5 6
12
p
D 3 3 4 43 4 3 4 PAPER 2
=
q
5 6
8
1. Represent the following information in the form of
matrices.
16. 2 3 6 –3 4 3 14 9 4 – 3 3 7 4 4 = (a) Wilson and Yasmins’s hobbies are collecting Chapter 2
+
11 5
–8 7
2 1
SPM stamps. Wilson has 126 local stamps and 98
2018 27 10 –1 6
A 3 5 13 4 C 3 –3 8 4 oversea stamps. Yasmin has 185 local stamps
and 103 oversea stamps.
Chapter 2
B 3 5 2 4 D 3 5 –9 4 (b) The table below shows the in-store and online
sales of three items in Yasmin’s shop.
13 10
8 14
Instant Tinned
17. It is given that K is a 2 × 2 matrix and Biscuit noodles food
SPM 1 4 m
–1
2018 K = 3 4 . Find the value of In-store 50 32 15
(–1)(4) – (–3)(2) –2 n
m and of n. Online 35 28 10
A m = 3, n = –1 C m = 3, n = 1
B m = –3, n = –1 D m = –3, n = 1
3
4
6
18. The table below shows Gopal’s and Fuad’s 2. It is given that matrix P = –4 –8 1 7 and p ij is
3
Mathematics marks in two tests. 0 2 9
an element of matrix P.
Test 1 Test 2 (a) State the order of matrix P.
Gopal 78 83 (b) State p 13 and p 32 .
(c) Find the value of p 21 + 4p 33 .
Fuad 54 67
Calculate the difference of their marks for each 5 –3
test. HOTS 3. It is given that matrix Q = 3 4 4 and matrix
Analysing 3 4
Test Test Test Test R = 3 5 –3 4 . If matrix Q and matrix R are equal,
1 2 1 2 0.75 2m
A [ 24 16 ] C [ 5 13 ] find the value of m.
Test Test Test Test
1 2 1 2 6 2 7
B [ 29 11 ] D [ 11 13 ] 4. It is given that M = 3 4 , N = 3 4 and P = 3 4 .
1
–3
5
Express each of the following as a single matrix.
19. The table below shows the amount of flour and red (a) M + N
bean paste required to make buns and breads. (b) (M + N) + P
(c) P – 3M
Red bean
Flour (d) 2N + M – 5P
paste
Bun 2 kg 1 kg 5. Find the product of each of the following.
7
Bread 3 kg 1 kg (a) 3 4 [–3 2]
–4
It is given that the prices of flour and red bean paste (b) 3 1 3 9 2 4
43
are RM2.50 per kilogram and RM9.00 per kilogram 2 5 –7 4
respectively. Calculate the costs, in RM, to make the (c) [8 7] 3 –1 4 4
buns and breads. HOTS Analysing 6 –3
4
4
A 3 18.00 buns C 3 18.50 buns 6. Lily and Jamilah want to buy sausages and nuggets


27.00 breads
27.00 breads
to sell on school canteen day. Table 1 shows the
4
4
B 3 20.50 buns D 3 14.00 buns quantity of sausages and nuggets they want to buy.


Table 2 shows the selling prices of sausages and
16.50 breads
29.50 breads
nuggets in two shops.
33
02 Focus Math F5_E2021.indd 33 18/02/2021 2:32 PM

Mathematics Form 5 Chapter 2 Matrices

Sausage Nugget 12. The table below shows the information of purchase
of exercise books by Madam Muna.
Lily 5 7
Jamilah 8 4 Type of Purchase Price per
exercise books quantity book (RM)
Table 1
Thick x 2.50
Shop A Shop B
Thin y 0.70
Sausage 12 15
Nugget 16 14 It is given that the purchase quantity of thick exercise
books is 4 more than the purchase quantity of thin
Table 2 Chapter 2
exercise books. The total payment of the exercise
(a) Calculate Lily’s and Jamilah’s total expenses to books purchased is RM26.
buy sausages and nuggets in each shop. (a) Write two linear equations in terms of x and y to
Chapter 2
(b) Decide purchase in which shop is more represent the information above.
profitable for Jamilah. Explain your answer. (b) Hence, by using matrix method, calculate the
HOTS
Evaluating value of x and of y. HOTS
Applying
7. It is given that F = 3 10 3 4 , G = 3 –8 5 1 4 and 13. Mr Jefri and Mr Siva each took their family to watch
–2 4
6
SPM
I = 3 1 0 4 . Find the value of each of the following. 2017 a movie. Mr Jefri spent RM72 for 4 tickets and
0 1
3 sets of food. Mr Siva spent RM52 for 3 tickets and
(a) IF + G 2 sets of food. Using the matrix method, calculate
(b) (G – F)I the prices, in RM, of a ticket, x, and a set of food, y.
8. Find the inverse matrix for each of the following 2 3 –5 –3
matrices. 14. It is given that S = 3 –4 –5 4 , T = p 3 q 2 4 and
SPM 1 0
(a) 3 2 3 4 2018 I = 3 0 1 4 .
4
6
(b) 3 13 7 4 (a) Find the value of p and of q if ST = I. linear
following
(b) Write
the
simultaneous
–4 –2
equations in the form of matrix:
x + 3y = 7
2
9. S 3 2 9 4 3 1 0 4 and S is a 2 × 2 matrix. Find –4x – 5y = –11
=
0 1
1 3
matrix S. Hence, by using matrix method, calculate the
value of x and of y.
1 –2 p 3 –5 1 0
10. (a) Given that 3 43 4 3 4 ,
=
14 –4 3 4 q 0 1
find the value of p and of q. 15. Mr Koh, Mr Omar and Madam Selvi invested in
SPM share market respectively. Mr Koh used RM5 600
(b) The inverse matrix of 3 2 –4 4 is p 3 5 4 4 . Find 2019 to invest 4 000 units of Share A and 1 000 units of
1
5
q 2
the value of p and of q. Share B. Mr Omar invested 3 000 units of Share A
while Madam Selvi invested 2 000 units of Share
1 1 n B. The difference between the investment costs of
11. (a) Given that 3 4 is the inverse matrix of
m –2 4 3 000 units of Share A and 2 000 units of Share
3 4 3 4 . Find the value of m and of n. B is RM2 000. It is given that the price per unit of
2 1
Share A is higher than the price per unit of Share B.
(b) Solve the following simultaneous linear Calculate the price per unit of Share A and the price
equations using matrix method. per unit of Share B.
4 x + 3y = 16
2 x + y = 7



34





02 Focus Math F5_E2021.indd 34 18/02/2021 2:32 PM

ANSWERS





Chapter Try This! 1.2
1 Variation 1. (a) The time taken is decreased.
(b) The time taken is doubled.
Try This! 1.1 2. The value of xy is not a constant, thus y does not vary inversely
1. (a) The price of the durian is reduced by 20%. as x.
2
2
(b) The price of the durian is doubled. The value of x y is a constant, thus y varies inversely as x , that
y is y ∝ 1 2 .
2. The value of is not a constant, thus y does not vary directly x
x
as x. 3.
y y
The value of is a constant, thus y varies directly as √x, that
√x
is y ∝ √x. 60
3. 50
V
40
6
30
5
20
4
10
3 1 –
O x
2 0.1 0.2 0.3 0.4 0.5 0.6
1 1
The graph of y against x is a straight line starting from the
O 10 20 30 40 50 60 A origin. Hence, y varies inversely as x.
4. (a) f = 4
The graph of V against A is a straight line passing through the 3 √g
origin. Hence, V varies directly as A. 128
(b) f = 2
4. (a) p = 1.5q g
2
3
(b) p = 48√q 5. (a) 1.024
5. d = 25t (b) 0.4
6. (a) 283.5 6. (a) y = 1
(b) 1.4 4x 2
(b) 2.5
3
7. (a) A = 25√B
(b) 512 7. (a) G = 45
√H
8. (a) m ∝ pq (b) 3.75
(b) G ∝ E F
2
(c) V ∝ r h 8. (a) T = 540
2
N
2
9. (a) y = xz 2 (b) 104
3
(b) 0.9 9. 25
10. (a) 18
(b) 1.8 10. 2
11. x = 121.5, y = 0.6
Try This! 1.3
12. (a) E = 9.5MH
(b) 57 1. (a) e ∝ 1
f√g
13. (a) I = 0.04pt Y
(b) Mr Kamal can reduce the term of loan. To maintain the (b) X ∝
Z
2
interest charged, the term of loan is reduced so that the
amount of loan is increased. (c) p ∝ q
3 √r
161




11 ANS Focus KSSM Math F5_E2021.indd 161 18/02/2021 3:15 PM

Mathematics Form 5 Answers

2. (a) m = 6n 4. (a) e = t
5√p 139
(b) 3 cm
3
(b) m = 15n
p 5. y = 4.5wx 2
3. (a) a = 8√b 6. y ∝ 1
c x
(b) 2 7. (a) The number of bags is halved.
(b) The number of bags is doubled.
1.2Q
4. (a) P = 6
R 2 8. (a) g = 3 √h
(b) x = 3, y = 0.8 (b) 10
5. (a) 4 9. (a) m = 24 2
(b) 0.04 n
(b) 0.5
6. 33 10. (a) V = 9
3b √W
7. (a) t =
1000m (b) 225
(b) 7.5 hours
11. (a) A = 20
8. (a) 4 (b) 8 √L
(b) The time needed varies inversely as the number of workers.
The time decreases if the number of workers increases. (c) 25
x
SPM Practice 1 12. (a) y ∝
z
√W
PAPER 1 (b) V ∝ U
1. B 2. D 3. C 4. C 5. D 3.2Q 3
13. (a) P =
6. A 7. B 8. A 9. C 10. A R
(b) 1.6
11. D 12. B 13. C 14. A 15. A
16. B 17. B 18. C 19. C 20. B 14. (a) m = 4n
√p
PAPER 2 (b) 81
1 15. (a) S = 72
1. T√U
3
(b) 4.8
2
2. (a) m = 1.5n
(b) 2.16 16. (a) 5.6
3. (a) E = 3.6 √F (b) If G is doubled, E is halved. This is because E varies
3
(b) 125 inversely as G.



























162





11 ANS Focus KSSM Math F5_E2021.indd 162 18/02/2021 3:15 PM

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