MATEMATIK TINGKATAN 5 BUKU B

MODULE

GLOBAL MEDIASTREET SDN. BHD.
MATEMATIK
DWIBAHASA

CIRI EKSKLUSIF

Aktiviti Berteraskan DSKP
Praktis Bab
Sudut KBAT
PAK-21

KOD QR BBUKU TINGKATAN

Nota Buku B (Bab Genap) 5
Video
Jawapan Buku A (Bab Ganjil) KSSM

Zakry bin Ismail

KANDUNGAN

BAB 2 1
4
MATRIKS 23
 2.1 Matriks
  2.2  Operasi Asas Matriks
Praktis Bab

BAB 4 GLOBAL MEDIASTREET SDN. BHD. 27
41
MATEMATIK PENGGUNA: PERCUKAIAN
4.1 Percukaian

Praktis Bab 4

BAB 6

NISBAH DAN GRAF FUNGSI TRIGONOMETRI
6.1  Nilai Sinus, Kosinus Dan Tangen bagi Sudut θ, 0°  θ  360° 47
6.2  Graf Fungsi Sinus, Kosinus dan Tangen 50
Praktis Bab 6 55

BAB 8 59
72
PEMODELAN MATEMATIK
8.1  Pemodelan Matematik

Praktis Bab 8

NOTA JAWAPAN

Bab 2 Matriks
GLOBAL MEDIASTREET SDN. BHD.
2.1 Matriks BAB 2

1 Lengkapkan jadual di bawah. VIDEO
Complete the following table.
Peringkat matriks
CONTOH Matriks Bilangan baris Bilangan lajur Order of matrix
Matrix Number of rows Number of columns
2×3
� 73 –61 85 � 2 3

(a) [3 –8] 1 2 1×2

(b) 2 3 1 3×1
� –1 �
6

(c) � –05 74 � 2 2 2×2


(d) a b 3 2 3×2
� c d �
e f

(e) � 85 –96 11 � 2 3 2×3

(f) – 11 7 –3
� 4 3 –9 � 3 3 3 × 3
4 6 22

2 Tuliskan satu contoh matriks bagi setiap peringkat yang berikut.
Write an example of matrix for each of the following orders.

CONTOH (a) 2 × 3 (b) 3 × 3
� da be cf �
1×2 a b c
� d e f �
Penyelesaian g h i
� a �
b

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–9 0 3
3 Diberi matriks P = �–2 4 –1� , kenal pasti unsur yang berikut dalam matriks P.
5 –6 7
BAB 2
–9 0 3
GLOBAL MEDIASTREET SDN. BHD.G iven the matrix P = � –2 4 –1 �, identify the following element in the matrix P.

5 –6 7

CONTOH (a) p12 = 0 (b) p23 = –1 (c) p32 = –6 (d) p33 = 7

p22

Penyelesaian
p22 = 4

10 5
4 Diberi matriks Q = �–9 13 � , cari nilai bagi setiap yang berikut.
7 –3

1 0 5
Given th e matrix Q = �–9 13 � , find the value for each of the following.

7 –3

CONTOH (a) q 11 + q22 == 10 + 13 (b) q31 + q32 = 7 + (–3)
23 = 4
q12 + q21

Penyelesaian
5 + (–9) = –4

(c) q22 – q31 == 13 – 7 (d) q12 – q21 = 5– (–9) (e) q11 – q22 = 10 – 13
6 = 14 = –3

5 Tentukan sama ada pasangan matriks yang berikut adalah sama.
Determine whether the following pair of matrices are equal.

CONTOH ( a) C = � –89 � , D = [–9 8]
C≠D
1 0.5
A = � 2 � , B = � 3 �
–1.5 – 2

Penyelesaian
A=B

� –11 – 1 –11 –0.25 3 –1 3 –1
(b) E = 3.25 4 � , = � 341 5 � (c) G = � –3 ab � , H = �–3 a + b �
5 F 6 2 6 2

E=F G≠H

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6 Diberi A = B, hitung nilai x dan y.
Given A = B, find the values of x and y.

CONTOH

A = � –x7 � , B = � 4y �

Penyelesaian
x = 4, y = –7
GLOBAL MEDIASTREET SDN. BHD. ( a) A = � 9x –61 � , B = � –95 6 �
BAB 2 y

x = –5, y = –1

( b) A = � 9 –3 y x + 3 –01 0 � , B = � 63 10 –010 � 6 9 6 –y + 3
5 5 (c) A = � 2x 10 � , B = � 12 10 �
–1 6 –1 6
x + 3 = 10 , 9 – y = 6
x = 7 y = 3 2x = 12 , –y + 3 = 9
x = 6 y = –6

7 Cari nilai m dan n bagi setiap yang berikut.
Find the values of m and n for each of the following.

CONTOH (a) � m7– 2 � = �3 1m–+n2 �
m – 2 = 3m + 12 , 7 = 1 – n
A = 1+0m � = 10 2m = –14 n = –6
�3 � 13 � m = –7

4 – 2n n – 8

Penyelesaian

3 + m = 13 , 4 – 2n = n – 8
m = 10 3n = 12
n = 4

(b) [ 2 m 2n – 3] = [ 2 6 – m –n ] ( c) � 3m4– 7 n5 � = � 45 4 5 �
–n
m = 6 – m , 2n – 3 = –n
2m = 6 3n = 3 3m – 7 = 5 , n = 4 – n
m = 3 n = 1 3m = 12 2n = 4
m = 4 n = 2

(d) � 6 3 9 � = � 5m6– 8 01.755 36n � 16 – 2m 20
3m 4 6 (e) � 8 � = � 8 �
15 15 2n + 1
16 – 2m = 20 , 15 = 2n + 1
3m = 5m – 8 , 9 = 3n 2m = –4 2n = 14
2m = 8 n = 3 m = –2 n = 7
m = 4

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Latihan Bestari 2.1

BAB 21 Diberi matriks P = � 05 –81 –163 �, tentukan peringkat 2 x – 1
4 Diberi matriks P = � 7 6 � dan matriks
GLOBAL MEDIASTREET SDN. BHD.mpG2i2av.ternikthsaPt .mSaetrtiexruPs=ny� a 50, k–e81n al–1p63 a�s,tdi eutenrsmuirnept1h2,ep13 dan 5 – y 3
order of 2 3
Q = � 7 6 � , tentukan nilai x dan y jika P = Q.
x 3
matrix P. Then, identify the elements of p12, p13 and p22. 2 x – 1 2 3
Given that matrix P = � 7 6 � and matrix Q = � 7 6 � ,
2 Diberi matriks Q = � 1 42 ––176 � , hitung nilai q12 + q21. 5 – y 3 x 3
Given that matrix Q = � 142 ––176 � , calculate the value of
q12 + q21. determine the values of x and y if P = Q.

3 Diberi P = � 2x–1– 3 4y–+2 5 � dan Q = �– 91 –2 � . Jika
2y – 9
P = Q, hitung nilai x dan y.
� 2x–1– 3 4y–+2 5 � � –91 –2
Given that P = and Q = 2y – 9 � . If

P = Q, calculate the values of x and y.

2.2 Operasi Asas Matriks

1 Selesaikan operasi tambah matriks yang berikut.
Solve the following addition operations of matrices.

CONTOH (a) � –123 � + � –137 �

�– 13 4 � + � 16 ––72 � �1 –23 + (1–37) �
5 +
=

Penyelesaian = � 150 �

�– 13 4 � + � 61 ––72 � = � –13++16 4 + ((––27)) �
5 5 +
= � 23 –33 �

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–1 1 GLOBAL MEDIASTREET SDN. BHD. 3 –5 1 –2
(b) � 2 � + � 0 � BAB 2(c) � –4 –1� + �–2 3 �

–9 –2 6 2 4 –5
–1 + 1 3 + 1 –5 + (–2)

= � 2 + 0 � = � –4 + (–2) –1 + 3 �
–9 + (–2) 6 + 4 2 + (–5)
0 4 –7

= � 2 � = � –6 2 �
–11 10 –3

3 4 1 –6 12 6 (e) � p −2q � + � 4p −7q �
(d) � 4 7 6 � + � 9 7 3 � = � p + 4p −2q + (−7q) �
5 10 2 11 13 1 = �5p −9q�
3 + (–6) 4 + 12 1 + 6
= � 4 + 9 7 + 7 6 + 3 �
5 + 11 10 + 13 2 + 1
–3 16 7
= � 13 14 9 �
16 23 3

(f) � – 3p –2q3 � + � 4 3p ––42q � (g) � 03 –4 –15q � + �1 26p –8 1–06q �
2p q
� –3p + 3 2–q3++((––42q)) �
= + 4p = � 03++126p –42p++(–q8 ) –15q++(–160)q �

= � 3 6p ––25q� = � 129p 2p–1+2 q –5q5 �

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2 Selesaikan operasi tolak matriks yang berikut.
Solve the following subtraction operations of matrices.

CONTOH

[ 7 11 ] – [ –2 5 ]

Penyelesaian
� 7 11 � – [ –2 5� = � 7 –(–2) 11 – 5�
=�9 6�
BAB 2 (a) �–34� – � 17 �
= �– 34––17 �
GLOBAL MEDIASTREET SDN. BHD. = � – 211 �

(b) � –5 8 –1 � – �3 4 –13 � (c) � –42 58 � – �– 71 4 –69 �
= [ –5 – 3 8 – 4 –1 – (–13) ] 4 – 7 8 5––(–69) �
= [ –8 4 12 ] = � –2 – (–14)

= � –123 –171 �

(d) � 07 58 –29 � – 17 5 8 a 2a
� 1 3 –17� �7b � � b �
6 –14 16 5 7 4 (e) 9 – 3

7 – 17 5 – 5 –9 – 8 a – 2a
= � 0 – 1 8 – 3 2 – (–17)� = � 7b – b �
9–3
6 – 5 –14 – 7 16 – 4 –a
= � 6b �
– 10 0 –17 6
= � –1 5 19 �

1 –21 12

(f) �– a9 171 c� – �– a6 3bc � (g) � a6 33c 1131 � – � 9 0b 75c –52 �

= � –9a––(a– 6) 7 – b � = �a 6––90b 3 – 5 11 – –(–52) �
11c – 3c 3c – 7c 13

= � –03 78–cb � = � a –6 9b ––42c 183 �

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3 Selesaikan operasi yang berikut. GLOBAL MEDIASTREET SDN. BHD.(a) [ 5 –2 4 ] + [ 4 –7 6 ] – [ 5 3 –9 ]
Solve the following operations. BAB 2 = [ 9 –9 10 ] – [ 5 3 –9 ]
= [ 4 –12 19 ]
CONTOH
(c) � 1 66 –7 –145 � – � 1 90 5 48 � + �2 45 8 44 �
� –72 � + � –23 � – � –61 � 21 1 8
Penyelesaian
� –72 � + � –23 � – � –61 � = � –95 � – � –61 � = �– 63 –12 –69 � + � 2 45 8 44 �
= � –34 � 20 8

(b) �– 1177 268� – � 252 –09 � + � 34 75 � = � 3 11 –4 –105 �
= �– 1329 2158 � + � 34 57 � 28
= � – 1366 3205 �

2 4 –11 –4 (e) � 1329 � – � –158 � + � –106 �
(d) � 7 � + � 36 � – �–18� = � 1277 � + � –106 �
–2 20 25 = � 2217 �
13 –4
= � 43 � – � –18�
18 25

1 7
= � 61 �
–7

–7 3 6 3 4 5 4 2 6 _ –1 4 7 6 5 8
(f ) � 5 –9 � + � 5 7� – �3 8� (g) � 9 1 44 � � 2 5 8 � + � 2 8 9�
5 6 2 9 9 3 3 6 2 0 2 4 2 5 3
5 –2 –1 6 5 8
–1 6 4 5 = � 7 –4 36� + � 2 8 9 �
= � 10 –2 � – � 3 8 � 3 4 –2 2 5 3
7 15 9 3

– 5 1 11 3 7
= � 9 4 45�
= � 7 –10� 5 9 1
–2 12

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4 Cari nilai x dan y.BAB 2 (a) [ 3 y – 1 2 ] + [ x 5 6 ] = [ 9 8 8 ]
Find the values of x and y. [3 + x y + 4 8 ] = [ 9 8 8 ]
GLOBAL MEDIASTREET SDN. BHD. 3 + x = 9 , y + 4 = 8
CONTOH x = 6 y = 4

� 4x �+ � –4y2 � = � –716 � (c) � 64 –121x –79 � = � 3– y2 –94 x –513� + �1 62 122 24 �
Penyelesaian � 46 –121x –79 � 4 –4x + 12 –79 �
� 4x+–42y � = � –716 � = �3 y + 12 11
4 + 4y = –16 , x – 2 = 7
4y = –20 x = 9 –2x = –4x + 12 , 6 = 3y + 12
y = 5 2x = 12 3y = –6
x = 6 y = –2
(b) � 27x 171 � = � 89 –3y9 � – � 22 22 �
� 27x 171 � = � 67 3–y1–12 �
2x = 6 , 7 = 3y – 2
x = 3 3y = 9
y = 3

2x 3 x 6 21 9 –15 3 6 – 2x 2
(d) � 5 4 � + �9 –2� = � 3y – 1 2 � (e) � 3y – 1� + �5 � = � –22 � – � 4 �

–7 11 2 0 –5 11 –27 7 –25 –5
3x 9 21 9 –12 4 – 2x
� 14 2 � = �3y – 1 2 � � 3y + 4� = � –26 �
–5 11 –5 11 –20 –20
3x = 21 , 14 = 3y – 1
x = 7 3y = 15 –12 = 4 – 2x , 3y + 4 = –26
y = 5 2x = 16 3y = –30
x = 8 y = –10

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5 Cari hasil darab matriks yang berikut. GLOBAL MEDIASTREET SDN. BHD. –5
Find the products of the following matrices. BAB 2(a) 3 � 6 �

CONTOH 3
3(–5)
2� 71 ––73 �
= � 3(6) �
Penyelesaian 3(3)
2 � 71 ––73 � = � 22((17)) 22((––73)) � –15
= �1 24 ––164 �
= � 18 �
9

(b) –4 � –1171 1132 � (c) 1 8 –8
= � ––44((–171)1 ) ––44((1132)) � 4 � –16 24 �
= � –4648 ––4528 � –4 6
= � 4141(41(––(1846)) ) 414114(((–2648))) �


2 –2
= �––14 63
2 �

(d) 0.5 � –92 2 54 –312 � (e) � – 76 –514 �6
= � 0.05.(5–(92)2 ) 0.5(4) 0.05.(5–(31)2) � = � –76((66)) –51(46()6) �
0.5(5) = � –4326 –3804 �

= � –41.51 22.5 1–.65 �

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6 Diberi A = � 25 31 �, B = � 32 ––96 � dan C = �– 61 –13 � , ungkapkan setiap yang berikut sebagai matriks tunggal.
Given that A = � 52 13 � , B = � 32 ––69 � dan C = � –61 –13 �, express each of the following as single matrix.
BAB 2
CONTOH (a) 3A + 2C
GLOBAL MEDIASTREET SDN. BHD.
4A – B 3� 25 31 �+ 2 �– 61 –13 �
= � 1 65 93 �+ � – 122 –26 �
Penyelesaian = � 11 38 –113 �
4� 25 31 � – � 32 ––96 � = �2 80 4
12 � – � 32 ––96 �

= �1 67 2110 �

(b) 3C – 2B (c) –3B + 4A
3 �– 61 –13 � – 2 � 32 ––69 � –3 � 32 ––96 � + 4 � 25 31 �
= �– 138 –39 � – � 64 ––1182 � = � – –96 2178 � + � 280 142 �
= � – 194 231 � = � 1 21 2329 �

(d) A – 2B + 3C (e) –B + C – 2A
� 52 31 � – 2 � 32 ––69 � + 3 �– 61 –13 � – � 32 ––96 � + � – 61 –13 � – 2 � 25 31 �
= � 25 13 � – � 64 ––1128 � + �– 138 –39 � = �– –32 96 � + � – 61 –13 � – � 140 62 �
= � –– 12 2113 � + � –1 38 –39 � = � – 44 130 � – � 1 40 26 �
= � –1 46 244 � = � –01 4 41 �

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7 Selesaikan setiap persamaan yang berikut.
Solve each of the following equations.

CONTOH GLOBAL MEDIASTREET SDN. BHD.(a) 5[ 2x–y]–1 [ 4x 12 ] = [ –8 –2y ]
BAB 22
–5 � 3xy � + 4� 250 � = � 55 �
[ 10x −5y] − [2x 6 ] = [ −8 −2y ]
Penyelesaian [ 8x −5y − 6 ] = [ −8 −2y ]
� ––155xy � + � 8200 � = � 55 �
� ––155xy++2800 � = � 55 � 8x = –8 , –5y – 6 = –2y
x = –1 3y = –6
y = –2

–15y + 80 = 5 , –5x + 20 = 5
15y = 75 5x = 15
y = 5 x = 3

2 –3 –14 (c) – 1 � –4x6 –41 2� + 3� ––1y –22x � = 4 � 0y 13 �
(b) –4 � –1 � + 2� 2x � = �–4x � 2
–2y –3y 8
–8 –6 –14 �– 32 x –62 � + � – –33y –66x � = � 40y 142 �
� 4 � + � 4x � = �–4x � � –2x0– 3y –21–26x � = � 40y 142 �
8y –6y 8
–14 –14
�4 + 4x � = �–4x � –2 – 6x = 4 , –2x – 3y = 4y
6x = –6 7y = –2(–1)
2y 8 x = –1 7y = 2

4 + 4x = –4x , 2y = 8 y = 2
8x = –4 y = 4 7

x = – 1
2

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8 Tentukan sama ada pendaraban AB bagi setiap pasangan matriks A dan B yang berikut boleh didarab.
Determine whether the multiplication AB of each of the following pairs of matrices, A and B can be multiplied.

BAB 2CONTOH (a) A = [ –7 6 11 ] ; B = � 59 �

GLOBAL MEDIASTREET SDN. BHD.A = [ −1 4 2 ] ; B = [ 3 −4 ]

Penyelesaian Peringkat matriks A Peringkat matriks B
1 × 3 2×1
Peringkat matriks A
Order of matrix A Peringkat matriks B Maka, A dan B tidak boleh didarab.
1 × 3 Order of matrix B

1×2

Maka, A dan B tidak boleh didarab.
Thus, A and B cannot be multiplied.

(b) A = � 30 73 –25 � , B = � 64 –43 � (c) A = � –21 43 � , B = � 97 24 –61 �
8 9

Peringkat matriks A Peringkat matriks B Peringkat matriks A Peringkat matriks B
2 × 3 3×2 2 × 2 2×3

Maka, A dan B boleh didarab. Maka, A dan B boleh didarab.

9 Cari hasil darab bagi setiap matriks yang berikut. (a) [ –2 3]� 56 �
Find the product for each of the following matrices. = [−2(5) + 3(6)]
= [8]
CONTOH
(c) [ –1 22 3 ]� 74 –91 �
� 31 24 �� 42 13 � –13 2

Penyelesaian = [ –1(7) + 22(4) + 3(–13) –1(–1) + 22(9) + 3(2) ]
= [ 42 205 ]
� 13 24 �� 42 13 � = � 31((44)) + 2(2) 1(3) + 42((11)) �
+ 4(2) 3(3) +
= � 280 153 �

(b) � ––68 97 � � 03 –41 �

= � ––68((33)) + 7(0) –6(–1) + 97((44)) �
+ 9(0) –8(–1) +
= �– –2148 3444 �

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10 Cari nilai m dan n bagi setiap yang berikut.
Find the values of m and n for each of the following.

CONTOH GLOBAL MEDIASTREET SDN. BHD.
BAB 2
[ m −3 ] � 52 –1n � = [ 3 −6 ]

Penyelesaian

[ m(2) + (−3)(5) m(1) + (−3)(−n) ] = [ 3 –6 ]

2m – 15 = 3 , m + 3n = –6
2m = 18 9 + 3n = –6
m = 9
3n = –15
n = –5

(a) � ––24n �[ 5 –3 ] = � ––2300 13m8 �
� – –42(n5()5 ) ––24n((––33)) � = � ––3200 31m8 �

12 = 3m , –10n = –30
m = 4 n = 3

(b) � 52 �[ 3 –2 n ] = � –156 3m4– 1 5225 �
� 25((33)) 52((––22)) 52((nn)) � = � –156 3m4– 1 5225 �

–10 = 3m – 1 , 2n = 22
3m = –9 n = 11
m = –3

(c) � mn 4–1 � �– 13 25 �= � –– 84 238 �

� mn((––33)) + (–1)1 mn(5(5) )++(–41(2)()2) � = � –– 84 238 �
+ 4(1)

–3m – 1 = –4 , –3n + 4 = –8
3m = 3 3n = 12
m = 1 n = 4

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(d) � – m2 –53 � � ––74 � = � – –26n �
� –2m(–(–7)7)++(–53(–)(4–)4 ) � = � – –26n �
BAB 2
–7m – 20 = –6 , 14 + 12 = –2n
GLOBAL MEDIASTREET SDN. BHD. 7m = –14 2n = –26
m = –2 n = –13

(e) �–– 35 –21 � � –0m 62 � = � 160 ––42n6 �
�– 3–(5–(m–m) +) +(–21()0()0 ) –3–(56()6+) +(–21()2()2 )� = � 160 ––42n6 �

5m = 10 , –20 = –4n
m = 2 n = 5

11 Ungkapkan setiap yang berikut sebagai satu matriks tunggal.
Express each of the following as a single matrix.

CONTOH (a) � –51 32 � � 10 10 �
= � –51 23 �
� 52 –13 � � 10 01 �

Penyelesaian
� 52 –13 �� 10 01 � = � 25 –13 �

(b) � –87 –43 � � 01 01 � (c) � 10 10 � � 184 –126 �
= � –87 –43 � = � 184 –126 �

(d) � 10 10 � � –69 130 � (e) � 10 01 � � 51 ––155 �
= � –69 130 � = � 51 ––155 �

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12 Ungkapkan setiap yang berikut sebagai satu matriks tunggal.
Express each of the following as a single matrix.

CONTOH GLOBAL MEDIASTREET SDN. BHD.(a) 4� 10 10 � � 31 37 � + � –62 31 �
BAB 2 = 4� 13 37 � + � – 62 13 �
� 45 ––12 �� 10 01 � � –23 � = � 142 1228 � + �– 62 31 �
= � 1100 3113 �
Penyelesaian

� 54 ––12 �� –23 � = � 45((22)) + ((––12))((––33)) �1
+
= � 1161 �

(b) � 42 –62 �� 10 01 � – � 01 01 �� 30 ––51 � (c) � 10 01 � � 27 –45 � � 31 �– � 42 �
= � 24 –62 � – � 30 ––15 � = � 27 –45 � � 31 �– � 42 �
= �– 41 37 � = � –1193 � – � 42 �
= � –1155 �

13 Tentukan sama ada matriks yang berikut mempunyai matriks songsang atau tidak.
Determine whether the following matrices have inverse matrices.

CONTOH (a) � –25 –104 �
–5(–4) – 10(2) = 0
�– 42 –63 �
Penyelesaian Tidak mempunyai matriks songsang
4(–3) – 6(–2) = 0

Tidak mempunyai matriks songsang
Does not has an inverse matrix

(b) � 82 –164 � 1 1 �
8(–4) – 16(2) = –64 (≠ 0) (c) � 2
–2 –4
Mempunyai matriks songsang
1 (–4) – 1(–2) = 0
2

Tidak mempunyai matriks songsang

 Global Mediastreet Sdn. Bhd. (762284-U) 15

14 Cari matriks songsang bagi setiap matriks yang berikut dengan menggunakan rumus.

Find the inverse matrix for each of the following matrices by using formula.

CONTOH
BAB 2 (a) Q = � –25 –38 �
P = � 13 –24 �
GLOBAL MEDIASTREET SDN. BHD. Q–1 = 1 � –58 –23 �
Penyelesaian 2(–8) – 3(–5)

P–1 = 3(2) – 1 � –21 34 � = – � –58 –23 �
(–4)(1)

= 1 � –21 43 � = � –85 –32 �
10

= 1 2 �
� 5 5
– 1 3
10 10

(b) R = � 76 87 � (c) S = � 53 ––64 �

R–1 = 7(7) 1 8(6) � –76 –78 � S–1 = 3(–6) 1 �– –56 43 �
– – (–4)(5)

= � –76 –78 � = 1 � –– 56 43 �
2
= � – –352 223 �

(d) T = � 64 –32 � (e) U = � 32 –51 �

T–1 = 1 3(4) � –– 42 –63 � U–1 = 2(5) 1 �– 53 12 �
6(–2) – – (–1)(3)

= – 1 � ––24 –63 � = 1 � –53 21 �
24 13

= � 16112 –1841 � = � –151333 1
13 �
2
13

 Global Mediastreet Sdn. Bhd. (762284-U) 16

15 Selesaikan.
Solve.

CONTOH GLOBAL MEDIASTREET SDN. BHD. (a) Diberi � ––41 72 � K = � 10 01 �, cari matriks K.
BAB 2Given that M �–– 14 27 � K = � 10 10 �, find matrix K.
Diberi M � 34 79 � = � 01 01 �, cari matriks M.
Given that M � 43 79 � = � 10 10 �, find matrix M. 1 � 12 ––47 �
–
Penyelesaian K = –4(2) 7(–1)

M = 4(7) 1 9(3) � –73 –49 � = – 1 � 12 ––74 �
– 1

= 1 � –73 –49 � = � –– 12 47 �
1

= � –73 –49 �

(b) Matriks songsang bagi B = � 56 23 � ialah 1 � n2 –63 �, (c) Diberi matriks P = � –65 –34 � dan matriks
m Q = m �– –64 –n5 � dengan keadaan PQ = � 01 01 �, cari
cari nilai m dan n. nilai m dan n.
Given the matrix P = � – 65 –34 � and matrix Q = m �–– 64 –n5 �
The inverse matrix of B= � 56 32 � is 1 � n2 –63 � , find the
m

values of m and n.

B–1 = 1 � –25 –63 � such that PQ = � 01 10 �, find the values of m and n.
6(2) – 3(5)
1
= – 1 � –25 –63 � Q = –5(–4) – 3(6) � ––64 ––53 �
3
1 � ––64 ––35 �
m = –3, n = –5 = 2

m= 1 ,n= –3
2

 Global Mediastreet Sdn. Bhd. (762284-U) 17

16 Tuliskan persamaan linear serentak yang berikut dalam bentuk matriks.
Write the following simultaneous linear equations in the matrix form.

CONTOH
BAB 2 (a) –3x + 5y = 7
x + 5y = 15 2x – 4y = –9
GLOBAL MEDIASTREET SDN. BHD.3 �– 23 –54 � � xy �= � –79 �
5 x – y = 1

Penyelesaian
� 531 –51 � � xy �= � 115 �

(b) x – 5 = 6y (c) 13y – 6x = 7
12x – 11y = 9 3
2 y + 5 = 2x
x – 6y = 5
12x – 11y = 9 –6x + 13y = 7
�1 12 ––161 � � xy �= � 59 � 3
2x – 2 y = 5

– 6 13 � � xy �= � 57 �
� 2 3
– 2

17 Dengan menggunakan kaedah matriks, cari nilai x dan y yang memuaskan persamaan berikut.
By using the matrix method, find the values of x and y that satisfy the following equations.

CONTOH (a) � 36 –51 �� xy �= � –38 �

� 13 –21 � � xy �= � –65 � � xy � = 3(5) – 1 � –56 13 � � –38 �
(–1)(6)
Penyelesaian
1 � –56((33))++13((––88)) �
� xy � = 3(2) – 1 � –21 13 �� –65 � = 21
(–1)(1)
1 �– 74 2�
= 1 � –21((66))++13((––55)) � = 21
7
� – 72 1� = � 13 �
= 1 –2
7
= � –13 � 1
∴x = 3 , y = –2

∴ x = 1, y = –3

 Global Mediastreet Sdn. Bhd. (762284-U) 18

(b) � 42 –23 � � xy �= � –51 � (c) � – 25 –46 � � xy �= � –36 �
� xy � = 2(–6) –1(4)(–5) �– 56 –24 �� –36 �
� xy � = 4(2) – 1 GLOBAL MEDIASTREET SDN. BHD.� –22 34 � � –51 �
(–3)(2) BAB 2

= 1 � –22((55))++34((––11)) � = 1 � –65((––66)) + –2(43()3) �
14 8 +

= 1 �– 71 4� = 1 � – 2244 �
14 8
= � 12 � = � –33 �
–1
1 x = 3, y = –3
x = 2 ,y = –1

(d) � 33 40 2250 �� xy �= � 6625.5 � (e) � 2 –43 � � xy �= � –108 �
1 4
� xy � = 1 � – 2304 –3205 � � 6625.5 � 4
30(20) – (25)(34) � xy � = 1 4 � 4 32 � � –108 �
�– 3 –1
= – 1 � 2–03(642(6.52).5+) (+–2350)((6655)) � 2(4) – �(1)
250
3 � 4(–8) + 4 (10)
= – 1 � – –137755 � = 28 3 �
250 –1(–8) + 2(10)

= � 01..75 � = 3 �– 536 �
28 28
x = 1.5, y = 0.7
= � –32 �

x = –2, y = 3

 Global Mediastreet Sdn. Bhd. (762284-U) 19

18 Dengan menggunakan kaedah matriks, selesaikan persamaan linear serentak yang berikut.
By using the matrix method, solve the following simultaneous linear equations.

BAB 2CONTOH (a) 2x + 3y = 5
4x + 7y = 13
GLOBAL MEDIASTREET SDN. BHD.2x – 3y = –4
6x – 8y = –9 � 42 37 � � xy � = � 153 �

Penyelesaian � xy � = 1 � –74 –23 �� 153 �
–
� 62 ––38 � � xy � = � ––49 � 2(7) 3(4)

� xy � 1 � ––68 32 �� ––49 � = 1 � 7–(54)(5+) (+–32)((1133)) �
– (–3)(6) 2
=
2(–8) = 1 � –64 �
2
= 1 � ––86((––44)) + 32((––99)) �
2 + = � –32 �

= 1 � 56 � x = –2, y = 3
2

= � 52 �
3
5
x = 2 , y = 3

(b) 6x – 5y = –2 (c) 4x + 9y = 15
7x + 6y = 45 3x + 8y = 12

� 76 –65 � � xy � = � –452 � � 43 98 � � xy � = � 1152 �
� xy � = 1 � xy � = 1
6(6) – (–5)(7) � –67 65 � � –452 � 4(8) – 9(3) � –83 –49 � � 1152 �

= 1 � –67((––22))++56((4455)) � = 1 � 8–(31(51)5+) +(–49()1(21)2) �
71 5

= 1 �2 28143 � = 1 � 132 �
71 5
= � 15352 �
= � 34 �

x = 3, y = 4 12 3
5 5
x = , y =

 Global Mediastreet Sdn. Bhd. (762284-U) 20

Latihan Bestari 2.2

1 Diberi matriks Q = � ––38m –32 � tidak mempunyaiGLOBAL MEDIASTREET SDN. BHD. 3 Dengan menggunakan kaedah matriks, hitung nilai x
matriks songsang, cari nilai m. BAB 2dan y.
Given that Q = � ––38m –32 � has no inverse matrix, find the By using matrix method, find the values of x and y.
4x + 2y = –6
value of m. –6x – 5y = 3

2 Diberim �– 56 n � �– 25 4 � = � 01 01 �,carinilaimdann. 4 Di sebuah kedai buku, Ali membayar RM12 untuk 4
2 –6 batang pembaris dan 3 bilah gunting. Abu membayar
Given that m � – 56 n2 �� – 25 –46 � = � 10 01 �, find the values RM19 untuk 6 batang pembaris dan 5 bilah gunting.
Dengan menggunakan kaedah matriks, cari harga,
of m and n. dalam RM, bagi sebatang pembaris dan sebilah
gunting.
In a book shop, Ali paid RM12 for 4 rulers and 3 scissors.
Abu paid RM19 for 6 rulers and 5 scissors. By using matrix
method, find the price, in RM, of a ruler and a scissor.

SUDUT KBAT

1 Jumlah harga bagi 1 kg daging ayam dan 1 kg daging lembu ialah RM25. Beza harga antara 2 kg daging ayam
2 4
dengan 1 kg daging lembu ialah RM40. Dengan menggunakan kaedah matriks, hitung harga 1 kg daging ayam dan
1 kg daging lembu.

The total price of 1 kg chicken and 1 kg of beef is RM25. The difference in price between 2 kg of chicken and 1 kg of beef is RM40.
2 4

By using matrix method, calculate the price of 1 kg of chicken and the price of 1 kg of beef.

Anggap RMx ialah harga 1 kg daging ayam
Anggap RMy ialah harga 1 kg daging lembu

1 x + 1 y = 25
2 4
2x – y = 40
� 12 –141 �� xy � =� 4205 �
2 –141
2
� xy � = 1� – 1
�� 4250 �
1 (–1) – 1 (2) –2
= 2 4
+�–1214(�4(04)0) �
–1(25) +

–1 � –2(25)

= � 3350 �

Harga 1 kg daging ayam ialah RM35 manakala 1 kg daging lembu ialah RM30.

 Global Mediastreet Sdn. Bhd. (762284-U) 21

2 Jadual di bawah menunjukkan bilangan dan harga seunit pencetak dakwat yang dibeli oleh Ahmad.
The table below shows the number and the unit price of ink printer purchased by Ahmad.

BAB 2 Jenama Bilangan pencetak dakwat Harga seunit (RM)
Brand Number of ink printer Price per unit (RM)
GLOBAL MEDIASTREET SDN. BHD.
P x 12.00

Q y 16.00

Diberi jumlah pencetak dakwat yang dibeli ialah 9 dan jumlah harga ialah RM124. Dengan menggunakan kaedah
matriks, hitung nilai x dan y.
Given the total number of ink printer purchased is 9 and the total price is RM124. By using matrix method, calculate the value of x

and y.

x+y=9
12x + 16y = 124

� 112 116 � � xy � = � 1 924 �
� xy � = 1
1(16) – 1(12) � – 1162 –11� �1 924 �

= 1 � 1–61(29()9+) +(–11()1(2142)4) �
4
� 2106 �
= 1
4
= � 45 �
x = 5, y = 4

3 Terdapat dua jenis bungkusan alat tulis di sebuah kedai buku. Bungkusan A mempunyai 4 batang pensel dan
3 batang pen dengan harganya RM4. Bungkusan B mempunyai 8 batang pensel dan 11 batang pen dengan harganya
ialah RM11. Dengan menggunakan kaedah matriks, cari harga, dalam RM, sebatang pensel dan sebatang pen.
There are two types of packaging of stationery in a book shop. Packaging A has 4 pencils and 3 pens with the price of RM4. Packaging

B has 8 pencils and 11 pens with the price of RM11. By using matrix method, find the price of a pencil and a pen.

Anggap RMx ialah harga sebatang pensel
Anggap RMy ialah harga sebatang pen

4x + 3y = 4 = 1 � 1121 �
8x + 11y = 11 20

� 48 131 � � xy � = � 141 � = � 00..6550 �

� xy � = 1 � –118 –43�� 141 �
4(11) – 3(8)

= 1 � 11–(84()4+) +(–43()1(11)1) �
20

Harga sebatang pensel ialah RM0.55 manakala sebatang pen ialah RM0.60.

 Global Mediastreet Sdn. Bhd. (762284-U) 22

Praktis BAB 2
GLOBAL MEDIASTREET SDN. BHD.
BAB 2SOALAN OBJEKTIF

2 –4 10 6 Cari nilai n dalam persamaan matriks berikut.
1 � 3 � – 2� 10 � = C � –17 � Find the value of n in the following matrix equation.

–2 6 –14 � 35 –61 � –2� 14 –n2 � = � –35 –34 �
10 –10
D �–17 � A 1 C 5
A � –17 � 14 B 4 D 8
14
10 11 23 7 Jika [ 2 ] �5 3x]x� �3xx= �[3=3[]3, t3h]e,nmxa=ka x=
C � –14 26� C 2
B � 17 � If [ 2 5
–14 3 22
11 –23 A 1
3 5 –1 3 D �–14 22 � 3
2 4 �–4 6 � + � 2 –2� = –1 22 1
B 2 D 3
0 5 –1 2
11 23 8 Diberi/ Given

A �–14 22� � n4 � – 6 � –41 � = � 5 1m �
3 22
11 23

B � –14 22�
–1 22

3 � –15 � + � 32 � – 1 � –24 � = Cari nilai m dan n.
2 Find the values of m and n.
A � –24 � C � 20 �
D � –02 � A m = 4, n = 5
B � –22 � B m = –4, n = 5
C m = 4, n = –5
D m = –4, n = –5

4 � 42 –02 � � –03 –51 �= 9 Matriks P = � –21 –m12 � tidak mempunyai matriks
� – 61 2 –14 � ––162 –14 songsang, cari nilai m.
–2 3 The matrix P = � –21 –m12 � has no inverse matrix, find the
A � C �

B �– –162 –14 � D � ––162 –14 � value of m. C 6
2 –2 A –6 D 5
B –5

5 P + � –23 –15 � = � 63 –12 � 10 Antara berikut, yang manakah tidak mempunyai
Cari matriks P. matriks songsang?
Which of the following matrices does not have the inverse
Find the matrix P.
matrix?
A � 19 03 � C � –91 –03 �
A � –14 82 � C � ––64 2 �
� –91 3 D � –19 30 � 3
0 � � –33 2 D � –45 –32 �
B B 2 �

 Global Mediastreet Sdn. Bhd. (762284-U) 23

SOALAN SUBJEKTIF

1 (a) Diberi matriks P = �– 38m –23 � tidak mempunyai 2 (a) Cari matriks songsang bagi M = � 21 ––43 �.
matriks songsang, cari nilai m. Find the inverse matrix of M = � 12 ––43 �.
(b) Dengan menggunakan kaedah matriks, hitung
Given that matrix P = � –38m –23 � has no inverse
matrix, find the value of m. nilai x dan y yang memuaskan persamaan serentak
berikut.
(b) Dengan menggunakan kaedah matriks, hitung By using matrix method, calculate the values of x and y
nilai x dan y yang memuaskan persamaan serentak that satisfy the following simultaneous equation.
BAB 2 berikut.
x – 3y = –5
GLOBAL MEDIASTREET SDN. BHD. By using matrix method, calculate the values of x and y 2x – 4y = 8
that satisfy the following simultaneous equation.
(a) M–1 = 1(–4) 1 � ––42 31 �
3x + 2y = –10 – (–3)(2)
–4x – 6y = 40
= 1 � ––42 31 �
(a) 8(–3) – 2(–3m) = 0 2
–24 + 6m = 0
6m = 24 = –2 3
m = 4 � 21 �
–1
( b) �� xy –3 4� = –263 �(– � x6y )�– 1=2( �– –44100) �� –46 –32�� –4100 � 2
= –110 � –64(–(–1100))++(–3(24)0(4)0) �
= –110 � –8200 � (b) �� 21xy � = ––341 �(� – xy4 �) =–1( �– –835) �(2) � ––24 31 �� –85 �
= � –28 �
x = 2, y = –8 = 1 � ––42((––55))++(13()8()8) �
2
1 �14 84 �
= 2

= � 292 �
x = 22, y = 9

 Global Mediastreet Sdn. Bhd. (762284-U) 24

3 Ami dan Diana pergi ke pasar untuk membeli ketam 4 Sebuah pasar raya menjual dua jenis susu tepung, M
dan udang. Ami membeli 10 kg ketam dan 10 kg dan N yang terdiri daripada tin besar dan tin kecil.
udang dengan harga RM720. Diana membeli 5 kg Harga jualan setin susu tepung dalam tin besar bagi
ketam dan 9 kg udang dengan harga RM540. Dengan kedua-dua jenama ialah RMx. Harga jualan setin susu
menggunakan kaedah matriks, cari harga, dalam RM, tepung dalam tin kecil bagi kedua-dua jenama ialah
bagi 1 kg ketam dan 1 kg udang. RMy. Jumlah tin susu tepung yang dijual pada bulan
Ami and Diana went to the market to buy crabs and prawns. Ogos ditunjukkan dalam jadual di bawah.
Ami buys 10 kg crabs and 10 kg prawns for RM720. Diana A supermarket sells two types of powdered milk, M and N in
buys 5 kg crabs and 9 kg prawns for RM540. By using matrix large cans and small cans. The price of a large can for both
method, find the price, in RM, of 1 kg crab and 1 kg prawn. brands is RMx. The price of a small can for both brands is
RMy. The total number of powdered milk sold in August are
shown in table below.
Anggap x ialah harga 1 kg ketamGLOBAL MEDIASTREET SDN. BHD.
Anggap y ialah harga 1 kg udang BAB 2Jenama Tin besar Tin kecil
Brand Large can Small can
10x + 10x = 720
5x + 9y = 540 m 30 25
� � 1 xy50 � =19100 � ( �9 xy) �–1 =10� ( 57524)00� � –95 –1100 � � 572400 �
n 34 20

= 1 � 9–(752(702)0+) (+–1100()(554400)) � Jumlah jualan bagi jenama m ialah RM6 250 dan
40 jenama n ialah RM6 500. Hitung harga, dalam RM,
setin susu tepung dalam tin besar dan tin kecil.
= 1 � 11 088000 � The total sales for brand m is RM6 250 and brand n is
40 RM6 500. Calculate the price, in RM, of a large can and a
small can of powdered milk.

= � 4257 � 30x + 25y = 6 250
34x + 20y = 6 500
Harga 1 kg ketam ialah RM27 manakala 1 kg udang � �3 3 xy40 � =22053 0� ( �2 xy0 �) –1=2� 566 (5320450)0� � –2304 –3205 �� 6 6 520500 �
ialah RM45.

= – 1 �–2 304(6(6225500))++(–302(56)(560500)0) �
250

= – 1 � –– 1377 550000 �
250

= � 17500 �

Harga setin susu tepung dalam tin besar ialah RM150
manakala tin kecil ialah RM70.


 Global Mediastreet Sdn. Bhd. (762284-U) 25

5 John mempunyai x keping duit syiling dan Arif 6 Kelas 5A terdiri daripada 30 orang murid. Bilangan
mempunyai y keping duit syiling. Jika John murid lelaki adalah 12 orang kurang daripada dua kali
memberikan 20 keping duit syiling kepada Arif, maka bilangan murid perempuan. Hitung bilangan murid
bilangan duit syiling mereka menjadi sama banyak. lelaki dan bilangan murid perempuan di dalam kelas
Jika Arif memberikan 50 keping duit syiling kepada itu.
John, nisbah duit syiling John kepada Arif ialah Class 5A has a total of 30 pupils. The number of boys is 12 less
3 : 2. Cari nilai x dan y. than two times the number of girls. Calculate the number of
John has x coins and Arif has y coins. If John gives 20 coins to boys and the number of girls.
Arif, then both of them have the same number of coins. If Arif
gives 50 coins to John, the ratio of John’s coins to Arif ’s coins
is 3 : 2. Find the value of x and y.
BAB 2 Anggap x ialah bilangan murid lelaki
Anggap y ialah bilangan murid perempuan
GLOBAL MEDIASTREET SDN. BHD. x + y = 30 … (1)
x – 20 = y + 20 x + 12 = 2y
x – y = 40 … (1) x – 2y = –12 … (2)
� � 11xy � – =12 �1 �( xy – 2� )=1– �1 –3(1012 )� � ––21 –11 � � –3102 �
x + 50 = 3
y – 50 2

2x + 100 = 3y – 150 = – 1 �– –21(3(300))++(–1(1–)(1–21)2)�
2x – 3y = –250 … (2) 3
�� 12xy � =––131 �(� – xy3 �) =–1( �–– 4120)5(0 2�) � ––23 11 � �– 42050 � 1 � ––4428�
= – 3

= – � ––23((4400)) + 11((––225500)) � = � 1 164 �
+
= –� ––337300 � Bilangan murid lelaki ialah 16 orang manakala murid
perempuan ialah 14 orang.

= �3 37300 �

x = 370, y = 330

 Global Mediastreet Sdn. Bhd. (762284-U) 26

MODULE MODULE PERFECT MATEMATIK TINGKATAN 5

Siri Module PERFECT dihasilkan sesuai untuk penggunaan guru dan murid di dalam kelas
atau di rumah. Kandungan buku ini adalah mesra murid dan mudah untuk digunakan. Latihan
dan aktiviti yang ditulis adalah berdasarkan buku teks dan Dokumen Standard Kurikulum dan
Pentaksiran (DSKP) terkini.

Latihan dan aktiviti yang pelbagai bentuk membolehkan murid menguasai sepenuhnya
kemahiran dalam topik yang dipelajari. Selain itu, penerapan soalan berunsurkan Kemahiran
Berfikir Aras Tinggi (KBAT), Pembelajaran Abad ke-21, peta pemikiran dan STEM akan dapat
meningkatkan daya pemikiran murid secara tidak langsung. Terdapat juga bahan-bahan yang
dimuatkan dalam kod QR bertujuan sebagai rujukan tambahan untuk murid..
GLOBAL MEDIASTREET SDN. BHD.
Judul-judul dalam siri ini

Subjek TINGKATAN
1 2 3 45
Bahasa Melayu
Bahasa Inggeris
Matematik
Sains
Sejarah
Geografi

Semenanjung Malaysia (WM) : RM5.50

Sabah & Sarawak (EM) : RM5.90

ISBN 978-967-0057-66-8

15, Jalan 9/152, Taman Perindustrian OUG, 9 789670 057668
Batu 6 1/2,Jalan Puchong, 58200 Kuala Lumpur.
Tel: +603 7783 6309 Faks : +603 7783 9089
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