24[A Math CD]

1 © Penerbitan Pelangi Sdn. Bhd.

Paper 1

1. The matrix is of 3 rows and 1 column.Hence, m = 3 and n = 1

Answer: B

2. P23 referring to the element in the 2nd row and the 3rd column.Hence, P23 = 6

Answer: D

3. Number of columns = 3

Answer: D

4. 3 × 2 matrix is a matrix which has 3 rows and 2 columns.

Answer: C

5. Compare the two matrices.p = 3, q = −5

Answer: A

6. Compare the two matrices.m = −2, n = 6

Answer: C

7. 5–3

+ 4–1

= 5 + 4–3 – 1

= 9– 4

Answer: A

8. 1 02 –3 +

3 –5–1 0

= 1 + 3 0 + (–5)

2 + (–1) –3 + 0

= 4 –51 –3

Answer: D

9. (5 0 –3) − (–1 3 2)= (5 – (–1) 0 – 3 –3 – 2)= (6 –3 –5)

Answer: B

10. –24

− 51

+ 7– 6

= –2 – 5 + 74 – 1 + (– 6)

= 0–3

Answer: C

11. 2 1 30 4

= 2 × 1 2 × 32 × 0 2 × 4

= 2 60 8

Answer: C

12. 3 –1 42 1

= 3 × (–1) 3 × 4

3 × 2 3 × 1

= –3 126 3

Answer: B

CHAPTER

24 MatricesCHAPTER

2

Mathematics SPM Chapter 24

© Penerbitan Pelangi Sdn. Bhd.

13. −2 0 21 –1

= –2 × 0 –2 × 2–2 × 1 –2 × (–1)

= 0 – 4–2 2

Answer: A

14. 1—2

4 02 –2

=

1—2

× 4 1—2

× 0

1—2

× 2 1—2

× (–2)

= 2 01 –1

Answer: D

15. (1 2) 03

= (1 × 0 + 2 × 3)= (6)

Answer: A

16. 13

(2 0)

= 1 × 2 1 × 03 × 2 3 × 0

= 2 06 0

Answer: D

17. 1 20 1

0 12 1

= 1 × 0 + 2 × 2 1 × 1 + 2 × 10 × 0 + 1 × 2 0 × 1 + 1 × 1

= 4 32 1

Answer: A

18. 2 10 3

12

= 2 × 1 + 1 × 20 × 1 + 3 × 2

= 46

Answer: D

Paper 2

1. (a) Not an identity matrix

(b) An identity matrix

(c) An identity matrix

(d) Not an identity matrix

2. (a) 2 34 –5

1 00 1 =

2 34 –5

(b) 1 00 1

4 –20 1 =

4 –20 1

3. (a) Inverse matrix of 3 51 2

= 1––––––––––––3 × 2 − 5 × 1

2 –5–1 3

= 1–––––6 – 5

2 –5–1 3

= 2 –5–1 3

(b) Inverse matrix of 5 23 2

= 1––––––––––––5 × 2 − 2 × 3

2 –2–3 5

= 1––––––10 – 6

2 –2–3 5

= 1—4

2 –2–3 5

4. (a) Inverse matrix of 2 – 42 –3

= 1––––––––––––––––2 × (–3) – (– 4) × 2

–3 4–2 2

= 1––––––– 6 + 8

–3 4–2 2

= 1—2

–3 4–2 2

(b) Inverse matrix of 5 –24 –2

= 1–––––––––––––––––5 × (–2) – (–2) × 4

–2 2–4 5

= 1––––––––10 + 8

–2 2–4 5

= − 1—2

–2 2–4 5

3



5. (a) Determinant of matrix 6 43 2 = 6 × 2 − 4 × 3

= 0

Hence, 6 43 2 has no inverse.

(b) Determinant of matrix 3 –16 –2

= 3 × (−2) − (−1) × 6 = 0

Hence, 3 –16 –2 has no inverse.

6. (a) 2 51 3

3 –5–1 2

= (2)(3) + (5)(–1) (2)(–5) + (5)(2)(1)(3) + (3)(–1) (1)(–5) + (3)(2)

= 1 00 1

(b) 2 –13 –1

–1 1–3 2

= (2)(–1) + (–1)(–3) (2)(1) + (–1)(2)(3)(–1) + (–1)(–3) (3)(1) + (–1)(2)

= 1 00 1

7. 4 31 1

xy

= 113

xy

= 1 –3–1 4

113

= (1)(11) + (–3)(3)(–1)(11) + (4)(3)

= 21

Hence, x = 2 and y = 1.

8. 4 –15 –1

mn

= 1316

mn

= –1 1–5 4

1316

= (–1)(13) + (1)(16)(–5)(13) + (4)(16)

= 3–1

Hence, m = 3 and n = −1.

Paper 1

1. 48

− –32

+ 12

64

= 48

− –32

+ 32

= 4 + 3 + 38 – 2 + 2

= 108

Answer: B

2. (4 x) x–6

= (16)

(4(x) + x(–6)) = (16) 4x − 6x = 16 −2x = 16

x = 16––––2 = −8

Answer: D

3. 2 15 4

3–2

= (2)(3) + (1)(–2)(5)(3) + (4)(–2)

= 6 – 215 – 8

= 47

Answer: A

4. 13 –1

20 45 2

–23

= 1(3)(–2) + (–1)(3)

2(0)(–2) + (4)(3)(5)(–2) + (2)(3)

= 1–9

212–4

Answer: B

4



5. (3 –5 0) – (2 4 –3) + 2(–1 6 4)= (3 –5 0) – (2 4 –3) + (–2 12 8)= (3 – 2 + (–2) –5 – 4 + 12 0 – (–3) + 8)= (–1 3 11)Answer: A

6. (–2 0 1) 1 03 5

–1 2= ((–2)(1) + (0)(3) + (1)(–1) (–2)(0) + (0)(5) + (1)(2))= (–3 2)

Answer: D

7. 1 –34 3

2 50 –2

= (1)(2) + (–3)(0) (1)(5) + (–3)(–2)(4)(2) + (3)(0) (4)(5) + (3)(–2)

= 2 118 14

Answer: B

8. 2 4 53 2 −

–1 60 3

= 8 106 4 −

–1 60 3

= 8 + 1 10 – 66 – 0 4 – 3

= 9 46 1

Answer: B

9. 12

4 –26 k

– 1 3–7 –1

= 1 –410 5

2 –1

3 12

k – 1 3

–7 –1 = 1 –4

10 5

2 – 1 –1 – 3

3 + 7 12

k + 1 = 1 –4

10 5

Hence, 12

k + 1 = 5

12

k = 4

k = 8

Answer: C

10. 4–2 3 11 5 –3 −

2 –1 43 2 1

= –8 12 44 20 –12 −

2 –1 43 2 1

= –8 – 2 12 – (–1) 4 – 44 – 3 20 – 2 –12 – 1

= –10 13 0

1 18 –13

Answer: A

11. 3 k5 –1 − 3

k 2–1 0 =

–6 –38 –1

3 k5 –1 −

3k 6–3 0 =

–6 –38 –1

3 – 3k k – 65 + 3 –1 – 0 =

–6 –38 –1

Hence, k − 6 = −3 k = −3 + 6 = 3

Answer: C

12. m4

– 2 3–2

= –8n

m4

– 6–4

= –8n

m – 64 – (–4)

= –8n

Hence, m – 6 = –8 m = –2 n = 8

Answer: B

13. 4 –17 2 + 3

2 –21 0 −

– 4 5–2 1

= 4 –17 2 +

6 –63 0 −

– 4 5–2 1

= 4 + 6 + 4 –1 – 6 – 57 + 3 + 2 2 + 0 – 1

= 14 –1212 1

Answer: C

5



14. (1 p) p3

= (2)

((1)(p) + (p)(3)) = (2)

Hence, p + 3p = 2 4p = 2

p = 2—4

= 1—2

Answer: A

Paper 2

1. (a) F 2 –15 – 4 =

1 00 1

F is the inverse matrix of 2 –15 – 4 .

F = 1–––––––––––––––––2 × (– 4) – (–1) × 5

– 4 1–5 2

= − 1—3

– 4 1–5 2

(b) 2 –15 – 4

pq

= 1034

pq

= − 1—3

– 4 1–5 2

1034

= − 1—3

– 40 + 34–50 + 68

= − 1—3

–618

= 2–6

Hence, p = 2 and q = −6.


= 1––––––––––––3 × 5 – 2 × 4

5 –2

– 4 3

= 1—7

5 –2– 4 3

(b) 3 24 5

mn

= 417

mn

= 1—7

5 –2

– 4 3417

= 1—7

20 – 34–16 + 51

= 1—7

–1435

= –25

Hence, m = −2 and n = 5.

3. (a) Inverse matrix of 3 –82 –7

= 1–––––––––––––––––3 × (–7) – (–8) × 2

–7 8–2 3

= − 1—5

–7 8–2 3

Hence, p = − 1—5

and t = 8.

(b) 3 –82 –7

xy

= 6555

xy

= − 1—5

–7 8–2 3

6555

= − 1—5

– 455 + 440–130 + 165

= − 1—5

–1535

= 3–7

Hence, x = 3 and y = −7.

4. (a) Inverse matrix of 3 –52 1

= 1––––––––––––––3 × 1 – (–5) × 2

1 5–2 3

= 1–––13 1 5

–2 3

Hence, p = 13 and q = 5.

(b) 3 –52 1

xy

= 512

xy

= 1–––13

1 5–2 3

512

= 1–––13 5 + 60

–10 + 36

= 1–––13 65

26

= 52

Hence, x = 5 and y = 2.

6



5. (a) E = 2 –51 4

F = Inverse matrix of E

= 1––––––––––––––2 × 4 – (–5) × 1

4 5–1 2

= 1–––13

4 5–1 2

Compare with F = p 4 5q 2 .

Hence, p = 1–––13 and q = −1.

(b) 2 –51 4

mn = 16

–5

mn = 1–––13

4 5–1 2

16–5

= 1–––13 64 – 25

–16 – 10

= 1–––13 39

–26

= 3–2


6. (a) Inverse matrix of 2 –14 3

= 1––––––––––––––2 × 3 – (–1) × 4

3 1–4 2

= 1–––10 3 1

–4 2

Compare with m 3 1–4 n

.

Hence, m = 1–––10 and n = 2.

(b) 2 –14 3

xy = 7

–16

xy = 1–––10

3 1–4 2

7–16

= 1–––10 21 + (–16)

–28 + (–32)

= 1–––10 5

–60

= 1—2–6

Hence, x = 1—2

and y = −6.


= 1––––––––––––––––4 × (–1) – (–3) × 2

–1 3–2 4

= 1—2

–1 3–2 4

= – 1—2

3—2

–1 2

Compare with k 3—2

–1 2.

Hence, k = − 1—2

.

(b) 4 –32 –1

pq = 27

11

pq = 1—

2 –1 3–2 4

2711

= 1—2 –27 + 33

–54 + 44

= 1—2 6

–10

= 3–5

Hence, p = 3 and q = −5.

Paper 1

1. 2 –16 3 +

–3 40 –1 −

–2 5–4 7

= 2 – 3 + 2 –1 + 4 – 56 + 0 + 4 3 – 1 – 7

= 1 –210 –5

Answer: C

2. M + –32

= 06

M = 06

− –32

= 0 + 36 – 2

= 34

Answer: B

7



3. –2 05 4 − N =

5 2–3 0

N = –2 05 4 −

5 2–3 0

= –2 – 5 0 – 25 + 3 4 – 0

= –7 –28 4

Answer: C

4. T − 2 –31 0 =

4 02 –1

T = 4 02 –1 +

2 –31 0

= 4 + 2 0 – 32 + 1 –1 + 0

= 6 –33 –1

Answer: A

5. (3 2) − (–1 5) + 1—2 (4 –2)

= (3 2) − (–1 5) + (2 –1)= (3 + 1 + 2 2 – 5 – 1)= (6 – 4)

Answer: D

6. –14

− 1—2 8

–2 + 3

–5

= –14

− 4–1

+ 3–5

= –1 – 4 + 34 + 1 – 5

= –20

Answer: A

7. 3 –25 1 − 2

0 –3–4 1

= 3 –25 1 −

0 –6–8 2

= 3 – 0 –2 + 65 + 8 1 – 2

= 3 413 –1

Answer: A

8. 2 –20 3 + 2

n 02 3 =

6 –24 9

2 –20 3 +

2n 04 6 =

6 –24 9

Hence, 2 + 2n = 6 2n = 6 − 2 n = 4—

2 = 2

Answer: B

9. 2 10 –4

–3 52 –1

= (2)(–3) + (1)(2) (2)(5) + (1)(–1)(0)(–3) + (–4)(2) (0)(5) + (–4)(–1)

= –4 9–8 4

Answer: D

10. –3 01 2

4–1

= (–3)(4) + (0)(–1)(1)(4) + (2)(–1)

= –122

Answer: C

11. –23

(–1 – 4)

= (–2)(–1) (–2)(– 4)(3)(–1) (3)(– 4)

= 2 8–3 –12

Answer: B

12. 2 m0 –1

1 00 2

= (2)(1) + (m)(0) (2)(0) + (m)(2)(0)(1) + (–1)(0) (0)(0) + (–1)(2)

= 2 2m0 –2

Answer: B

13. (2 –1) 34

= ((2)(3) + (–1)(4))= (2)

Answer: A

8



14. –32

(1 –2)

= (–3)(1) (–3)(–2)(2)(1) (2)(–2)

= –3 62 –4

Answer: D

15. (h 2)3 0–1 4 = (2h 8)

((h)(3) + (2)(–1) (h)(0) + (2)(4)) = (2h 8)

Hence, 3h − 2 = 2h 3h − 2h = 2 h = 2

Answer: A

16. k – 23

(0 1) = 0 –50 3

(k – 2) × 0 (k – 2) × 10 3

= 0 –50 3

Hence, k − 2 = −5 k = −5 + 2 = −3

Answer: D

17. (p 5) –12

= (6)

((p)(–1) + (5)(2)) = (6)

Hence, −p + 10 = 6 p = 10 − 6 = 4

Answer: B

18. 2m

(–3 1) = –6 26 –2

(2)(–3) (2)(1)(m)(–3) (m)(1)

= –6 26 –2

Hence, m = −2

Answer: C

19. (p 2p) 2–3

= (20)

((p)(2) + (2p)(–3)) = (20)Hence, 2p − 6p = 20 −4p = 20 p = 20–––

– 4 = −5Answer: B

Paper 2

1. (a) E = 4 35 4

F is the inverse matrix of E.

F = 1––––––––––––4 × 4 – 3 × 5

4 –3–5 4

= 1—1

4 –3–5 4

= 4 –3–5 4

(b) 4 35 4

mn

= 46

mn

= 4 –3–5 4

46

= 16 – 18–20 + 24

= –24


2. (a) E = 4 21 1

F is the inverse matrix of E.

F = 1––––––––––––4 × 1 – 2 × 1

1 –2–1 4

= 1—2

1 –2–1 4

Compare with 1—q

1 –2p 4 .

Hence, p = −1 and q = 2.

(b) 4 21 1

xy

= –10–2

xy

= 1—2

1 –2–1 4

–10–2

= 1—2 –10 + 4

10 – 8

= 1—2 –6

2

= –31

Hence, x = −3 and y = 1.

9



3. (a) G = 3 –41 2

Inverse matrix of G

= 1––––––––––––––3 × 2 – (–4) × 1

2 4–1 3

= 1–––10

2 4–1 3

(b) 3 –41 2

xy

= 74

xy

= 1–––10 2 4–1 3

74

= 1–––10 14 + 16–7 + 12

= 1–––10 305

= 31—2

Hence, x = 3 and y = 1—2 .

4. (a) P = 4 31 h

Determinant of P = 4 × h – 3 × 1 = 4h − 3

P has no inverse. Hence, 4h − 3 = 0 h = 3—4

(b) (i) P = 4 31 2

Inverse matrix of P

= 1––––––––––––4 × 2 – 3 × 1

2 –3–1 4

= 1—5 2 –3–1 4

(ii) 4 31 2

ef

= 214

ef

= 1—5 2 –3–1 4

214

= 1—5 42 – 12–21 + 16

= 1—5 30–5

= 6–1

Hence, e = 6 and f = −1.


= 1––––––––––––––––4 × (–4) – (–5) × 3

–4 5–3 4

= 1––––1

–4 5–3 4

= 4 –53 –4

Compare with 4 hk –4 .

Hence, h = −5 and k = 3.

(b) 4 –53 –4

xy

= 3225

xy

= 4 –53 –4

3225

= 128 – 12596 – 100

= 3–4


6. (a) F = m 110 n

Inverse matrix of F

= 1––––––––mn – 10

n –1–10 m

Compare with 1–––10 n –1

–10 5.

Hence, m = 5 and mn − 10 = 10 5n = 10 + 10

n = 20–––5 = 4

(b) F xy

= 46

5 110 4

xy

= 46

xy

= 1–––10 4 –1

–10 546

= 1–––10 16 – 6

– 40 + 30

= 1–––10 10

–10

= 1–1


10



7. (a) F = n 34 9

Inverse matrix of F

= 1––––––––9n – 12

9 –3–4 n

Compare with 1—k

9 –3–4 1 .

Hence, n = 1 and 9n − 12 = k 9(1) − 12 = k k = −3

(b) F xy

= 1–2

1 34 9

xy

= 1–2

xy

= 1––––3

9 –3–4 1

1–2

= − 1––3 9 + 6

– 4 – 2

= − 1––3 15

–6

= –52


8. (a) Inverse matrix of –1 –16 8

= 1––––––––––––––––1 × 8 – (–1) × 6

8 1–6 –1

= – 1––2

8 1–6 –1

(b) –1 –16 8

hk

= –215

hk

= − 1––2

8 1–6 –1

–215

= − 1––2 –16 + 15

12 – 15

= − 1––2 –1

–3

= 1—23—2

Hence, h = 1––2 and k = 3––

2 .

9. (a) 1 –3–2 8

8 32 1

= 8 – 6 3 – 3–16 + 16 –6 + 8

= 2 00 2

(b) 8 32 1

hk

= 72

1 –3–2 8

8 32 1

hk

= 1 –3–2 8

72

2 00 2

hk

= 7 – 6–14 + 16

2h2k

= 12

Hence, 2h = 1 and 2k = 2 h = 1—

2 k = 1

10. (a) 2 –28 –12

6 –14 –1

= 12 – 8 –2 + 248 – 48 –8 + 12

= 4 00 4

(b) 6 –14 –1

mn

= 911

2 –28 –12

6 –14 –1

mn

= 2 –28 –12

911

4 00 4

mn

= 18 – 2272 – 132

4m4n

= – 4– 60

Hence, 4m = −4 and 4n = −60 m = −1 n = −15

11. (a) 3 14 –1

F = 3 14 –1

Hence, F = 1 00 1

11



(b) Inverse matrix of 3 14 –1

= 1––––––––––––––3 × (–1) – 1 × 4

–1 –1–4 3

= − 1––7

–1 –1–4 3

3 14 –1

de

= 1124

de

= − 1––7

–1 –1–4 3

1124

= − 1––7 –11 – 24

–44 + 72

= − 1––7 –35

28

= 5–4

Hence, d = 5 and e = −4.

12. (a) M = 4 –68 k

Determinant of M = 4 × k − (−6) × 8 = 4k + 48

When M has no inverse, 4k + 48 = 0 4k = − 48

k = – 48––––4 = −12

(b) N = 9 24 1

Inverse matrix of N

= 1––––––––––––9 × 1 – 2 × 4

1 –2–4 9

= 1 –2–4 9

(c) 9 24 1

mn

= 31

mn

= 1 –2–4 9

31

= 3 – 2–12 + 9

= 1–3


13. (a) AH = 1 00 1

A 3 –24 1

= 1 00 1

A is the inverse matrix of 3 –24 1

.

A = 1––––––––––––––3 × 1 – (–2) × 4

1 2–4 3

= 1–––11 1 2

–4 3

(b) H xy

= 12

3 –24 1

xy

= 12

xy

= 1–––11 1 2

–4 312

= 1–––11 1 + 4

– 4 + 6

= 1–––11 5

2

= 5––112––11

Hence, x = 5–––11 and y = 2–––11

.

14. E = m 32 1

(a) Determinant of E = m × 1 − 3 × 2 = m − 6

When E has no inverse, then m − 6 = 0 m = 6(b) E = 8 3

2 1 Inverse matrix of E

= 1––––––––––––8 × 1 – 3 × 2

1 –3–2 8

= 1—2 1 –3

–2 8

12



(c) 8 32 1

xy

= 42

xy

= 1—2 1 –3

–2 842

= 1—2 4 – 6

–8 + 16

= 1—2 –2

8

= –14



= 1––––––––––––3 × 4 – 1 × 8

4 –1–8 3

= 1—4 4 –1

–8 3

(b) 3 18 4

mn

= –34

mn

= 1—4 4 –1

–8 3–34

= 1—4 –12 – 4

24 + 12

= 1—4 –16

36

= – 49


16. (a) Inverse matrix of 3 –1

2 1—3

= 1–––––––––––––––3 × 1—

3 – (–1) × 2

1—3 1

–2 3

= 1—3

1—3 1

–2 3

(b) 3 –1

2 1—3

pq

= 218

pq

= 1—3

1—3 1

–2 3

218

= 1—3 7 + 8

– 42 + 24

= 1—3 15

–18

= 5– 6

Hence, p = 5 and q = − 6.

Documents

24[A Math CD]