4.10 SPM Practice (Long Questions)

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Question 5:
(a) Given  1 14 ( 2 s −4 t )( t −1 4 2 )=( 1 0 0 1 ), find the value of s and of t.
(b) Write the following simultaneous linear equations as matrix form:
3x – 2y = 5
9x + y = 1
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) 1 14 ( 2 s −4 t )( t −1 4 2 )=( 1 0 0 1 ) 1 14 ( 2t+4s −2+2s −4t+4t 4+2t )=( 1 0 0 1 ) −2+2s 14 =0   2s=2      s=1 4+2t 14 =1 4+2t=14 2t=10 t=5

(b) ( 3 −2 9 1 )( x y )=( 5 1 )   ( x y )= 1 21 ( 1 2 −9 3 )( 5 1 )   ( x y )= 1 21 ( ( 1 )( 5 )+( 2 )( 1 ) ( −9 )( 5 )+( 3 )( 1 ) )   ( x y )= 1 21 ( 7 −42 )   ( x y )=( 1 3 −2 ) ∴x= 1 3 ,  y=−2



Question 6:
It is given that matrix P=( 6 −3 −5 2 ) and matrix Q= 1 m ( 2 3 5 n )  such that PQ=( 1 0 0 1 ).
(a) Find the value of m and of n.
(b) Write the following simultaneous linear equations as matrix form:
6x – 3y = –24
–5x + 2y = 18
Hence, using matrix method, calculate the value of x and y.

Solution:
(a) m=6( 2 )−( −3 )( −5 )   =12−15 m=−3 n=6

(b) ( 6 −3 −5 2 )( x y )=( −24 18 )   ( x y )= 1 12−15 ( 2 3 5 6 )( −24 18 )   ( x y )= 1 −3 ( ( 2 )( −24 )+( 3 )( 18 ) ( 5 )( −24 )+( 6 )( 18 ) )   ( x y )= 1 −3 ( 6 −12 )   ( x y )=( −2 4 ) ∴x=−2,  y=4

4.10 SPM Practice (Long Questions)

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Question 3:
It is given that Q  ( 3 2 6 5 ) = ( 1 0 0 1 ) , where Q is a 2 x 2 matrix.
(a) Find Q.
(b) Write the following simultaneous linear equations as matrix equation:
3u + 2v = 5
6u + 5v = 2
Hence, using matrix method, calculate the value of u and v.

Solution:
(a) 
Q = ( 3 2 6 5 ) − 1 Q = 1 3 ( 5 ) − 2 ( 6 ) ( 5 − 2 − 6 3 ) Q = 1 3 ( 5 − 2 − 6 3 ) Q = ( 5 3 − 2 3 − 2 1 )

(b)
( 3 2 6 5 ) ( u v ) = ( 5 2 ) ( u v ) = 1 3 ( 5 − 2 − 6 3 ) ( 5 2 ) ( u v ) = 1 3 ( ( 5 ) ( 5 ) + ( − 2 ) ( 2 ) ( − 6 ) ( 5 ) + ( 3 ) ( 2 ) ) ( u v ) = 1 3 ( 21 − 24 ) ( u v ) = ( 7 − 8 ) ∴ u = 7 , v = − 8


Question 4:
It is given that Q ( 3 − 2 5 − 4 ) = ( 1 0 0 1 )  , where Q is a 2 × 2 matrix.
(a) Find the matrix Q.
(b) Write the following simultaneous linear equations as matrix equation:
3x – 2y = 7
5x – 4y = 9
Hence, using matrix method, calculate the value of x and y.

Solution:
(a)
Q = 1 3 ( − 4 ) − ( 5 ) ( − 2 ) ( − 4 2 − 5 3 ) = − 1 2 ( − 4 2 − 5 3 ) = ( 2 − 1 5 2 − 3 2 )

(b)
( 3 − 2 5 − 4 ) ( x y ) = ( 7 9 ) ( x y ) = − 1 2 ( − 4 2 − 5 3 ) ( 7 9 ) ( x y ) = − 1 2 ( ( − 4 ) ( 7 ) + ( 2 ) ( 9 ) ( − 5 ) ( 7 ) + ( 3 ) ( 9 ) ) ( x y ) = − 1 2 ( − 10 − 8 ) ( x y ) = ( 5 4 ) ∴ x = 5 , y = 4

4.9 SPM Practice (Short Questions)

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4.9.2 Matrices, SPM Practice (Short Questions)
Question 5:
Given that ( 3   x ) ( x − 1 ) = ( 18 ) ,  find the value of x.

Solution:
( 3   x ) ( x − 1 ) = ( 18 )
[3 × x + x (–1)] = (18)
3x – x = 18
2x = 18
x = 9



Question 6:
( 3 4 − 2 3 ) ( 5 − 2 ) =

Solution:
( 3 4 − 2 3 ) ( 5 − 2 ) = ( ( 3 ) ( 5 ) + ( 4 ) ( − 2 ) ( − 2 ) ( 5 ) + ( 3 ) ( − 2 ) ) = ( 15 − 8 − 10 − 6 ) = ( 7 − 16 )



Question 7:
( 2 4 − 3 4 0 1 ) ( 1 − 3 ) =

Solution:
Order of the product of the two matrices
= ( 3 × 2 ) ( 2 × 1 ) = ( 3 × 1 )

( 2 4 − 3 4 0 1 ) ( 1 − 3 ) = ( 2 ( 1 ) + 4 ( − 3 ) ( − 3 ) ( 1 ) + 0 ( − 3 ) ( 4 ) ( 1 ) + 1 ( − 3 ) ) = ( 2 − 12 − 3 + 0 4 − 3 ) = ( − 10 − 3 1 )



Question 8:
( 1  −1   2 )( 5 1 −3 2 0 4 )=

Solution:
Order of the product of the two matrices
= ( 1 × 3 ) ( 3 × 2 ) = ( 1 × 2 )

( 1  −1   2 )( 5 1 −3 2 0 4 )=
= (1×5 + (–1)(–3) + (2)(2)  1×1 + (–1)(0) + (2)(4))
= (5 + 3 + 4   1 + 0 + 8)
= (12   9)

4.5 Multiplication of Two Matrices (Sample Question 2)

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Example 2:
Find the values of m and n in each of the following matrix equations.
( a ) ( 3 m ) ( 1       n ) = ( 3 12 − 2 − 8 ) ( b ) ( m 2 − 3 1 ) ( 2 n ) = ( 12 4 + 2 n ) ( c ) ( m − 3 − 1 1 ) ( − 1 2 4 n ) = ( − 14 − 11 5 3 )
 
Solution:
(a) ( 3 m ) ( 1   n ) = ( 3 12 − 2 − 8 ) ( 3 3 n m m n ) = ( 3 12 − 2 − 8 ) m = − 2 ,   3 n = 12 n = 4

(b) ( m 2 − 3 1 ) ( 2 n ) = ( 12 4 + 2 n ) ( 2 m + 2 n − 6 + n ) = ( 12 4 + 2 n ) − 6 + n = 4 + 2 n n = − 10 2 m + 2 n = 12 2 m + 2 ( − 10 ) = 12 2 m − 20 = 12 2 m = 32 m = 16

(c) ( m − 3 − 1 1 ) ( − 1 2 4 n ) = ( − 14 − 11 5 3 ) ( − m + ( − 12 ) 2 m + ( − 3 n ) 1 + 4 − 2 + n ) = ( − 14 − 11 5 3 ) − m − 12 = − 14 − m = − 2 m = 2 − 2 + n = 3 n = 5


9.6 SPM Practice (Long Questions)

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Question 6:
Diagram below shows the locations of points P, Q, R, A, K and C, on the surface of the earth. O is the centre of the earth.


(a) Find the location of A.

(b) Given the distance QR is 3240 nautical miles, find the longitude of Q.

(c) Calculate the distance, in nautical miles of KA, measured along the common parallel latitude.

(d) An aeroplane took off from A and flew due west to K along the common parallel of latitude. Then, it flew due south to Q. The average speed of the aeroplane was 550 knots.
Calculate the total time, in hours, taken for the whole flight.


Solution:
(a)
Longitude of A = (180o – 15o) = 165o E
Latitude of A = 50o N
Therefore, position of A = (50o N, 165o E).

(b)
∠QOR= 3240 60 = 54 o ∴Longitude of Q=( 165 o − 54 o )E  = 111 o E

(c)
Distance of KA
= 54 x 60 x cos 50o
= 2082.6 nautical miles

(d)
Total distance=AK+KQ   =2082.6+( 50×60 )   =5082.6 nautical miles Total time = 5082.6 550  =9.241 hours

9.6 SPM Practice (Long Questions)

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Question 5:
A (53o N, 84o E), B (53o N, 25o W), C and D are four points on the surface of the earth. AC is the diameter of the parallel of latitude 53o N.

(a) State the location of C.

(b) Calculate the shortest distance, in nautical mile, from A to C measured along the surface of the earth.

(c) Calculate the distance, in nautical mile, from A due east B measured along the common parallel of latitude.

(d) An aeroplane took off from B and flew due south to D. The average speed of the flight was 420 knots and the time taken was 6½ hours.
Calculate
(i) the distance, in nautical mile, from B to D measured along the meridian.
(ii) the latitude of D.


Solution:
(a)
Latitude of C = 53o N
Longitude of C = (180o – 84o) E = 96o E
Therefore location of C = (53o N, 96o E)

(b)
Shortest distance from A to C
= (180 – 53 – 53) x 60
= 74 x 60
= 4440 nautical miles

(c)
Distance from A to B
= (84 – 25) x 60 x cos 53o
= 59 x 60 x cos 53o
= 2130.43 nautical miles

(d)
 ( i ) Distance travel from B to D =420×6 1 2 ←  Distance travelled  = average speed × time taken  =2730 nautical miles ( ii ) Difference in latitude between B to D = 2730 60 = 45.5 o ∴Latitude of D=( 53 o − 45.5 o )N = 7.5 o N

9.6 SPM Practice (Long Questions)

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Question 4:
P (25o N, 35o E), Q (25o N, 40o W), R and S are four points on the surface of the earth. PS is the diameter of the common parallel of latitude 25o N.

(a) Find the longitude of S.

(b) R lies 3300 nautical miles due south of P measured along the surface of the earth.
Calculate the latitude of R.

(c) Calculate the shortest distance, in nautical mile, from P to S measured along the surface of the earth.

(d) An aeroplane took off from R and flew due north to P. Then, it flew due west to Q.
The total time taken for the whole flight was 12 hours 24 minutes.

(i) Calculate the distance, in nautical mile, from P due west Q measured along the common parallel of latitude.
(ii) Calculate the average speed, in knot, of the whole flight.

Solution:
(a)
Longitude of S = (180o – 35o) W = 145o W

(b)
 ∠POR= 3300 60  = 55 o Latitude of R= ( 55−25 ) o  = 30 o S

(c)
Shortest distance from P to S
= (65 + 65) x 60
= 130 x 60
= 7800 nautical miles

(d)(i)
Distance of PQ
= (35 + 40) x 60 x cos 25o
= 75 x 60 x cos 25o
= 4078.4 nautical miles


( ii ) Total distance travelled RP+PQ =3300+4078.4 =7378.4 nautical miles Average speed = Total distance travelled Time taken = 7378.4 12.4 ← 12 hours 24 min=12+ 12 60 =12+0.4 =595.0 knot


9.6 SPM Practice (Long Questions)

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Question 3:
Diagram below shows the locations of points A (34o S, 40o W) and B (34o S, 80o E) which lie on the surface of the earth. AC is a diameter of the common parallel of latitude 34o S.


(a) State the longitude of C.

(b) Calculate the distance, in nautical mile, from A due east to B, measured along the common parallel of latitude 34o S.

(c) K lies due north of A and the shortest distance from A to K measured along the surface of the earth is 4440 nautical miles.
Calculate the latitude of K.

(d) An aeroplane took off from B and flew due west to A along the common parallel of latitude. Then, it flew due north to K. The average speed for the whole flight was 450 knots.
Calculate the total time, in hours, taken for the whole flight.

Solution:
(a)
Longitude of C = (180o – 40o) E = 140o E

(b)
Distance of AB
= (40 + 80) x 60 x cos 34o
= 120 x 60 x cos 34o
= 5969 nautical miles

(c)
∠AOK= 4440 60  = 74 o Latitude of K= ( 74−34 ) o N  = 40 o N

(d)
Total distance travelled BA+AK =5969+4440 =10409 nautical miles Total time taken = Total distance travelled Average speed = 10409 450 =23.13 hours


7.5 SPM Practice (Long Questions)

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Question 7:
Diagram below shows 4 cards labelled with letters.


Two cards are chosen at random from the box, one by one, without replacement.
(a) List the sample space.
(b) Find the probability that
(i) at least one card chosen is labelled C,
(ii) both cards chosen are labelled with same letter.

Solution:
(a)
Sample space, S
= {(C, O1), (C, O2), (C, L), (O1, C), (O1, O2), (O1, L), (O2, C), (O2, O1), (O2, L), (L, C), (L, O1), (L, O2)}
n(S) = 12

(b)(i)
At least one card chosen is labelled C
= {(C, O1), (C, O2), (C, L), (O1, C), (O2, C), (L, C)}
Probability= 6 12 = 1 2

(b)(ii)
Both cards chosen are labelled with same letter
= {(O1, O2), (O2, O1)}
Probability= 2 12 = 1 6


Question 8 (6 marks):
A bag contains five cards, labelled with the letters I, J, K, M and U.
One card is picked at random from the bag and the letter is recorded. Without replacement, another card is picked at random from the bag and the letter is also recorded.

(a) Complete the sample space in the answer space.

(b) By listing all the possible outcomes of the event, find the probability that

(i) the first card picked is labelled with a vowel.

(ii) the first card picked is labelled with a consonant and the second card picked is labelled with a vowel.

Answer:
{(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), (   , ), (   , ), ( , ), ( , ), (   , ), (   ,   ), (   , ), (   ,   ), (   , ), (   ,   )}


Solution:
(a)
S = {(I, J), (I, K), (I, M), (I, U), (J, I), (J, K), (J, M), (J, U), (K, I), (K, J), (K, M), (K, U), (M, I), (M, J), (M, K), (M, U), (U, I), (U, J), (U, K), (U, M)}

(b)(i)
{(I, J), (I, K), (I, M), (I, U), (U, I), (U, J), (U, K), (U, M)}
Probability of first card with vowels = 8 20 = 2 5

(b)(ii)
{(J, I), (J, U), (K, I), (K, U), (M, I), (M, U)}
Probability = 6 20 = 3 10

7.5 SPM Practice (Long Questions)

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Question 5:
A fair dice is tossed. Then a card is picked at random from a box containing a yellow, a red and a purple card.

(a) By using the letter Y to represent the yellow card, the letter R to represent the red card and the letter P to represent the purple card, complete the sample space in the diagram in the answer space.

(b) By listing down all the possible outcomes of the event, find the probability that
(i) a number less than six and a red card are chosen.
(ii) a number greater than three or a purple card is chosen.

Answer:
(a)

Solution:
(a)

(b)
( i ) Sample space of a number less than six and a red card are chosen { ( 1,R ),( 2,R ),( 3,R ),( 4,R ),( 5,R ) } Probability= 5 18 ( ii ) Sample space of a number greater than three or a purple card is chosen {( 4,Y ),( 5,Y ),( 6,Y ),( 4,R ),( 5,R ),( 6,R ),( 4,P ),( 5,P ),( 6,P ) ,( 1,P ),( 2,P ),( 3,P )} Probability= 12 18  = 2 3



Question 6:
A coin is tossed and a dice is rolled simultaneously. By listing the sample of all the possible outcomes of the event, find the probability that
(a) a head and an odd number are obtained.
(b) a head or a number greater than 4 are obtained.

Solution:
 ( a ) Sample space of a head and an odd number { ( H,1 ),( H,3 ),( H,5 ) } Probability= 3 12   = 1 4 ( b ) Sample space of a head or a number greater than 4 { ( H,1 ),( H,2 ),( H,3 ),( H,4 ),( H,5 ),( H,6 ),( T,5 ),( T,6 ) } Probability= 8 12   = 2 3