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4.10 SPM Practice (Long Questions)

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Question 9:
(a) Given 1s(−42−53)(t−25−4)=(1001), find the value of s and of t.

(b) Using matrices, calculate the value of x and of y that satisfy the following matrix equation:
(−42−53)(xy)=(12)


Solution:
(a)1s(t−25−4)=(−42−53)−1=1(−4)(3)−(2)(−5)(3−25−4)=1−2(3−25−4)∴

(b) ( −4 2 −5 3 )( x y )=( 1 2 )   ( x y )= 1 −2 ( 3 −2 5 −4 )( 1 2 )   ( x y )= 1 −2 ( ( 3 )( 1 )+( −2 )( 2 ) ( 5 )( 1 )+( −4 )( 2 ) )   ( x y )= 1 −2 ( −1 −3 )   ( x y )=( 1 2 3 2 ) ∴x= 1 2 ,  y= 3 2



Question 10 (5 marks):
During the sport day, students used coupons to buy food and drink. Ali and Larry spent RM31 and RM27 respectively. Ali bought 2 food coupons and 5 drinks coupons while Larry bought 3 coupons and 1 drinks coupon.
Using the matrix method, calculate the price, in RM, of a food coupon and of a drinks coupon.

Solution:
Ali spent RM31. He bought 2 food coupons and 5 drinks coupons.
Larry spent RM27. He bought 3 food coupons and 1 drinks coupons.
x = price of one food coupon
y = price of one drinks coupon

( 2    5 3    1 )( x y )=( 31 27 )             ( x y )= 1 2( 1 )−5( 3 ) ( 1    −5 −3     2 )( 31 27 )             ( x y )= 1 2−15 ( 1( 31 )+( −5 )( 27 ) −3( 31 )+2( 27 ) )             ( x y )= 1 −13 ( −104 −39 )             ( x y )=( 8 3 ) x=8 and y=3 Thus, the price for a food coupon is RM8 and the price for drink coupon is RM3.