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)∴s=−2, t=3
(b)(−42−53)(xy)=(12) (xy)=1−2(3−25−4)(12) (xy)=1−2((3)(1)+(−2)(2)(5)(1)+(−4)(2)) (xy)=1−2(−1−3) (xy)=(1232)∴x=12, y=32
(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)∴s=−2, t=3
(b)(−42−53)(xy)=(12) (xy)=1−2(3−25−4)(12) (xy)=1−2((3)(1)+(−2)(2)(5)(1)+(−4)(2)) (xy)=1−2(−1−3) (xy)=(1232)∴x=12, y=32
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 53 1)(xy)=(3127) (xy)=12(1)−5(3)(1 −5−3 2)(3127) (xy)=12−15(1(31)+(−5)(27)−3(31)+2(27)) (xy)=1−13(−104−39) (xy)=(83)x=8 and y=3Thus, the price for a food coupon is RM8and the price for drink coupon is RM3.
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 53 1)(xy)=(3127) (xy)=12(1)−5(3)(1 −5−3 2)(3127) (xy)=12−15(1(31)+(−5)(27)−3(31)+2(27)) (xy)=1−13(−104−39) (xy)=(83)x=8 and y=3Thus, the price for a food coupon is RM8and the price for drink coupon is RM3.