Question 5:
(a) Given 114(2s−4t)(t−142)=(1001), find the value of s and of t.(a) Given 114(2s−4t)(t−142)=(1001), 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)114(2s−4t)(t−142)=(1001)114(2t+4s−2+2s−4t+4t4+2t)=(1001)−2+2s14=0 2s=2 s=14+2t14=14+2t=142t=10t=5(a)114(2s−4t)(t−142)=(1001)114(2t+4s−2+2s−4t+4t4+2t)=(1001)−2+2s14=0 2s=2 s=14+2t14=14+2t=142t=10t=5
(b)(3−291)(xy)=(51) (xy)=121(12−93)(51) (xy)=121((1)(5)+(2)(1)(−9)(5)+(3)(1)) (xy)=121(7−42) (xy)=(13−2)∴x=13, y=−2(b)(3−291)(xy)=(51) (xy)=121(12−93)(51) (xy)=121((1)(5)+(2)(1)(−9)(5)+(3)(1)) (xy)=121(7−42) (xy)=(13−2)∴x=13, y=−2
(a) Given 114(2s−4t)(t−142)=(1001), find the value of s and of t.(a) Given 114(2s−4t)(t−142)=(1001), 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)114(2s−4t)(t−142)=(1001)114(2t+4s−2+2s−4t+4t4+2t)=(1001)−2+2s14=0 2s=2 s=14+2t14=14+2t=142t=10t=5(a)114(2s−4t)(t−142)=(1001)114(2t+4s−2+2s−4t+4t4+2t)=(1001)−2+2s14=0 2s=2 s=14+2t14=14+2t=142t=10t=5
(b)(3−291)(xy)=(51) (xy)=121(12−93)(51) (xy)=121((1)(5)+(2)(1)(−9)(5)+(3)(1)) (xy)=121(7−42) (xy)=(13−2)∴x=13, y=−2(b)(3−291)(xy)=(51) (xy)=121(12−93)(51) (xy)=121((1)(5)+(2)(1)(−9)(5)+(3)(1)) (xy)=121(7−42) (xy)=(13−2)∴x=13, y=−2
Question 6:
It is given that matrix P=(6−3−52) and matrix Q=1m(235n) such that PQ=(1001).It is given that matrix P=(6−3−52) and matrix Q=1m(235n) such that PQ=(1001).
(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) =12−15m=−3n=6(a)m=6(2)−(−3)(−5) =12−15m=−3n=6
(b)(6−3−52)(xy)=(−2418) (xy)=112−15(2356)(−2418) (xy)=1−3((2)(−24)+(3)(18)(5)(−24)+(6)(18)) (xy)=1−3(6−12) (xy)=(−24)∴x=−2, y=4(b)(6−3−52)(xy)=(−2418) (xy)=112−15(2356)(−2418) (xy)=1−3((2)(−24)+(3)(18)(5)(−24)+(6)(18)) (xy)=1−3(6−12) (xy)=(−24)∴x=−2, y=4
It is given that matrix P=(6−3−52) and matrix Q=1m(235n) such that PQ=(1001).It is given that matrix P=(6−3−52) and matrix Q=1m(235n) such that PQ=(1001).
(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) =12−15m=−3n=6(a)m=6(2)−(−3)(−5) =12−15m=−3n=6
(b)(6−3−52)(xy)=(−2418) (xy)=112−15(2356)(−2418) (xy)=1−3((2)(−24)+(3)(18)(5)(−24)+(6)(18)) (xy)=1−3(6−12) (xy)=(−24)∴x=−2, y=4(b)(6−3−52)(xy)=(−2418) (xy)=112−15(2356)(−2418) (xy)=1−3((2)(−24)+(3)(18)(5)(−24)+(6)(18)) (xy)=1−3(6−12) (xy)=(−24)∴x=−2, y=4
Question 7:
(a) Find the inverse matrix of (3254).(a) Find the inverse matrix of (3254).
(b) Ethan and Rahman went to the supermarket to buy cucumbers and carrots. Ethan bought 3 cucumbers and 2 carrots for RM9. Rahman bought 5 cucumbers and 4 carrots for RM16.
By using matrix method, find the price, in RM, of a cucumber and the price of a carrot.
Solution:
(a)Inverse matrix of (3254)=112−10(4−2−53)=12(4−2−53)=(2−1−5232)(a)Inverse matrix of (3254)=112−10(4−2−53)=12(4−2−53)=(2−1−5232)
(b)3x+2y=9……………..(1)5x+4y=16……………(2)(3254)(xy)=(916) (xy)=(2−1−5232)(916) (xy)=((2)(9)+(−1)(16)(−52)(9)+(32)(16)) (xy)=(18−16−452+24) (xy)=(232)x=2, y=32∴Price of a cucumber=RM2 Price of a carrot=RM1.50
(a) Find the inverse matrix of (3254).(a) Find the inverse matrix of (3254).
(b) Ethan and Rahman went to the supermarket to buy cucumbers and carrots. Ethan bought 3 cucumbers and 2 carrots for RM9. Rahman bought 5 cucumbers and 4 carrots for RM16.
By using matrix method, find the price, in RM, of a cucumber and the price of a carrot.
Solution:
(a)Inverse matrix of (3254)=112−10(4−2−53)=12(4−2−53)=(2−1−5232)(a)Inverse matrix of (3254)=112−10(4−2−53)=12(4−2−53)=(2−1−5232)
(b)3x+2y=9……………..(1)5x+4y=16……………(2)(3254)(xy)=(916) (xy)=(2−1−5232)(916) (xy)=((2)(9)+(−1)(16)(−52)(9)+(32)(16)) (xy)=(18−16−452+24) (xy)=(232)x=2, y=32∴Price of a cucumber=RM2 Price of a carrot=RM1.50