Question 8:
The inverse matrix of (4−125) is t(51−2n).
(a) Find the value of n and of t.
(b) Write the following simultaneous linear equations as matrix equation:
4x – y = 7
2x + 5y = –2
Hence, using matrix method, calculate the value of x and of y.
Solution:
(a)t(51−2n)=(4−125)−1=1(4)(5)−(−1)(2)(51−24)=122(51−24)∴t=122, n=4
(b)(4−125)(xy)=(7−2) (xy)=122(51−24)(7−2) (xy)=122((5)(7)+1(−2)(−2)(7)+(4)(−2)) (xy)=122(35−2−14−8) (xy)=122( 33−22) (xy)=( 32−1)∴x=32, y=−1
The inverse matrix of (4−125) is t(51−2n).
(a) Find the value of n and of t.
(b) Write the following simultaneous linear equations as matrix equation:
4x – y = 7
2x + 5y = –2
Hence, using matrix method, calculate the value of x and of y.
Solution:
(a)t(51−2n)=(4−125)−1=1(4)(5)−(−1)(2)(51−24)=122(51−24)∴t=122, n=4
(b)(4−125)(xy)=(7−2) (xy)=122(51−24)(7−2) (xy)=122((5)(7)+1(−2)(−2)(7)+(4)(−2)) (xy)=122(35−2−14−8) (xy)=122( 33−22) (xy)=( 32−1)∴x=32, y=−1
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