Q. 41

Let f : A A. f–1 o g–1

B. f o g

C. g–1 o f–1

D. g o f

Answer :

Given that, f : A B and g : B C be the bijective functions.


Let A = {1,3,4}, B ={2,5,1} and C = {3,1,2}


f : A B is bijective function.


f = {(1, 2), (3, 5), (4, 1)


f-1 = {(2,1),(5,3),(1,4)}


g : B C is bijective function.


g = {(2, 3), (5, 1), (1, 4)}


g-1 ={(3,2),(1,5),(4,1)}


Now,


gof (1) = g(f(1)) = g(2) = 3


gof (3) = g(f(3)) = g(5) = 1


gof (4) = g(f(4)) = g(1) = 4


gof = {(1,3),(3,1),(4,4)} (1)


(gof)-1 = {(3,1),(1,3),(4,4)} (2)


fog (2) = f(g(2)) = f(3) = 5


fog (5) = f(g(5)) = f(1) = 2


fog (1) = f(g(1)) = f(4) = 1


fog = {(2,5),(5,2),(1,1)} (3)


(fog)-1 = {(5,2),(2,5),(1,1)} (4)


f-1og-1 (3) = f-1(g-1(3)) = f-1(2) = 1


f-1og-1 (1) = f-1(g-1(1)) = f-1(5) = 3


f-1og-1 (4) = f-1(g-1(4)) = f-1(1) = 4


f-1og-1 = {(3,1),(1,3),(4,4)} (5)


g-1of-1 (2) = g-1(f-1(2)) = g-1(1) = 5


g-1of-1 (5) = g-1(f-1(5)) = g-1(3) = 2


g-1of-1 (1) = g-1(f-1(1)) = g-1(4) = 1


g-1of-1 = {(2,5),(5,2),(1,1)} (6)


On comparing 1,2,3,4,5 and 6 we observe that 2 and 5 are same.


i.e (g o f)–1 = f-1og-1

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