Let `f = { (1,a), (2,c), (4,d), (3,b) and g^-1={(2,a), (4,b), (1,C), (3,d)}` then show that `(gof)^-1=f^-1og^-1`

Let f={(1,a),(2,c),(4,d),(3,b)} and g^(-1) = {(2,a),(4,b),(1,c),(3,d)} then show that (gof)^(-1) = f^(-1) og^(-1) .

Let X = { 1,2,3,4}, Y = {a,b,c,d} and f={(1,a), (4,b), (2,c),(3,d),(2,d)}. Then f is

If f={(1,a), (2, b), (1, b), (3,c), (1,c)}, g=(a, p), (b,r), (c,q), (c,p)} , then (gof)^(-1)=

If A={1, 2, 3}, B={a, b, c, d}, f={(1, a), (2, b), (3, d)} , then f is

If f={(a, 0), (b, 2), (c, -3)}, g={(a, -1), (b, 1), (c, 2)} then 2f-3g=

Let A={1,2,3},B={a,b,c),C={p,q,r}. If f:A to B, g: B to C are defined by f={(1,a),(2,c),(3,b)},g={a,q),(b,r),(c,p)} then show that f^(-1)og^(-1)=(gof)^(-1) .