Let `S = {1, 2, 3}`. Determine whether the functions `f : S->S` defined as below have inverses. Find `f^(-1)`, if it exists
.(a) `f = {(1, 1), (2, 2), (3, 3)}`
(b) `f = {(1, 2), (2, 1), (3, 1)}`
(C) `f = {(1, 3), (3, 2), (2, 1)}`

Explanation

`S = {1, 2, 3}`

`f : S->S` is given by

(a) `f={(1,1),(2,2),(3,3)}`

Now f is one-one and onro

∴ `f^(-1)={(1,1),(2,2),(3,3)}`

(b) `f = {(1, 2), (2, 1), (3, 1)}`

Since `f(2)=f(3)=1`

∴ f is not one-one, so it is not invertible.