Solution:
is invertible if is a bijection (i.e one-one onto function)
One-One function
Suppose be two arbitrary elements in
Therefore,
When
So, is one-one function.
Onto function
Suppose be an arbitrary element of (Co-domain)
Therefore,
On Solving, we obtain
As
Thus, Range of co-domain of Thus, is onto function.
Thus, is invertible.
Now we need to find ,
Suppose
Now, replace all with and all with .
Now, solve for y
On Solving, we obtain
Now replace with
As a result,