let f(x) = (ax + b) / (cx + d) , x [does not equal] -d/c
a. Show that f is one to one if ad-b[doesnt equal] 0.
b. Suppose that ad-bc[doesnt equal] 0. Find f^-1.
2. You need this condition so that the brackets (ad-bc) can be cancelled
in ImPerfectHacker's post, as otherwise x_1 and x_2 can have any values
(not necessarily equal) and still satisfy:
(ad-bc)x_1 = (ad-bc)x_2
as this would be 0 x_1 = 0 x_2 if ad-bc=0
If you don't want to do all that work, there's always the "cheating" way.
Notice that ad-bc is also the determinant of the matrix . This suggests f and that matrix share some nice properties; Especially, if s=ad-bc is not zero, the inverse matrix exists and defines the wanted mapping .