Bilinear transformation which maps $z=(1,i,-1)$ and $w=(2,i,-2)$ I have three equations after simplifying this a bit
$a+b-2c-2d=0$
$ai+b+c-di=0$
$-a+b-2c+2d=0$
How do I proceed after obtaining these equations? Please comment after every step or I will be lost.
If you care to know, this is from the chapter Complex Variables in my book.
EDIT- my book explains it like this
a+b-2c-2d=0 ----(1)
ai+b+c-di=0-----(2)
-a+b-2c+2d=0----(3)
(1)+(3) gives 2b-4c *\I get that but want to know why we added 1+3 and not 1+2*
i.e 2b-4c=b-2c
b-2c=0 ----(4)
(2)+i X (3) gives
(1+i)b+(1-2i)c+id=0 ----- (5) *\how does this step work.. totally lost in this step*

The important thing is that you have 3 equations and 4 unknowns. Thus you have to consider one of the unknowns, say $c$, as a parameter, and you will obtain a unique solution in the remaining variables $(a,b,d)$ depending on this parameter. In this way, any method will give you:
$$a=6ic, \ \ \ b=2c, \ \ \ d=3ic$$
Plugging these values into
$Z=\dfrac{az+b}{cz+d}$, and simplifying by $c$, one gets: $Z=\dfrac{6iz+2}{z+3i}$.
