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*

Add the first and the third equations to get 
$$(a+b-2c-2d)+(-a+b-2c+2d)=0\implies 2(b-2c)=0\implies b=2c$$
Plugging this in at the first equation, $a-2d=0\implies a=2d$.
Plugging these two into the second equation
$$2di+2c+c-di=0\implies di+3c=0$$
