how do u solve this system??

Question

(1-i)a+2b=0
-a-(1+i)b=0
a-(1+i)c=0
?(i^2=-1)a,b,c are parameters.compelet answer plz.

Answer 1

(1-i)a + 2b = 0 (A)
-a -(1+i)b = 0: a = -(1+i)b (B)
-(1-i)(1+i)b + 2b = 0: -(1-i)(1+i) = -2 so -2 b + 2b = 0

The problem you have is indeterminant such that any b works.
But once you select a b, a and c are determined by:
a = -2*b/(1-i)= -(1+i)b
c = a/(1+i) = -2*b/((1-i)*(1+i)) = -b

Answer 2

I don't think there's a unique solution. The first two equations are equivalent (the second is the first multiplied by (i-1)).

You can determine that b = -c by adding the second and third equations. And the first two equations do constrain a relative to b, but there's no single solution.

Answer 3

Equation 2 times - (1-i) is equation 1 so your system is

dependent.