Linear algebra - solve linear system with complex constants

1. Sep 30, 2007

Solve the following linear system:
ix + (1+i)y = i
(1-i)x + y - iz = 1
iy + z = 1

I am getting nowhere with this.
is there a trick to do these? I keep getting more and more variations of i. like i^2-1, and (1-i^2)-1

ix + (1+i)y=i
(1-i)x + y-iz=1
y + z = 1

ix + (1+i)y = i
i(1-i)x - (i^2-1)z = i-1 [iR2-r3]
[(1-i)(1+i)-i]x - (1+i)iz = (1+i)-i [(1+i)R2-R1]

okay.. and i simplified this, and got stuck.

ix + (1+i) = i
i(1-i)x + (1-i^2)z = i-1
[(1-i^2)-i]x - (1+i)iz = 1

any help would be great, thanks.

Last edited: Sep 30, 2007
2. Sep 30, 2007

Staff: Mentor

Since i is the sqrt(-1) then i^2 = -1. That will simplify all of your powers of i down to simple complex numbers. Other than that there is nothing different or unusual in solving this system.

3. Sep 30, 2007

HallsofIvy

Looks to me like it's straight forward. If y+ z= 1 then clearly z= 1- y. Put that into the second equation and you have (1-i)x+ y- i- iy= (1-i)x+ (1-i)y= 1- i so x+ y= 1. From that, y= 1-x so the first equation becomes ix+ (1+i)(1-x)= ix+ 1-x+i- ix= x+ 1-i= i. That should be easy to solve.