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Linear algebra - solve linear system with complex constants

  1. Sep 30, 2007 #1
    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. jcsd
  3. Sep 30, 2007 #2


    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.
  4. Sep 30, 2007 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
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