Linear algebra - solve linear system with complex constants

  • #1
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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:

Answers and Replies

  • #2
Dale
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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
HallsofIvy
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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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