- #1

braindead101

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**Solve the following linear system:**

ix + (1+i)y = i

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

iy + z = 1

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.

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