- #1
braindead101
- 162
- 0
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 = 1any help would be great, thanks.
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 = 1any help would be great, thanks.
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