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## Homework Statement

Calculate the voltage at each of the extraordinary nodes (V1 – V3), and the current through each of the six resistors.

http://i.imgur.com/VbZQH.png

## Homework Equations

KCL: ƩIn = 0

V=ir

I=[itex]A^{-1}[/itex]B

[itex]\Delta[/itex] = a11C11 + a12C12 + a13C13

[itex]A^{-1}[/itex] = [itex]\frac{adjA}{\Delta}[/itex]

adjA = [itex][C_{jk}]^{T}[/itex]

## The Attempt at a Solution

For I1+I2+I3=0:

[itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V1-V2}{0.5}[/itex] + [itex]\frac{V1-2}{0.1}[/itex] = 0

For I4+I5+I6=0:

[itex]\frac{V2-V1}{0.5}[/itex] + [itex]\frac{V2-V3}{0.5}[/itex] + [itex]\frac{V2}{0.1}[/itex] = 0

For I7+I8+I9=0:

-[itex]\frac{V1-4}{0.2}[/itex] + [itex]\frac{V3-V2}{0.5}[/itex] + [itex]\frac{V3-3}{0.1}[/itex] = 0

([itex]\frac{1}{0.2}[/itex] + [itex]\frac{1}{0.5}[/itex] + [itex]\frac{1}{0.1}[/itex])V1 - ([itex]\frac{1}{0.2}[/itex] + [itex]\frac{1}{0.5}[/itex])V2 + 0V3 = 40

([itex]\frac{1}{0.5}[/itex]+[itex]\frac{1}{0.5}[/itex]+[itex]\frac{1}{0.1}[/itex])V2 - ([itex]\frac{1}{0.5}[/itex])V1 + (-[itex]\frac{1}{0.5}[/itex])V3 = 0

(-[itex]\frac{1}{0.2}[/itex])V1 + (-[itex]\frac{1}{0.5}[/itex])V2 + [itex]\frac{1}{0.5}[/itex] + [itex]\frac{1}{0.1}[/itex])V3 = 10

And then I simplified that to:

17V1 - 7V2 + 0V3 = 40

-2V1 + 14V2 -2V3 = 0

-5V1 - 2V2 + 12V3 = 10

My book says I should use Cramer's rule or matrix inversion to solve this. I chose to use matrix inversion because I thought solving a simultaneous equation would take a while.

I ended up with I1 = 2.627, I2 = 2/3, I3 = 2.039

That's obviously wrong since the sum should equal zero. I'm not sure what I did wrong, I just followed the steps in my book.