I have done the nodal analysis for the following question, and everything appears to have been done correctly as far as I can see. However, when I simulate the results I get values for the circuit which are wildly different.
Where would the error have been introduced? All sign conventions are followed and I've doubled checked the working out.
Edit : Ix = (v1-v2)/2 | This was not included in the diagram
Best Answer
Your mistake is again the current direction.
For \$V_1\$ node you have wrote: $$+12A + \frac{V_1 - V_2}{2\Omega}....= 0$$
As we can see from a circuit diagram the \$12A\$ current is entering the node(incoming current). But we have a problem here because you decided to give a "+" for 2 ohms resistor current as well.
But this notation \$\frac{V_1 - V_2}{2\Omega}\$ imposes the current direction. And this current direction is from \$V_1\$ to \$V_3\$ (the current is flowing out of the node).
And here we have an error in your current direction assumption. And this is why you are getting the wrong answer.
The correct equation should look like this:
$$-12A + \frac{V_1 - V_2}{2\Omega} + \frac{V_1 - V_3}{4\Omega} = 0$$ $$\frac{V_2}{4\Omega} + \frac{V_2 - V_1}{2\Omega} + \frac{V_2 - V_3}{8\Omega} = 0$$ $$ 2\times\frac{V_1 - V_2}{2\Omega} + \frac{V_3 - V_1}{4\Omega} + \frac{V_3 - V_2}{8\Omega} = 0$$
Or like this (not recommended way):
$$+12A + \frac{V_2 - V_1}{2\Omega} + \frac{V_3 - V_1}{4\Omega} = 0$$ $$\frac{V_2}{4\Omega} + \frac{V_2 - V_1}{2\Omega} +\frac{V_2 - V_3}{8\Omega} = 0$$ $$ 2\times\frac{V_1 - V_2}{2\Omega} +\frac{V_3 - V_1}{4\Omega}+\frac{V_3 - V_2}{8\Omega} = 0$$