adjacent voltage sources in nodal analysis
The \$10\:\textrm{V}\$ source is basically just a dead short with \$10\:\textrm{V}\$ across it. Also, the current-dependent voltage source is also just a dead short. So, in one sense it's all just the same node (supernode.) So \$U_1=U_2+10\:\textrm{V}\$ and \$U_3=U_2+5\cdot i\$. You really only have one independent variable there.

This fact also essentially eliminates the \$6\:\Omega\$ resistor from consideration. If \$U_1\$ and \$U_2\$ and \$U_3\$ are all just the same node, then all the current through the \$6\:\Omega\$ resistor is just going out and than back into the same "node" (okay, supernode) so you don't even need to think about it. Just ignore it. It's "shorted out," in effect, and doesn't impact the nodal calculation. (Sure, there will be a current through it.)

And thanks for using the schematic editor for your diagram. I added the current you were talking about (you can scroll down a bit and find the array and text tools, if you look.) Reading your textual description of i, I get the following:







simulate this circuit &ndash; Schematic created using CircuitLab

At this point, I completely ignored the \$6\:\Omega\$ resistor and very quickly wrote out the following equation:

$$\begin{align*}
\frac{U_2+10\:\textrm{V}}{2\:\Omega}+\frac{U_2}{4\:\Omega}+\frac{U_2+5\cdot \frac{0\:\textrm{V}-\left(U_2+10\:\textrm{V}\right)}{2\:\Omega}}{3\:\Omega}&amp;=0\:\textrm{A}
\end{align*}$$

solving, I get \$U_2=\frac{40}{3}\:\textrm{V}\$.

Of course, that doesn't match anything you wrote out. So I turned i around the other way. That seemed to work:







simulate this circuit

$$\begin{align*}
\frac{U_2+10\:\textrm{V}}{2\:\Omega}+\frac{U_2}{4\:\Omega}+\frac{U_2+5\cdot \frac{\left(U_2+10\:\textrm{V}\right)-0\:\textrm{V}}{2\:\Omega}}{3\:\Omega}&amp;=0\:\textrm{A}
\end{align*}$$

Now \$U_2=-\frac{160}{23}\$ (and the rest falls out in obvious ways from there.)

Note that there was no needed matrix. Other than ground, you really only have one other independent node present. All you do is solve for the only scalar variable present.