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From schematics we have:

\$i_1=\frac{v_1-v_2}{3},i_2=\frac{v_1-v_3}{2},i_3=\frac{v_3}{6},i_x=\frac{v_2}{4}\$

for node 1:

\$10 = i_1 + i_2\$

\$5v_1-2v_2-3v_3=60\$ EQ1

for node 2:

\$4i_x + i_1 - i_x = 0\$

\$4v_1+5v_2=0\$ EQ2

for node 3:

\$i_2 - i_3 - 4i_x = 0\$

\$6v_1 - 3v_2 -8v_3 = 0\$ EQ3

from three equations we get

\$v = (\frac{400}{23},\frac{-320}{23},\frac{420}{23})\$

but correct solution is

\$v = (80, -64, 156)\$

Where is problem? My guess is that controlling current \$i_x\$ and dependent source are connected to same node.

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The algebra for equation 3 is the problem. It should reduce to: $$ 3V_1-6V_2-4 V_3=0 $$ That will give you the right answer

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  • 1
    \$\begingroup\$ Agreed. I found this error as well. I got \$ -12 v_2 \$ instead of \$ -3 v_2 \$ but yes, you can reduce that row. I think Matej forgot it was \$ 4i_x \$ and not \$ i_x \$ \$\endgroup\$ – KingDuken Aug 4 '16 at 15:08

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