Calculate Norton equivalent in a circuit with a voltage controlled voltage source
Here's your schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

Using nodal (KCL) I get:

$$\begin{align*}
\frac{V_u}{R_1}+\frac{V_u}{R_2}&amp;=\frac{V_0}{R_1}+\frac{V_+}{R_2}\\\\
\frac{V_+}{R_2}+\frac{V_+}{R_3}&amp;=\frac{V_u}{R_2}+\frac{A\cdot V_u}{R_3}
\end{align*}$$

Solve with sympy:

var('vu vplus r1 r2 r3 A v0')
eq1=Eq(vu/r1+vu/r2,v0/r1+vplus/r2)
eq2=Eq(vplus/r2+vplus/r3,A*vu/r3+vu/r2)
solve([eq1,eq2],[vu,vplus])[vplus].subs({r1:1,r2:3,r3:5,v0:5,A:2})
55/7
eq3=Eq(vplus/r2+vplus/r3,A*vu/r3+vu/r2+1)
solve([eq1,eq3],[vu,vplus])[vplus].subs({r1:1,r2:3,r3:5,v0:5,A:2})
75/7


Please note that equation 3 adds an amp of current injected into Port B. (You can see the +1 at the end as the only modification to equation 2.)

From the above, we know that \$V_\text{TH}=\frac{55}{7}\:\text{V}\$ and that \$R_\text{TH}=\frac{\Delta V}{\Delta I}=\frac{\frac{75}{7}\:\text{V}-\frac{55}{7}\:\text{V}}{1\:\text{A}-0\:\text{A}}=\frac{20}{7}\:\Omega\$. Converting to Norton is \$I_\text{N}=\frac{V_\text{TH}}{R_\text{TH}}=\frac{11}{4}\:\text{A}\$ and \$R_\text{N}=R_\text{TH}=\frac{20}{7}\:\Omega\$.