How to solve this nonlinear 3 by 3 system of equations?

Question

So I have to solve this nonlinear system of equations with 3 equations and 3 unknowns. The system is: $$\begin{cases} 2A+3B-1=2CA \\ 2B+2BC=-3A\\ B^{2}+A^{2}=-1 \end{cases}$$

No methods that I'm familiar with seem to help me solve this and I haven't come to the right idea in like 3 hours.

Answer 1

Elimination of the variables $A$ and $C$ yields \begin{align*} C & =\frac{ - 16B^3 + 48B^2 - 17B + 21}{24},\\ A & =\frac{ - 16B^3 - 17B - 3}{12}, \end{align*} satisfying the quartic equation (here in $B$) is $$16B^4 + 17B^2 + 6B + 9=0.$$ It has four solutions over $\Bbb C$.

Answer 2

Multiply the first by $B$ and the second by $A$ and eliminate $ABC$. You obtain a conic in $A,B$:

$$3A^2+4AB+3B^2-B=0$$

which simplifies to

$$4AB-B-3=0.$$

Then squaring,

$$16A^2B^2=-16(B^2+1)B^2=(B+3)^2,$$ which is a nasty quartic.

Note: if you deduce $A$ from $B$ using the before last equation, no extra root is introduced by the squaring.