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Considering I have the following two equations: $$-\beta\overline{\lambda}+a(r\beta+p\overline{r}\beta)-(rC+\beta\overline{A}) = 0$$ $$-r\lambda+a(pr^2-|\beta|^2)-(rA-\beta\overline{C}) = 1$$

where $\lambda,r,\beta, A, C$ are complex numbers, others are real numbers. How can I express $\lambda$ without in terms of $\beta$ and $r$?

Nothing
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  • As there are only two equations, you can't eliminate both of $\beta$ and $r$. –  Jun 23 '21 at 14:40
  • @YvesDaoust What if i use the fact that $|\beta|^2\geq 0 $ and do it into two separate case, i.e $\beta = 0$ and $\beta\neq 0$? – Nothing Jun 23 '21 at 14:55
  • @Nothing: It would probably help if you explained how the system came about and what you're ultimately trying to accomplish. – Blue Jun 23 '21 at 15:10
  • If $\beta=0$, there is no $\beta$ to eliminate anymore. If $\beta\ne0$, you have not progressed by a millimeter. –  Jun 23 '21 at 15:16
  • @Blue, actually everything starts from to solve $r+\beta j = -(\lambda + (A+Cj) - \alpha_1(pr+\beta j))^{-1}$, where here $j$ means the quaternion. (i.e $a+bj$). I tried to simplify this and equating both side and I get the above set of equations. – Nothing Jun 23 '21 at 15:30

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Conjugate the first equation and multiply it by $r$, multiply second equation by $\bar\beta$, then you will get the system of linear equations with one variable $\lambda$. I hope you will be able to find when the system is solvable and what is the solution.

Vasily Mitch
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