If I have a system of $8$ linear equations for the variables $\{\alpha ,\beta ,\gamma ,\delta ,\eta ,\lambda ,\xi ,\rho \}$ and with parameters $\{x,y,z\}$ reals and $x>0$. I want to work on nontrivial solutions of the system, therefore, I need to compute the determinant of the coefficient matrix (given below).
$$ \gamma +\delta +\beta \; e^{\frac{i x}{2}} (x-1)+\delta \; x=\gamma\; x+\alpha \; e^{-\frac{1}{2} (i x)} (x+1), \\ \gamma +\delta +x (\gamma +\lambda )=\eta +\lambda +x (\delta +\eta ), $$
$$ \eta (x-1)+\delta (x-1) e^{-i (x+y)}=(x+1) \left(\lambda +\gamma \;e^{i (x-y)}\right), \\\alpha +e^{-\frac{1}{2} i (x+2 y)} \left((x+1) \left(\delta +\beta \; e^{\frac{3 i x}{2}+i y}\right)-\gamma \; e^{2 i x} (x-1)\right)=\alpha \; x, $$
$$ e^{-i (x+y)} \left(\rho +\xi\; e^{2 i x} (x+1)-\rho \; x+\alpha (x-1) e^{\frac{3 i x}{2}+i y}-\beta (x+1) e^{\frac{1}{2} i (x+2 y)}\right)=0,\\ e^{-i (x+y)} \left(\xi \; e^{2 i x} (x-1)-\rho (x+1)+\eta (x-1) \left(-e^{i (y+z)}\right)+\lambda (x+1) e^{i (2 x+y+z)}\right)=0, $$ $$ \xi +\rho +\rho \; x+\lambda (x-1) e^{i (x+z)}=\xi \; x+\eta (x+1) e^{-i (x-z)},\\\rho \;x+\alpha e^{\frac{i x}{2}} (x+1)=\xi +\rho +\beta \; e^{-\frac{1}{2} (i x)} (x-1)+\xi \; x $$
My questions:
If I need to evaluate the problem only for large values of the parameter $x \to \infty$, is it possible to simplify this system of equation accordingly from the beginning so that computing the determinant is simpler?
Again, if I need to deal with those parts of determinant which are dependent of the parameters $y$ and $z$, is it possible to simplify this system of equation from the beginning accordingly? (note that $y$ and $z$ are only present in exponential form in the equations)
$$\scriptsize\left( \begin{array}{cccccccc} -e^{-\frac{1}{2} (i x)} (x+1) & e^{\frac{i x}{2}} (x-1) & 1-x & x+1 & 0 & 0 & 0 & 0 \\ 0 & 0 & x+1 & 1-x & -x-1 & x-1 & 0 & 0 \\ 0 & 0 & (x+1) \left(-e^{i (x-y)}\right) & (x-1) e^{-i (x+y)} & x-1 & -x-1 & 0 & 0 \\ 1-x & e^{i x} (x+1) & (x-1) \left(-e^{\frac{3 i x}{2}-i y}\right) & (x+1) e^{-\frac{1}{2} i (x+2 y)} & 0 & 0 & 0 & 0 \\ e^{\frac{i x}{2}} (x-1) & -e^{-\frac{1}{2} (i x)} (x+1) & 0 & 0 & 0 & 0 & (x+1) e^{i (x-y)} & (1-x) e^{-i (x+y)} \\ 0 & 0 & 0 & 0 & (1-x) e^{-i (x-z)} & (x+1) e^{i (x+z)} & (x-1) e^{i (x-y)} & (x+1) \left(-e^{-i (x+y)}\right) \\ 0 & 0 & 0 & 0 & (x+1) \left(-e^{-i (x-z)}\right) & (x-1) e^{i (x+z)} & 1-x & x+1 \\ e^{\frac{i x}{2}} (x+1) & -e^{-\frac{1}{2} (i x)} (x-1) & 0 & 0 & 0 & 0 & -x-1 & x-1 \\ \end{array} \right)$$