Is it possible to simplify the coefficient matrix for large values of $x$? If I have a system of $8$ linear equations for the eight variables $\{\alpha ,\beta ,\gamma ,\delta ,\eta ,\lambda ,\xi ,\rho \}$ and with the three 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).
My question:

Assuming that the determinant of the coefficient matrix is of a polynomial form in $x$ as $f(x)=c_1 x^8+c_2 x^7+c_3 x^6+...+c_0$; I need to evaluate the problem only for large values of $x \to \infty$, in other words, I only need to find the coefficient $c_1$. My question is if it is possible to simplify (manipulate) this system of equations (or equivalently, the coefficient matrix) from the beginning for large values of $x$ so that computing the determinant is simpler.

$$  \gamma +\delta +\beta \; e^{\frac{i x}{2}} (x-1)+\delta \; x-\gamma\;  x-\alpha \; e^{-\frac{1}{2} (i x)} (x+1)=0,  \\   \gamma +\delta +x (\gamma +\lambda )-\eta -\lambda -x (\delta +\eta )=0,  $$
$$  \eta  (x-1)+\delta  (x-1) e^{-i (x+y)}-(x+1) \left(\lambda +\gamma  \;e^{i (x-y)}\right)=0, \\\alpha +e^{-\frac{1}{2} i (x+2 y)} \left((x+1) \left(\delta +\beta \; e^{\frac{3 i x}{2}+i y}\right)-\gamma \; e^{2 i x} (x-1)\right)=\alpha \; x,  $$
$$  \rho +\xi\;  e^{2 i x} (x+1)-\rho \; x+\alpha  (x-1) e^{\frac{3 i x}{2}+i y}-\beta  (x+1) e^{\frac{1}{2} i (x+2 y)}=0,\\ \xi \; e^{2 i x} (x-1)-\rho  (x+1)+\eta  (x-1) \left(-e^{i (y+z)}\right)+\lambda  (x+1) e^{i (2 x+y+z)}=0, $$
$$  \xi +\rho +\rho \; x+\lambda  (x-1) e^{i (x+z)}-\xi \; x-\eta  (x+1) e^{-i (x-z)}=0,\\\rho  \;x+\alpha  e^{\frac{i x}{2}} (x+1)-\xi-\rho -\beta \; e^{-\frac{1}{2} (i x)} (x-1)-\xi \; x=0. $$
Coefficient Matrix
$$\scriptsize\left(
\begin{array}{cccccccc}
 -e^{-\frac{1}{2} (i x)} (x+1) & e^{\frac{i x}{2}} (x-1) & 1-x & x+1 & 0 & 0 & 0 & 0 \\
 0 & 0 & x+1 & 1-x & -x-1 & x-1 & 0 & 0 \\
 0 & 0 & (x+1) \left(-e^{i (x-y)}\right) & (x-1) e^{-i (x+y)} & x-1 & -x-1 & 0 & 0 \\
 1-x & e^{i x} (x+1) & (x-1) \left(-e^{\frac{3 i x}{2}-i y}\right) & (x+1) e^{-\frac{1}{2} i (x+2 y)} & 0 & 0 & 0 & 0 \\
 e^{\frac{i x}{2}} (x-1) & -e^{-\frac{1}{2} (i x)} (x+1) & 0 & 0 & 0 & 0 & (x+1) e^{i (x-y)} & (1-x) e^{-i (x+y)} \\
 0 & 0 & 0 & 0 & (1-x) e^{-i (x-z)} & (x+1) e^{i (x+z)} & (x-1) e^{i (x-y)} & (x+1) \left(-e^{-i (x+y)}\right) \\
 0 & 0 & 0 & 0 & (x+1) \left(-e^{-i (x-z)}\right) & (x-1) e^{i (x+z)} & 1-x & x+1 \\
 e^{\frac{i x}{2}} (x+1) & -e^{-\frac{1}{2} (i x)} (x-1) & 0 & 0 & 0 & 0 & -x-1 & x-1 \\
\end{array}
\right)$$
P.S. I have already asked this in MathStackExchange but did not receive any answers. Any comments are highly appreciated.
 A: Let $M:=M(x,y,z)$ be the $8\times8$ matrix in question. Let $m(x):=M(x,0,0)$.
We have
$$\det m(x)=
-256 e^{i x/2} x^2 \cos (2 x) \big((x^2+1)^2 \cos (2 x)-(x^2-1)^2\big).$$
So, $|\det m(x)|$ will be oscillating between the value $0$ (attained when $\cos (2 x)=0$ -- that is, when $x=(2k-1)\pi/4$ for natural $k$) and the value $512 x^2 (1 + x^4)\sim512 x^6$ (attained when $\cos (2 x)=-1$ -- that is, when $x=(2k-1)\pi/2$ for natural $k$). The value $0$ will also be taken by $\det m(x)$ at the occurring almost periodically roots of the equation $\cos (2 x)=r(x):=\dfrac{(x^2-1)^2}{(x^2+1)^2}$, as $r(x)$ is strictly increasing from $0$ to $1$ for $x$ increasing from $1$ to $\infty$, whereas $\cos (2 x)$ is periodically oscillating between $-1$ and $1$.
In general,
$$\det M(x,y,z)=32 x^2 e^{i (x-4 y+2 z)/2} \\ 
\times\big(-4 (x^2-1)^2 \sin ^2(x) [\cos (2 y)+\cos (z)] \\ 
+4 (x^2-1)^2 \cos (2 x)-4 (x^2+1)^2 \cos (4 x)-16 x^2\big).$$
So, by the triangle inequality,
$$|\det M(x,y,z)|\le32 x^2 \\ 
\times\big(4 (x^2-1)^2 \times2 \\ 
+4 (x^2-1)^2+4 (x^2+1)^2+16 x^2\big)=512 x^2 (1 + x^4).$$
Thus,

$|\det M(x,y,z)|\le512 x^2 (1 + x^4)$ for all $x,y,z$, whereas $|\det M(x,0,0)|$ will be oscillating between $0$ and $512 x^2 (1 + x^4)\sim512 x^6$.

In particular, it follows that your assumption

Assuming that the determinant of the coefficient matrix is of a polynomial form in $x$ as $f(x)=c_1 x^8+c_2 x^7+c_3 x^6+...+c_0$; I need to evaluate the problem only for large values of $x \to \infty$, in other words, I only need to find the coefficient $c_1$.

is quite counterfactual.

Here are the calculations, done in Mathematica (click on the image below to magnify it):

