I am working through some of the derivations in this paper
In the appendix (p23, Eq. B.17), we arrive at a set of quadratic equations for variables $E,L$ of the form:
$$f_1 E^2 - 2g_1 E L - h_1 L^2 - d_1 = 0$$
$$f_2 E^2 - 2g_2 E L - h_2 L^2 - d_2 = 0$$
where the $f_i,g_i,h_i,d_i$ can just be viewed as constants for this discussion
The author then eliminates $L$ to obtain a quadratic in $E^2$ as,
$$ (\rho + 4 \eta \sigma) E^4 - 2(\kappa \rho + 2 \epsilon \sigma) E^2 + \kappa^2 \tag{1}\label{1}$$
where $\epsilon, \eta, \kappa, \rho, \sigma$ are defined as determinants of matrices which are functions of the $f_i,g_i,h_i,d_i$ e.g.
$$ \kappa = \begin{vmatrix} d_1 & h_1 \\ d_2 & h_2 \end{vmatrix} $$
with the other parameters defined in the paper as Eqs B.19 - B.21.
My question is how does the author take this step and eliminate $L$ and arrive at Eq. 1? Is there a way to formulate this as some matrix problem? I fail to see where the determinants have come from!
Thanks in advance for any help
Multiply the first equation by $h_2$ and the second equation by $h_1$ and subtract gives \begin{eqnarray*} (f_1h_2-f_2h_1)E^2-2(g_1h_2-g_2h_1)EL-(d_1h_2-d_2h_1)=0. \end{eqnarray*} Similarly multiply the first equation by $g_2$ and the second equation by $g_1$ and subtract gives \begin{eqnarray*} (f_1g_2-f_2g_1)E^2-(h_1g_2-h_2g_1)L^2-(d_1g_2-d_2g_1)=0. \end{eqnarray*} Now rearrange the first equation (in this answer) to \begin{eqnarray*} 2(g_1h_2-g_2h_1)EL =(f_1h_2-f_2h_1)E^2-(d_1h_2-d_2h_1) \end{eqnarray*} and square this and eliminate $L^2$ by rearrange the second eqaution to $(h_1g_2-h_2g_1)L^2=?$.
This will give a quadratic in $E^2$ as required.