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If I have a system of four equations and i need to get ratio $E/A$ out of it. How can i do it? $\mathcal K$, $\mathcal L$ and $d$ are conctants, $A$ is independant variable where $B,C$ and $D$ are dependant variables.

\begin{align} A + B &= C + D & Ce^{\mathcal K d} + De^{-\mathcal K d} &= E e^{i \mathcal L d}\\ i \mathcal L A - i \mathcal L B &= \mathcal KC - \mathcal K D & \mathcal K C e^{\mathcal K d} - \mathcal K D e^{-\mathcal K d}&= i \mathcal L E e^{i \mathcal L d} \end{align}

EDIT: I know how to write this in a matrix form as well: \begin{align} \begin{pmatrix}-1 & 1 & 1 & 0 \\ i \mathcal L & \mathcal K & -\mathcal K & 0 \\ 0 & e^{\mathcal Kd} & e^{-\mathcal Kd} & -e^{i\mathcal Ld} \\ 0 & \mathcal Ke^{\mathcal Kd} & -\mathcal Ke^{-\mathcal Kd} & -i\mathcal Le^{i\mathcal Ld}\end{pmatrix} \begin{pmatrix}B \\ C \\ D \\ E\end{pmatrix}=\begin{pmatrix}A \\ i\mathcal LA \\ 0 \\ 0\end{pmatrix} \end{align}

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You should specify what $A,B,C,D,i,\mathcal{L},\mathcal{K}$ and $d$ is. –  Stefan Hansen Apr 25 '13 at 6:48
    
$\mathcal K$, $\mathcal L$ and $d$ are conctants, $A$ is independant variable where $B,C$ and $D$ are dependant variables. –  71GA Apr 25 '13 at 6:59
    
71GA: By variable you mean they vary in $\mathbb{R}$? Why did you use the tag matrices? –  Stefan Hansen Apr 25 '13 at 7:24
    
Yes they vary in $\mathbb{R}$ and i used matrices, because i know that my system can be written in a matrix form (check my edit). –  71GA Apr 25 '13 at 7:27

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