Can this systems of equations be solved? There is a system of equations as follow:
$$\left[\begin{matrix}x\\ y\\ \end{matrix} \right] = \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]\left[\begin{matrix}\frac{1}{x}\\ \frac{1}{y}\\ \end{matrix} \right]$$
Given $ \left[\begin{matrix}a& b\\b & c\\\end{matrix} \right]$, can this system of equations be solved for $x$ and $y$?

And how to solve the extended case?
$$\left[\begin{matrix}x_1\\ x_2\\ \vdots \\x_n \end{matrix} \right] =
 \left[\begin{matrix}a_{11}& a_{12} & \dots &a_{1n}\\a_{12} & a_{22} & \dots  &a_{2n}\\ \vdots & \vdots & &\vdots \\a_{1n} & a_{2n} & \dots &a_{nn}\end{matrix} \right]\left[\begin{matrix}\frac{1}{x_1}\\ \frac{1}{x_2}\\ \vdots \\ \frac{1}{x_n}\\ \end{matrix} \right]$$

This is basically asking for $\frac{a}{x}+\frac{b}{y}=x$ and $\frac{b}{x}+\frac{c}{y}=y$. From the first equation, $y=\frac{xb}{x^2-a}$. You can substitute this into the second equation to get: $$\frac{b}{x}+\frac{c(x^2-a)}{bx}=\frac{xb}{x^2-a}$$
This should be translatable into a polynomial. However, the smallest polynomial that you can multiply this by to remove all denominators is $bx(x^2-a)=bx^3-abx$. Doing this gives $$b^2(x^2-a)+c(x^2-a)^2=b^2x^2.$$ 
This results in a quartic polynomial, which is always solvable, though tedious. There's probably a shortcut that I'm missing, but at least this is a solvable problem.
