Can this systems of equations be solved?

Question

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 $y$?

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

Answer 1

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.

Answer 2

After pie314271's answer

The equation $$b^2(x^2-a)+c(x^2-a)^2=b^2x^2$$ reduces to $$a\left(a c- b^2\right)-2 a c x^2+c x^4=0$$ which is just a quadratic in $x^2$ and the solutions are given by $$x=\pm\sqrt{a\pm\frac{\sqrt{a}}{\sqrt{c}}\,b}$$ So, the whole set of solutions is $$\left\{x= -\sqrt{a-\frac{\sqrt{a} b}{\sqrt{c}}},y= \frac{\sqrt{c} \sqrt{a-\frac{\sqrt{a} b}{\sqrt{c}}}}{\sqrt{a}}\right\},\left\{x= \sqrt{a-\frac{\sqrt{a} b}{\sqrt{c}}},y= -\frac{\sqrt{c} \sqrt{a-\frac{\sqrt{a} b}{\sqrt{c}}}}{\sqrt{a}}\right\},\left\{x=-\sqrt{a+\frac{\sqrt{a} b}{\sqrt{c}}},y= -\frac{\sqrt{c} \sqrt{a+\frac{\sqrt{a} b}{\sqrt{c}}}}{\sqrt{a}}\right\},\left\{x= \sqrt{a+\frac{\sqrt{a} b}{\sqrt{c}}},y=\frac{\sqrt{c} \sqrt{a+\frac{\sqrt{a} b}{\sqrt{c}}}}{\sqrt{a}}\right\}$$