I am trying to solve this equation for $x$: $$M \cdot (1/x) = x$$
where $M$ is a non-negative, real-valued invertible $n x n$ matrix, and $x$ is a non-zero $n \times 1$ vector. The notation $1/x$ just denotes the element-wise reciprocal of $x$:
$$ 1/x = (1/x_1,1/x_2,...1/x_n)'$$
Is there a closed form solution to this equation? It seems so simple that I feel there must be, but I am stuck.
If it's not possible to give a closed form solution, is it at least possible to show that there must be some solution that is real and strictly positive? In other words, that there is at least one solution, $x^*=(x_1^*,...,x_n^*)$, where $x_i^*$ is real, and $x_i^*>0 \hspace{.2cm}\forall i \hspace{.2cm}$?
EDIT: To clarify, $M$ is a non-negative matrix.
Thanks!
Consider the case $2 \times 2$. Given an invertible matrix $$M = \left[ \begin{array}{cc} a & b\\ c & d\\ \end{array}\right],$$ we want to find a vector $(x,y) \in \mathbb{R}^2$ such that $$M = \left[ \begin{array}{cc} a & b\\ c & d\\ \end{array}\right] \left[ \begin{array}{c} \frac{1}{x}\\ \frac{1}{y}\\ \end{array}\right] = \left[ \begin{array}{c} x\\ y\\ \end{array}\right].$$
This gives the system $$\left\{ \begin{array}{c} \frac{a}{x} + \frac{b}{y} = x\\ \frac{c}{x} + \frac{d}{y} = y\\ \end{array}\right.$$ which can not have real solutions for some choices of the coefficients $a,b,c,d$. For example, consider $a,b,d < 0$ and $c = 0$.
If you are concerned about higher dimensions, it's still possible to come up with counterexamples, as showed in the comments.