Can some advise me on how to solve this system of equations?

Question

I have the following system of 3 equations and 3 unknowns: $$c_{0} = \frac{x_0}{x_0 + x_1},\ \ c_{1} = \frac{x_1}{x_1 + x_2},\ \ \ c_{2} = \frac{x_2}{x_2 + x_0},$$ where all $c_i\!\in\!(0,1)$ are known and all $x_i > 0$ are unknown. Am I right in that the solution of this system is the nullspace of the following matrix? $$\mathbf{A}=\left[\begin{matrix}(c_0-1)& c_0 & 0 \\ 0 & (c_1-1) & c_1 \\ c_2 & 0 & (c_2-1) \end{matrix}\right].$$ If so, I want to find the non-trivial solution, i.e. the basis for $null(\mathbf{A})$.

p.s. I have attempted to simplify $\mathbf{A}$ to its reduced row echelon form $rref(\mathbf{A})$. I know that $null(\mathbf{A}) = null(rref(\mathbf{A}))$, but I get a diagonal matrix for $rref(\mathbf{A})$. So does this mean that $null(\mathbf{A}) = \mathbf{0}$, and therefore, there are no solutions to the system?

Answer 1

We rewrite the equations as a system of linear equations $Ax=0$ with $$ A=\begin{pmatrix} c_0-1 & c_0 & 0 \\ 0 & c_1-1 & c_1 \\ c_2 & 0 & c_2-1& \end{pmatrix} $$ and $x=(x_0,x_1,x_2)^t$. The nullspace of $A$ is non-trivial if and only if $\det(A)=0$. We have $$ \det(A)=2c_0c_1c_2 - c_0c_1 - c_0c_2 + c_0 - c_1c_2 + c_1 + c_2 - 1. $$ The trivial solution $x=0$, i.e., $x_0=x_1=x_2=0$ is forbidden by you.

So let us suppose that $\det(A)=0$ and $2c_0c_1 - c_0 - c_1 + 1\neq 0$. Then we can express $c_2$ by $c_0$ and $c_1$ and then $\ker(A)$ is spanned by

$$ \begin{pmatrix} c_1-1\\ -c_0 \\- 2c_0c_1 + c_0 + c_1 - 1 \end{pmatrix}.$$ Similarly for the other cases.

Answer 2

If $(x_0,x_1,x_2)$ is a solution, so is $\lambda(x_0,x_1,x_2)$. So we may choose $x_0$ arbitrarily and draw

$$x_1=\frac{1-c_0}{c_0}x_0,\\x_2=\frac{c_2}{1-c_2}x_0.$$