I have the following system of equations:
$$ 0 = \left[ (p_{31} - p_{11})^2 - \omega_{22} (p_{21} - p_{11})^2 \right] x^2_1 + \left[(p_{32} -p_{12} )^2 - \omega_{22} (p_{22} - p_{12})^2 \right] x_2^2+ \left[(p_{33} - p_{13})^2 - \omega_{22} (p_{23} - p_{13})^2 \right]x_3^2$$
$$0 = \left[ (p_{41} - p_{11})^2 - \omega_{33} (p_{21} - p_{11})^2 \right] x^2_1 + \left[(p_{42} -p_{12} )^2 - \omega_{33} (p_{22} - p_{12})^2 \right] x_2^2+ \left[(p_{43} - p_{13})^2 - \omega_{33} (p_{23} - p_{13})^2 \right]x_3^2 $$
$$0 = \left[ (p_{51} - p_{11})^2 - \omega_{44} (p_{21} - p_{11})^2 \right] x^2_1 + \left[(p_{52} -p_{12} )^2 - \omega_{44} (p_{22} - p_{12})^2 \right] x_2^2+ \left[(p_{53} - p_{13})^2 - \omega_{44} (p_{23} - p_{13})^2 \right]x_3^2 $$
The values for $\omega_{22}, \omega_{33}, \text{ and } \omega_{44}$ are obtained as follows:
$$P_1 = \begin{bmatrix} p_{21} - p_{11}, & ,p_{22} - p_{12} , & p_{23} - p_{13} ; \\ p_{31} - p_{11} , & p_{32} - p_{13} , & p_{33} - p_{13} ; \\ \vdots & \vdots & \vdots ; \\ p_{51} - p_{11} & p_{52} - p_{15} & p_{53} - p_{13} ; \\ \end{bmatrix} $$
$$\Sigma = \begin{bmatrix} x_1 & 0 & 0 ; \\ 0 & x_2 & 0 ; \\ 0 & 0 & x_3 \\ \end{bmatrix},$$
$$B = P_1*\Sigma*P_1'$$ and $$\Omega_ = B/B_{11}.$$ That is, $\Omega$ contains all the elements of the matrix B relative to the first entry B.
Taking the values of $\omega_{22}$, $\omega_{33}$, and $\omega_{44}$, and the $p_{ij}$s as given, and knowing the $p_{ij}$'s are positive and satisfy $p_{i0} = 1 - p_{i2} - p_{i2} - p_{i3}$ for $i=1,2,3,4,5$, and constants $p_{i0}$. How can I find the rank of the matrix $A$, where $A$ is the matrix representing the system of equations at the top as $Ax=0$?
I have created some $\Omega$ matrices from a given $P_1$ and a diagonal matrix $\Sigma$ and in all the cases the rank of $A$ was $2$, but I wonder if this is always the case. I am not sure how to prove this. Any hint would be highly appreciated!