Solve linear algebra expressions I have three known, non-singular matrices $\mathbf{A} \in \mathbb{C}^{3\times 3}$, $\mathbf{B} \in \mathbb{C}^{3\times 3}$, and $\mathbf{C} \in \mathbb{R}^{3\times 3}$ and one unknown matrix $\mathbf{X} \in \mathbb{C}^{3\times 3}$ (also non-singular).  They are related to each other by the following two equations:
$\mathbf{A} = (\mathbf{C} + \mathbf{X})^H(\mathbf{C} + \mathbf{X})$
$\mathbf{B} = (\mathbf{C} + \mathbf{X})(\mathbf{C} + \mathbf{X})^H$
I am hoping to solve for $\mathbf{X}$.  Obviously, in general there are infinite solutions. (E.g. if $\mathbf{A} = \mathbf{B} = \mathbf{I}$ then any $\mathbf{X}$ that makes $(\mathbf{C}+\mathbf{X})$ unitary will work.)  So I will add the following constraints: $\mathbf{A}$ and $\mathbf{B}$ are not diagonal matrices. Also, $\mathbf{X}$ is not a hermitian matrix, and each of its elements is non-zero. (If it is necessary, it would also be possible for me to manipulate the situation to ensure a symmetric $\mathbf{C}$, but I am hoping that I do not have to do this.)
So, with these added constraints, is there a unique solution for $\mathbf{X}$? If the solution is not unique, are there a finite number of solutions?  In either case, is there a closed form description of the solution(s)?
Any insight into this situation would be helpful.  If possible, please formulate answers for non-mathematicians such as myself.
Much thanks.
 A: Sorry there was a typo. Fixed now. 
NOTE:
I think I have $A$ and $B$ reversed. Too much work to redo the math. I solved
$$
A = (C+X)(C+X)^H \\
B = (C+X)^H (C+X)
$$
Look at 
$$
A (C+X) - (C+X)B$$
This is just a linear equation in $X$
$$
AX - XB = CB-AC
$$
and can be solved in any number of ways (provided $A$ and $B$ do not share any eigenvalues).
Here is a worked out example. I am using real matrices for simplicity
Let
$$A=\pmatrix{-7&5&-5\cr -10&-4&9\cr -10&9&-6\cr },~~
B=\pmatrix{7&-4&-1\cr -1&-7&9\cr -4&-4&6\cr },~~
C=\pmatrix{3&-10&1\cr 2&-1&-1\cr 8&-4&5\cr }$$
Let 
$$X=\pmatrix{{\it x_{11}}&{\it x_{12}}&{\it x_{13}}\cr {\it x_{21}}&
 {\it x_{22}}&{\it x_{23}}\cr {\it x_{31}}&{\it x_{32}}&{\it x_{33}}
 \cr }$$
Then 
$$
A(C+X)-(C+X)B = 
\pmatrix{-\left(5\,{\it x_{31}}-5\,{\it x_{21}}-4\,{\it x_{13}}-
 {\it x_{12}}+14\,{\it x_{11}}+78\right)&{\cdot}&{\cdot}\cr 9\,
 {\it x_{31}}+4\,{\it x_{23}}+{\it x_{22}}-11\,{\it x_{21}}-10\,
 {\it x_{11}}+15&{\cdot}&{\cdot}\cr 4\,{\it x_{33}}+
 {\it x_{32}}-13\,{\it x_{31}}+9\,{\it x_{21}}-10\,{\it x_{11}}-100&
 {\cdot}&{\cdot}\cr }
= \pmatrix{0&0&0\cr 0&0&0\cr 0&0&0\cr }$$
I am only showing the first column
These are just linear equations can be readily solved to get
$$X=\pmatrix{-3&10&-1\cr -2&1&1\cr -8&4&-5\cr }$$
This shows that if there is a solution, this has to be it.
Substituting in the original equation, we see that $X$ does not solve the problem. So the original problem has no solution. I am trying to find the necessary condition for a solution to exist but can't find a simple characterization yet.
