How to solve the equation having multiple variables? I have the following set of equations:
$A_{1} = M_{1}*B_{11} + M_{2}*B_{12}~~~~~~~~~~~~......~~Eqn.(1)\\
A_{2} = M_{1}*B_{21} + M_{2}*B_{22}~~~~~~~~~~~~......~~Eqn.(2)\\
:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~:\\
A_{n} = M_{1}*B_{n1} + M_{2}*B_{n2}~~~~~~~~~~~......~~Eqn.(n)$
where $A_{1}$ to $A_{n}$ and $(B_{11}, B_{12})$ to $(B_{n1}, B_{n2})$ are the known values all belonging to $\mathbb{R}$. Also, $M_{1}$ and $M_{2}$ are the multivariate equations as shown below:
$M_{1} = (X_{9} X_{5} X_{1}) + (X_{9} X_{6} X_{3}) + (X_{10} X_{7} X_{1}) + (X_{10} X_{8} X_{3})$, and
$M_{2} = (X_{9} X_{5} X_{2}) + (X_{9} X_{6} X_{4}) + (X_{10} X_{7} X_{2}) + (X_{10} X_{8} X_{4})$
Here, $X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, X_{6}, X_{7}, X_{8}, X_{9}$ and $X_{10}$ are the variables in $\mathbb{R}$. We can also represent these equations in the matrix form as shown below:
$M_{1} = \begin{bmatrix}X_{9} & X_{10}\end{bmatrix}\begin{bmatrix}X_{5} & X_{6}\\X_{7} & X_{8}\end{bmatrix}\begin{bmatrix}X_{1} \\ X_{3}\end{bmatrix}\\
M_{2} = \begin{bmatrix}X_{9} & X_{10}\end{bmatrix}\begin{bmatrix}X_{5} & X_{6}\\X_{7} & X_{8}\end{bmatrix}\begin{bmatrix}X_{2} \\ X_{4}\end{bmatrix}$
Now, I can get the values of $M_{1}$ and $M_{2}$ by solving the linear equations $Eqn. (1)$ to $Eqn. (n)$ shown above. Let's say, after solving those equations, I get the value of $M_{1} = K_{1}$ and the value of $M_{2} = K_{2}$, where $K_{1}, K_{2} \in \mathbb{R}$. Thus, now we can write as follows:
$K_{1} = (X_{9} X_{5} X_{1}) + (X_{9} X_{6} X_{3}) + (X_{10} X_{7} X_{1}) + (X_{10} X_{8} X_{3})$, and
$K_{2} = (X_{9} X_{5} X_{2}) + (X_{9} X_{6} X_{4}) + (X_{10} X_{7} X_{2}) + (X_{10} X_{8} X_{4})$
Now, my question is that what do we exactly call these above type of equations and how do we get the values of the variables $X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, X_{6}, X_{7}, X_{8}, X_{9}$ and $X_{10}$ by solving these equations. 
What do I think is that there are no unique values for these variables, but we can get a solution set out of many existing ones. If I am right, we may take help of non-convex optimization methods. But, as I am new to this domain I don't know methods to solve these type of problems.
Thus, I request the answer to this question from the community.
 A: You basically need to solve
$$
\underbrace{(K_1,K_2)}_{v^T}=\underbrace{(X_9,X_{10})}_{z^T}
\underbrace{\pmatrix{X_5&X_6\\ X_7&X_8}}_{X}
\underbrace{\pmatrix{X_1&X_2\\ X_3&X_4}}_{Y},
$$
or in a more abstract form, $v^T=z^TXY$, where the $z$ and the matrices $X,Y$ on the RHS are unknown. This should be easy. When $v\ne0$, there are three (possibly overlapping) classes of solutions according to the ranks of $X$ and $Y$:


*

*$X$ and $Y$ are any two invertible matrices and $z^T = v^TX^{-1}Y^{-1}$;

*$Y$ is any invertible matrix, $X=xv^TY^{-1}$ for some vector $x\ne0$ and $z$ is any vector such that $z^Tx=1$;

*$Y=yv^T$ for some vector $y\ne0$, $X$ is any invertible matrix and $z$ is any vector such that $z^TXy=1$.


When $v$ is known to be zero, the above three classes of solutions are still valid and there is one additional class of solutions:


*

*$Y=y_1y_2^T$ for some vectors $y_1$ and $y_2$, $X$ is any matrix and $z$ is any vector such that $(z^TXy_1)y_2^T=0$ (so, if $y_2\ne0$, we need $z\perp Xy_1$, otherwise $z$ is arbitrary).

