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I am having a algebric problem in my thesis work. It is some how like this ...

I have to find $X$, $Y$, $X'$ and $Y'$, where these are unknown $2\times 2$-matrices and $A$, $B$, $C$, $I$, $J$, $K$ and $L$ are known $2\times 2$-matrices. \begin{align*} A \cdot X \cdot Y \cdot B &= I\\ A \cdot X \cdot Y' \cdot B &= J\\ A \cdot X \cdot Y \cdot C \cdot X' \cdot Y' \cdot B &= K\\ A \cdot X' \cdot Y' \cdot B &= L \end{align*} Real goal was to find $X$ and $Y$ matrices (individually), more equations are created to simplify problem and make knowns and unknowns equal.

It is somehow looks realistic, because right now I have 4 equations and 4 unknowns. Further equations can be generated by keeping 2 unknowns between $A$ and $B$.

Please can anyone say about it? Thanks

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    @user1551 If i am able to solve this theoretical problem, then i could be able to solve my practical problem.2012-11-26

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Let $U = XY$, $V = X Y'$, $W = X'Y'$. Your equations say $\eqalign{A U B &= I \cr A V B &= J \cr A U C W B &= K \cr A W B &= L\cr }$

which is four equations in three unknowns. Generically there will be no solutions. In fact, if $A$ and $B$ are invertible we must have $U = A^{-1} I B^{-1}$, $V = A^{-1} J B^{-1}$, $W = A^{-1} L B^{-1}$, and then the third equation says $I B^{-1} C A^{-1} L = K$, which may or may not be true.

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    @Salman : When you have found $U$, $V$ and $W$, then you can fix $X$ arbitrary but non-singular, solve for $Y$ and $Y'$ by letting $Y = X^{-1} U$, $Y' = X^{-1}V$ and then $X' = WY'^{-1}$.2012-11-26