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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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What do you know about $A$,$B$,$C$...$L$? because for instance, if $A$ and $B$ are zero matrices, your thing is easy to solve. It'll all depend on the properties of your matrices –  Patrick Da Silva Nov 26 '12 at 8:52
    
Is it a numerical problem (i.e. you look for a solution algorithm) or a theoretical problem (i.e. you look for a closed or semi-closed form solution)? Do you have some specific $A,B,C,I,J,K,L$ in mind? –  user1551 Nov 26 '12 at 8:58
    
@PatrickDaSilva A,B,C,I,J,K and L are 2x2 non-singular matrices follows properties of ABCD Chain Matrix. –  Salman Nov 26 '12 at 9:10
    
@user1551 If i am able to solve this theoretical problem, then i could be able to solve my practical problem. –  Salman Nov 26 '12 at 9:12

1 Answer 1

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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I am having no problem ,until this step. But problem is finding X and Y matrices (Individually), not there product –  Salman Nov 26 '12 at 8:58
    
@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 $X' = WY'^{-1}$. –  Patrick Da Silva Nov 26 '12 at 9:29

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