How do I solve $\begin{pmatrix} 2 &1 \\ 1& 2 \end{pmatrix}X-X\begin{pmatrix} 1 &-1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} 1 &1 \\ 1&-1 \end{pmatrix}$. I think i need to use the inverse but I do not know really how.
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You can just perform matrix multiplication: for $$ X=\left( \begin{array}{cc} x_{1,1} & x_{1,2} \\ x_{2,1} & x_{2,2} \\ \end{array} \right) $$ you get $$ \left( \begin{array}{cc} x_{1,1}-x_{1,2}+x_{2,1}-1 & x_{1,1}+x_{1,2}+x_{2,2}-1 \\ x_{1,1}+x_{2,1}-x_{2,2}-1 & x_{1,2}+x_{2,1}+x_{2,2}+1 \\ \end{array} \right)=0 $$ which gives $$ \left\{x_{1,1}=\frac{7}{5},x_{1,2}=-\frac{1}{5},x_{2,1}=-\frac{3}{5},x_{2,2}= -\frac{1}{5}\right\} $$
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$\begingroup$ Oh thanks :D. How did you multiply them soo fast? $\endgroup$– GhostJan 4, 2017 at 18:56
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$\begingroup$ I would lie If I said I made it by hand :D $\endgroup$ Jan 4, 2017 at 18:57
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$\begingroup$ This is a particular case of Sylvester equation (en.wikipedia.org/wiki/Sylvester_equation) $\endgroup$ Jan 4, 2017 at 19:05
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$\begingroup$ the knowledge that this is a Sylvester equation does not help to solve it anyhow differently/faster. The solution above cannot be significantly simplified (which is in particular clear from the answer which is quite ugly). $\endgroup$ Jan 4, 2017 at 19:07