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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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    Oh thanks :D. How did you multiply them soo fast? – 2017-01-04
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    I would lie If I said I made it by hand :D – 2017-01-04
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    This is a particular case of Sylvester equation (https://en.wikipedia.org/wiki/Sylvester_equation) – 2017-01-04
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    and how do i solve that system of equations? – 2017-01-04
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    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). – 2017-01-04
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    Three things. 1) I just wanted to say that this question belongs to a larger domain 2) I agree that for $2 \times 2$ matrices, it is not interesting, but in the case you deal with $n \times n$ matrices with $n \geq 4$, you will be happy to have the general formula (as given in Wikipedia article). 3) Moreover, an increasing number of software implement Sylvester equation as a specific function in their new releases (Mathematica, Matlab...). – 2017-01-04