Matrix Equation Imagine the question: If $K$ and $L$ are $2\times 2$ matrices (knowing all of their components) and $KM=L$, solve for the matrix $M$.
One simple solution is to set the components of $M$ as $x,y,z,w$ and find the product $KM$ etc. You will end up with two linear systems, from which you can calculate the unknown components.
I'm not looking for this solution though... I could multiply the equation with $K^{-1}$. So here comes the trouble for me: 
a) How can I find the matrix $K^{-1}$?
b) Will the original equation become $M=K^{-1}L$ or $M=LK^{-1}$? (because generally $AB\neq BA$)
 A: You can multiply both hands of $KM=L$ by $K^{-1}$ on the left to obtain
$$
M=K^{-1}L
$$
only when $K^{-1}$ exists.
It is very basic in the theory of matrices that $K^{-1}$ exists if and only if the determinant of $K$ is non-zero. This fact is true for square matrices of any degree (not just $2\times2$).
Once you know that $\det(K)\neq0$, the computation of $K^{-1}$ is standard.
When $K=\left(\begin{array}[cc]{} a & b\\ c & d\end{array}\right)$ we know that
$$
\det(K)=\delta=ad-bc
$$
and
$$
K^{-1}=\frac1\delta\left(\begin{array}[cc]{} d & -b\\ -c & a\end{array}\right)
$$
A: Yes you can multiply by $K^{-1}$ if $\det{K}\neq 0$.
In general, if $K$ is a $n\times n$ invertible matrix, we have the following formula:$$K^{-1}=\dfrac{1}{\det{K}}\text{Co}(K)^T$$ where $\text{Co(K)}$ is the comatrix of $K$ and $A^T$ is the transpose of $A$.
You can also find $K^{-1}$ by using $KK^{-1}=I_2$ (the identity $2\times 2$ matrix) and do the first methode you used to find $M$.
Then if you have $KM=L$ then $(K^{-1}K)M=K^{-1}L$ and so $M=K^{-1}L$.
