Inverse of a matrix belonging to $GL(2,\mathbb{Z}_{11})$ $GL(2,\mathbb{Z}_{11})$ is a general linear group whose elements are $2\times 2$ matrices where every element of matrices belongs to the set $\mathbb{Z}_{11}$ (so modulo 11 arithmetic would take place) and determinant should non-zero for matrices.
So I have been given a matrix 
$$
A = \begin{pmatrix}
2 & 6 \\ 
3 & 5 
\end{pmatrix},
$$
where A belongs to $GL(2,\mathbb{Z}_{11})$ and I have to find it's inverse. So what confused me is the way the inverse would be solved ? What is process to find it's inverse ?
 A: Especially for a $2\times2$ matrix, there is an easy formula for the inverse:
$$
\left( \matrix{a & b \\ c& d} \right )^{-1} = \frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \right )
$$
In your case, that gives us 
$$
\left( \matrix{2 & 6 \\ 3& 5} \right )^{-1} = \frac{1}{2\cdot5-3\cdot6}\left( \matrix{5 & 5 \\ 8 & 2} \right ) = \frac{1}{3}\left( \matrix{5 & 5 \\ 8 & 2} \right )
= 4\cdot\left( \matrix{5 & 5 \\ 8 & 2} \right )
= \left( \matrix{9 & 9 \\ 10 & 8} \right )
$$
You can check it by just multiplying them.
To address a very valid point raised in the comments, let me explain why I write $\tfrac13=4$. It is important to note that by "$3$" and "$4$" I do not mean the natural numbers, but their residue class in $\Bbb Z_{11}$. And since $3\cdot 4 = 12=1 + 11$, we can see that $3\cdot4=1$ holds in $\Bbb Z_{11}$. Since $\Bbb Z_{11}$ is a field, we may divide by any nonzero element and the formula $\tfrac13=4$ is justified. Now, how to figure out that $\tfrac13=4$ holds in $\Bbb Z_{11}$? Since $11$ is rather small, in this case I did it by trial and error; just multiply $3$ with a couple of numbers and see if the residue modulo $11$ is $1$. When the prime number in question is much larger, the extended Euclidean algorithm is your best way to find the inverse of an element, however I feel that this may be a topic for another question.
