You can also compute it with the following method. First note that $50 = 32 + 16 + 2$. Let $$A=\left(\begin{array}{cc}2&0\\2&1\end{array}\right).$$
Then $$A^{50} = A^{32} A^{16} A^2$$.
We can compute $A^{2^n}$ easier than $A^{50}$,
$$A^2 = \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)$$
$$A^4 = \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)\left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right)= \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right)$$
$$A^8 = \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right) \left( \begin{array}{cc} 16 & 0\\ 30 & 1 \end{array} \right) = \left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)$$
$$A^{16} = \left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)\left( \begin{array}{cc} 256 & 0\\ 510 & 1 \end{array} \right)=\left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)$$
and
$$A^{32} = \left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)\left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right)=\left( \begin{array}{cc} 4294967296& 0\\ 8589934590 & 1 \end{array} \right).$$
Finally $$A^{50} =\left( \begin{array}{cc} 4294967296& 0\\ 8589934590 & 1 \end{array} \right) \left( \begin{array}{cc} 65536& 0\\ 131070& 1 \end{array} \right) \left( \begin{array}{cc} 4 & 0\\ 6 & 1 \end{array} \right).$$
This isn't the best way to go about it, but it is a lot easier than multiplying $A$ to itself 50 times. You should probably persue @science's method to be honest.