prove that matrix multiplication in a specific matrix gives equal numerators . I have a matrix, 
$$ A =\begin{pmatrix}
 1/2 & 1/2 \\
 1/4 & 3/4
\end{pmatrix}$$
and using MATLAB I observed that
$$ A^2 =\begin{pmatrix}
  3/8 & 5/8 \\
 5/16 & 11/16
\end{pmatrix}.$$
The equality of numerators in the diagonal seems to be a rule, so I want to prove that all $A^n$ are of form 
$$ A^n =\begin{pmatrix}
  \text{something} & a/2^{2n-1} \\
 a/2^{2n} & \text{something else}
\end{pmatrix}$$ 
I think I should use induction, but it seems that I should first prove that the elements in the matrix have a general formula or a relationship that would help me prove this.
 A: Start by noting that your matrix is of the form $\begin{pmatrix} a & 1-a \\ b & 1-b\end{pmatrix}$ for $0\le a,b \le 1$. Let $S$ be the set of all such matrices. It's simple to check that  $A,B \in S \implies AB \in S$. 
Conseqently $A^n = \begin{pmatrix} \alpha_n & 1-\alpha_n \\ \beta_n & 1-\beta_n\end{pmatrix}$ for some $(\alpha_n, \beta_n) \in [0,1] \times [0,1]$.
Proposition: For $n\ge 1, \alpha_{n+1} = \frac{1 - \beta_n}{2} , \beta_{n+1} = \frac{1 + \beta_n}{4}$.
Proof: Proceed inductively. You have already demonstrated the base case. We assume the hyposthesis for $n$. Note that the hyposthesis yields that $\alpha_n = 1 - 2\beta_n $.
Now, from matrix multiplication, 
$$\alpha_{n+1} = \frac{\alpha_n + \beta_n}{2} = \frac{1 - 2\beta_n + \beta_n}{2} = \frac{1 - \beta_n}{2}\\
\beta_{n+1} = \frac{\alpha_n + 3\beta_n}{4} = \frac{1 - 2\beta_n + 3\beta_n}{4} = \frac{1 + \beta_n}{4}$$
This establishes the hypothesis for all finite $n$. The result you desired is a simple consequence.

Incidentally, the set $S$ above is the set of $2\times 2$ right stochastic matrices. There's a vast theory of the same, and of time homogeneous Markov chains that you may want to look at if you're interested in such beasts.
