Given is markov chain - Determine the probability $f_1(n)$ 
Given is markov chain $\left\{X_n\right\}_{n \in \mathbb{N}}$ with
  transition probabilities
$$M= \begin{pmatrix} 
1/2   &   1/2   &   0   &   0   &   0   &   0   \\  
1/4   &   3/4   &   0   &   0   &   0   &   0   \\  
1/4   &   1/4   &   1/4 &   1/4 &   0   &   0   \\  
1/4   &     0   &   1/4 &   1/4 &   0   &   1/4 \\  
0     &     0   &   0   &   0   &   1/2 &   1/2 \\ 
0     &     0   &   0   &   0   &   1/2 &   1/2 
\end{pmatrix}$$
Determine the probability $f_1(n)$ where you return to state $1$ after
  $n$ steps  (for the first time).

I'm not sure how you can solve this because if I understood it correctly, we have $n$ steps and we are looking for a probability, so we have two unknowns...
Anyway, I think the correct way of calculating it is (don't miss the little exponent $n$ of that huge matrix!)
$$f_1(n) = \begin{pmatrix} 
1/2   &   1/2   &   0   &   0   &   0   &   0   \\  
1/4   &   3/4   &   0   &   0   &   0   &   0   \\  
1/4   &   1/4   &   1/4 &   1/4 &   0   &   0   \\  
1/4   &     0   &   1/4 &   1/4 &   0   &   1/4 \\  
0     &     0   &   0   &   0   &   1/2 &   1/2 \\ 
0     &     0   &   0   &   0   &   1/2 &   1/2 
\end{pmatrix}^n \cdot 
\begin{pmatrix}
1\\ 
0\\ 
0\\ 
0\\ 
0\\
0
\end{pmatrix}$$
Here I'm stuck again for the same reason, I see no way of calculating the probability...? : /
 A: Note that if the chain starts at state $1$ then it can return to the state in one step, i.e., $f_1(1) = 1/2$. Consider now that the chains goes to state $2$, once $X_n = 2$, the number of steps until the chain returns to state $1$ is distributed geometrically with $p=1/4$, hence 
$$
f_1(n) = \frac{1}{2}I\{n=1\}+\frac{1}{2}\left(\frac{3}{4} \right)^{n-2}\frac{1}{4}I\{n\ge2\}
$$
A: You need this:
$$
f_1(n) = \begin{pmatrix}
1, &
0, &
0, &
0, &
0, &
0
\end{pmatrix} \begin{pmatrix} 
1/2   &   1/2   &   0   &   0   &   0   &   0   \\  
1/4   &   3/4   &   0   &   0   &   0   &   0   \\  
1/4   &   1/4   &   1/4 &   1/4 &   0   &   0   \\  
1/4   &     0   &   1/4 &   1/4 &   0   &   1/4 \\  
0     &     0   &   0   &   0   &   1/2 &   1/2 \\ 
0     &     0   &   0   &   0   &   1/2 &   1/2 
\end{pmatrix}^n \cdot 
$$
What tells you it's done that way is that the rows add up to $1$ and the columns don't.
Then you can see that you have a Markov chain in which, once you're in state $1$ or state $2,$ you can never get to other states than $1$ and $2.$ And that means you only need to pay attention to the first two rows and the first two columns.
