Let's assume that there is a markov chain with a transition matrix $P$: $\begin{bmatrix} 0 &0 &0 &\frac{1}{2} & \frac{1}{2} & 0\\ 0& 0& 0& \frac{1}{2}& \frac{1}{2} & 0\\ 0& 0& \frac{1}{2}& 0 & \frac{1}{2} & 0\\ \frac{1}{3}& \frac{1}{3} & 0 & 0 & 0 & \frac{1}{3}\\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$ Write probability distribution after seven steps assuming that inital vector is: $\bar{p}(0)=(0,0,0,1,0,0)$
So my goal is to get $\bar{p}(7)=(x_{1},x_{2},x_{3},x_{4},x_{5},x_{6})$ according to the matrix I can say that $\bar{p}(7)=(x_{1},x_{2},0,0,0,\frac{1}{3})$ because, there is no way to go from state $4$ to states $5,4,3$. I have a little trouble with $x_{1},x_{2}$, but I can do that in two ways.
First: I can raise first and second column of the matrix to the power of seven and use: $\bar{p}(7)= \bar{p}(0) \cdot P^{7}$
Second: I can use: $\bar{p}(n)= \bar{p}(n-1) \cdot P$, and find $\bar{p}(1),\bar{p}(2),..,\bar{p}(6)$ and count it from $\bar{p}(7)=\bar{p}(6) \cdot P$
I think that there is much simpler solution that my approaches and I would be very grateful to show me how to do that quicker.
You must raise $P$ to the power 7. This is not as easy as you seem to think. For instance, state 4 does not go to states 5,4,3 directly, but it can go indirectly after many steps.
Your matrix has some particularities. No state goes into state 3; state 7 is isolate. But this is not enough.
I have computed $P^7$ using Maple, and the answer is horrible.