The matrix I mentioned below is irreducible and primitive or isn't? 
$$
\left[
\begin{array}{@{}ccccc@{}}
0.9& 0.1& 0& 0& 0& 0& \\
0& 0.9& 0.1& 0& 0& 0& \\
0& 0& 0.9& 0& 0& 0.1& \\
0& 0& 0& 0.9& 0.1& 0& \\
0& 0& 0& 0.1& 0.9& 0& \\
0.1& 0& 0& 0& 0& 0.9& \\
\end{array}
\right]
$$

I think this matrix is irreducible and primitive but after I looked properties of irreducible i was confused and don't know how can to prove that the matrix irreducible and primitive
 A: As every state here is pathwise connected to any other state, the matrix is irreducible. Also, as every state is accessible from itself, it has to be aperiodic and hence the matrix is primitive.
A: You have to look at the powers of your matrix. A matrix is primitive when for some power $k\in \mathbb{N}_0$, all entries of the matrix are strictly positive. A matrix is irreducible when for each entry, there is a power $k\in \mathbb{N}_0$ such that that entry for the power of the matrix is strictly positive. Primitive matrices are always irreducible but not the other way around.
Take for instance a $n\times n$-matrix with $0$ on the diagonal, $1$ everywhere else. That matrix is primitive. It is also therefore irreducible. Check this for yourself by computing powers.
On the other hand a $n\times n$-matrix with 1's on the $(k,k+1)$ for $k\in{1,\ldots,n}$ and $(n,1)$ entries of the matrix but zeros everywhere else is only irreducible. Check again by computing powers.
A: As for seeing the matrix is primitive, recall the definition. A matrix is primitive if there is $k \gt 0$ such that $P^k$ has all strictly positive entries. As my friend said Primitive matrices are always irreducible!
When we draw the directed graph corresponding to the matrix, then your matrix is irreducible if it is possible to travel from any node to any other node and Your example not fails this condition because as it is possible to travel away from node.
