could we also calculate the number of permutations we have to do,to get the matrix $I$? I am given the matrix $$B= \begin{pmatrix}
0 & 0 &0  & 1\\ 
 0& 0 & 1 &0 \\ 
1 &0  &0  &0 \\ 
0 & 1 & 0 & 0
\end{pmatrix}$$
and I want to find its order. So, I have to find the minimum $n \in \mathbb{N}$ such that $B^n=I$.
Instead of this way,could we also calculate the number of permutations we have to do,to get the matrix $I$?
 A: Rather than the number of permutations, you have to compute the order of the permutation which corresponds to this matrix. The permutation in question is 
$$\begin{pmatrix} 
1 & 2 & 3 & 4\\
3 & 4 & 2 & 1
\end{pmatrix},$$
which is the cycle $(1324)$, which has order $4$. Hence, the $n$ in question is $4$.
A: One way of doing this would be to let the matrix act on the vector $\langle 1, 2, 3, 4 \rangle$, recognizing that the matrix will simply permute the entries of this vector.  
When we let the matrix act on the vector once, we get the vector $\langle 4, 3, 1, 2 \rangle$.  The resulting permutation is a $4$-cycle in $S_4$, which has order $4$.  Therefore, $A^4 = I$.

Side note:
In general, if you are given the set $M$ of all $n \times n$ matrices whose rows are composed of the standard basis vectors $e_1, e_2, ..., e_n$, then there is an isomorphism $\phi:M \rightarrow S_n$.  See this article.
A: The characteristic polynomial is:
$$p(\lambda)=\det(B-\lambda I)=\left| \begin{matrix}
-\lambda & 0 &0  & 1\\ 
 0& -\lambda & 1 &0 \\ 
1 &0  & -\lambda  &0 \\ 
0 & 1 & 0 & -\lambda
\end{matrix}\right|=\lambda^4-1$$
So by Cayley–Hamilton theorem:
$$p(B)=0$$
$$B^4-I=0$$
$$B^4=I$$
