Computing the inverse of a permutation I didn't understand the permutation and of course, I got this question wrong.

Compute the inverse of the following permutation:
  $$
\begin{pmatrix} 
1&2&3&4&5&6\\
2&5&4&1&6&3
\end{pmatrix}
$$

Could you please help me with this question and provide an explanation?
Furthermore, is it possible to compute the direct permutation, not the inverse one? If yes, how?
Thank you in advance!
 A: The permutation
$$
\begin{pmatrix} 
1&2&3&4&5&6\\
2&5&4&1&6&3
\end{pmatrix}
$$
means that you have six objects, standing in a line, and the object in position 1 has moved to position 2, the object in position 2 has moved to position 5, and so on, up to the object in position 6 that has moved to position 3.
The inverse permutation is the permutation that moves the objects back where they started. The original permutation took the object that was in position 2 and  moved it to position 5, and the inverse permutation must move it back, so the inverse permutation  must move the object at position 5 back to position 2.  
Can you do the rest?
A: Hint: Let $\pi$ be your permutation. You know that $\pi^{-1}$ satisfies the condition $\pi^{-1}\circ\pi=id$. Now, you know that $\pi(1)=2$, and you know that $$(\pi^{-1}\circ \pi)(1)=id(1)$$
Now, try to simplify this equation. What do you get?
A: Another Hint
\begin{pmatrix} 
1&2&3&4&5&6\\
2&5&4&1&6&3
\end{pmatrix}
reads "1" goes to "2", "2" goes to "5", ...
if you want to invert that permutation "do" the reverse.
\begin{pmatrix} 
1&2&3&4&5&6\\
 &1& & &2& 
\end{pmatrix}
because "2" goes to "1", "5" goes to "2", ..., the rest is for you :-)
And I believe you mean the cycle notation for the permutation, if you are talking of a direct permutation?
