Why adjugate matrix 2x2 is different from 3x3 and others? I've understand the simple way of calculating the adjugate matrix.
In short:
1). We need to calculate all cofactors:
Using the next formula: $A_{ij} = (-1)^{i+j} M_{ij}$
2). Transpose it.
It's also can be read from Wikipedia:
https://en.wikipedia.org/wiki/Adjugate_matrix

In linear algebra, the adjugate, classical adjoint, or adjunct of a
  square matrix is the transpose of the cofactor matrix.

If to view examples, such short algorithm is correct for squared matrices 3x3 and larger...
But, for 2x2 is just a rule:
M = [ a b ]
    [ c d ]

adj( M ) = [  d -b ]
           [ -c  a ]

What do I dislike?
I dislike that, because it's some kind of prayer. And many forums in web are saying:

Just remember it? right?

Damn! People, it's a MATH, it's not some humanity science like history. It's A MATH, math consists of logic and explanation as any other technical stuff.
I don't want just to remember stupidly some formula, I want understand why is exactly using such adjugate calculation especially for matrix 2x2?
I don't understand the next... For the matrix 3x3 we calculate all cofactors and then transpose it, for e.g.:
Original:
1   2   3
2   5   4
5   2   3

Cofactor matrix:
  7  14 -21
  0 -12  8
 -7   2  1

Transposed cofactor matrix:
  7   0 -7
 14 -12  2
-21   8  1

As you can see the transpose process for 3x3 didn't exchange the value a11 and a33, but why there is such an exchange for 2x3 matrix? I don't understand... I want to get logical explanation, not just a strict prayer.
I want to hear explanation why? and such explanation must be logical.
 A: Consider what happens when you're getting the matrix of minors for a $3\times 3$: You are getting the value for the $a_{ij}$ element by calculating the determinant of the non $i,j$ rows (I won't go through this since, judging by your question, you know how to do this already).
Now consider the same logic applied to a $2\times 2$ matrix: Do the same steps, except instead of having a $2\times 2$ matrix to calculate the determinant from, you have a $1\times 1$. In this case, the determinant is the single element in that matrix. From this, you can do the same steps as you would for a $3\times 3$.
I'll run through an example here (I'll compare with a $3\times 3$ matrix since that's the simplest matrix where these rules are first introduced):
Matrix $A = \begin{bmatrix} a & b\\ c & d\end{bmatrix}$.
Matrix of minors $= \begin{bmatrix} d & c\\ b & a\end{bmatrix}$ - for element $(1,1)$ (i.e. $a$), "cover" row $1$ and column $1$ and you're left with $d$ so that equals the determinant for that element for the matrix of minors (as with a $3\times 3$).
Cofactor matrix, $C = \begin{bmatrix} d & -c\\ -b & a\end{bmatrix}$ - exact same as for a $3\times 3$.
$\operatorname{adj}A = C^T = \begin{bmatrix} d & -b\\ -c & a\end{bmatrix}$ - again, exact same as for a $3\times 3$.
This is where you get the "Just remember it" from. It follows the exact same steps as for any $n\times n$ matrix.
Then you calculate $A^{-1}$ as usual: $A^{-1} = \frac{1}{\det A}\operatorname{adj}A$.
Hope this helps.
