Why in these matrices are $AB=BA$ not equal? What is the logic behind them? We know that in matrices AB=BA.Why in this Matrices $A=\begin{bmatrix} -1 & 3\\
 2 & 0\end{bmatrix}$, $B=\begin{bmatrix} 1 & 2\\
-3 & -5\end{bmatrix}$ are not equal to $AB=BA$. WHY? This is matrix  of order $2\times 2$ for both $A$ and $B$.

You can think of matrices as linear transformations.  For example, let $T$ be the map which rotates a point ninety degrees counterclockwise.  It is the function from the plane $\mathbb{R}^2$ to itself given by the formula $(x,y) \mapsto (-y,x)$.  It is represented by the matrix $$A  =\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
in the sense that if you want to compute what $T$ does to a point $v = (x,y)$, just compute the matrix product $Av^{\textrm{tr}}$, where here $v^{\textrm{tr}}$ means $\begin{pmatrix} x \\ y \end{pmatrix}$.  
On the other hand, consider the function $S$ which takes a point $(x,y)$ and reflects it across the $x$-axis, giving you the point $(x,-y)$.  It is represented by the matrix $$B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ Under the interpretation of matrices as functions from the plane to itself, the composition $T \circ S$ (that is, $S$ followed by $T$) corresponds to the matrix product $AB$, and $S \circ T$ corresponds to the matrix product $BA$.
But it is intuitively obvious that $T \circ S$ and $S \circ T$ are not the same function: rotating a point ninety degrees counterclockwise, and then reflecting it across the $x$-axis, usually does not do the same thing as first reflecting the point across the $x$-axis, and then rotating it ninety degrees.  Since $T \circ S \neq S \circ T$, we cannot expect to have $AB = BA$, and you can indeed check that these matrices are not equal.
