Prove that if $A,B$ are equivalent, then $A^T,B^T$ are 
I know the definition.
Def of Equivalent Matrices :
Two matrices A and B are said to be equivalent if one can be obtained from the other by a sequence of elementary row and column operations.
But how to prove it? Anybody please help. 

Since $A,B$ are equivalent, it holds that there is an invertible $n\times n$  matrix $P$ and an invertible $m\times m$ matrix $Q$ such that:
$$ B = Q^{-1} A P  .$$


*

*From the above equality, we have:
$$ B^T = \left(Q^{-1} A P  \right)^T  = P^T A^T \left(Q^{-1}\right)^T = K^{-1} A^T \left(Q^{-1}\right)^{T}, \quad \text{where } K =\left(P^T\right)^{-1}.$$

*It is not true that $A^2, B^2$ are always equivalent. 

Proposition: Two matrices are equivalent if and only if their rank is the same.

Consider the matrices $A = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$ and $B= \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix},$ which are equivalent (they have the same rank equal to $1$; if you want to follow the previous definition pick $P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $ and $Q = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$). 
But $A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}$ and $B^2 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.$ We notice that $\operatorname{rank}A^2 = 0$, but $\operatorname{rank}B^2 = 1.$


*

*It is not always true. Consider the previous matrices $A,B$. 
Notice that $$AB = \begin{bmatrix} 0 & 0 \\  0 & 0 \end{bmatrix} \text{  and } BA = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}.$$ 



UPDATE: Since you refer to non - singular matrices, I suppose that both matrices $A,B$ are $n$ by $n$ and wlog, suppose that $A$ is non - singular, which means $\det A \neq 0.$


*

*It is true that $B = Q^{-1} A P \implies \det (B) = \det(Q^{-1}) \cdot \det A \cdot \det P \neq 0.$


What we need is that $\det(AB) = \det A\cdot \det B = \det(BA) \neq 0,$ hence $\operatorname{rank}(AB) = \operatorname{rank}(BA).$


*

*The second case has already been covered.

