If $A$ is a non- singular matrix and $AB= I$ , can we comment $B$ is inverse of $A$? The complete definition of inverse of a matrix is- if $AB=BA=I$, then $B$ is inverse of $A$.
But l want to  know that from $AB=I$, can we comment $B$ is inverse of $A$? Is it possible that $AB=I$ but $BA$ is not equal to $I$? 
[ I personally think it is impossible since the inverse of a matrix is unique. Please clarify if I am wrong.]
 A: 
The complete definition of inverse of a matrix is- if AB=BA=I, then B is inverse of A.

This actually is the definition of the inverse for any ring. If only one of $AB=I$ or $BA=I$ holds then we talk about partial inverses.

But l want to know that from AB=I, can we comment B is inverse of A? Is it possible that AB=I but BA is not equal to I?

Yes and no. Depends on the underlying ring.
Let $A,B\in\mathbb{M}_n(R)$ for some commutative ring $R$. If $AB=I$ then
$$1=\det(I)=\det(AB)=\det(A)\det(B)$$
showing that $\det(A)$ is invertible in $R$ and so $A$ is invertible. It's a theorem that $A$ is invertible if and only if $\det(A)$ is invertible. In particular $BA=I$.
For non-commutative rings the determinant is not well defined and moreover it may happen that $AB=I$ while $BA\neq I$. So over non-commutative rings there are partial inverses that are not inverses.
A: Edit: as noted in the comments, this is not necessarily true for matrices with coefficients in non-commutative rings. However, the following is a positive result in the subcase where $A, B \in M_n(\mathbb{k})$ with $\mathbb{k}$ a field. In particular, this holds in $M_n(\mathbb{C})$ 
Let $A, B \in M_n(\mathbb{k})$. If $A$ is non singular, it has an inverse. Now, left multiplying by $A^{-1}$, we go from
$$
AB = I
$$
to
$$
B = A^{-1}
$$
Let's weaken the hypothesis and only assume that $AB = I$, $A$ not necessarily invertible. In other words, the linear mappings induced by $A$ and $B$, defined as $f(x) = Ax$ and $g(x) = Bx$ verify $fg \equiv id$, so $g$ must be injective. Since we are in finite dimension, a linear mapping is injective iff it is bijective. Therefore $g$ is invertible, or equivalently, $B$ is invertible. By the original argument, $B = A^{-1}$.
A: It is not difficult to prove that if [B] is the inverse of [A] then [A][B]=[I} and ALSO (quite generally) [B][A]=[I].  That is, the operation of matrix multiplication of one square matrix by its inverse matrix inverse is always commutative - one of the few cases where this is generally true for matricies.
