Answer: D
As the matrices are non singular so their determinant $\neq 0.$ Hence, the inverse matrix always exist.
But for a group to be abelian it should follow commutative property. As, matrix multiplication is not commutative, $\langle A,*\rangle$ is a group but not an abelian group.
Two nonsingular matrix can not give a singular matrix after product. Therefore it satisfies closure property also.
Proof: Let $A$ and $B$ are nonsingular and $C$ is singular.
Claim: For any $A$, $B$ and $C$:- $AB = C$ is NOT possible.
I will prove using contradiction. Let it be possible to have $AB = C.$
$\begin{align} AB &= C \\ \implies \det(AB)&=\det(C)\end{align}$
$\qquad\;\; \det(AB)=0 \text{ [}\because \det(C)=0 \;\text{as}\; C\; \text{is singular}]$
$\implies \mid A \mid. \mid B \mid =0$
$\implies \mid A \mid =0 \text{ or } \mid B \mid =0$
Which is contradiction as $A$ and $B$ both are non singular.