Do the set of all non-singular matrices forms a group under multiplication. ?
I don't think it does as the multiplication operator is not even defined for every two element in the set .
Just need a confirmation.
EDIT 1 : Non-singular means that determinant is not equal to zero.
Best Answer
If by "non-singular" you mean that the determinant is $\neq 0$, then the answer is yes, since you are dealing with square matrices and $\det (AB)=\det A \det B$. If you mean that the rank is maximal and you are considering $m\times n$-matrices, then the answer is no because multiplication is not even defined unless $m=n$.