Here is one of many possible proofs.
Notice that, by the last equation, we can perform the same elementary row operations
on B and BA simultaneously, and retain the equality.
Further note that row operations of the first kind: adding a multiple of one
row to another, dont change the determinant, on either side of the equation.
We first notice that the claim is easy if the matrix B is diagonal (has
all off diagonal elements equal to zero.)
For B diagonal we have (BA)_{ij }= (B_{ii})(A_{ij}).
We can factor B_{ii} from each row in evaluating the determinant,
getting BA=
(_{i}B_{ii})A;
but the product of the diagonal elements is the determinant of a diagonal matrix,
so that we get BA=BAin
this special case.
If we can reduce B to diagonal form by elementary row operations of the first
kind obtaining a diagonal matrix B, and perform the identical operations on
BA to form (BA) we then have, as desired: BA=(BA)=BA=BA.
You can always reduce any matrix B to diagonal form by elementary row
operations of the first kind unless B is singular and its determinant
is 0. In that case the columns of B do not span the entire space, so that the
columns of BA dont do so either, and BA must also be singular.