 Solution

Write $B=[b_1,\dots ,b_n]$. Since $AB=[A{b}_1,\dots ,A{b}_n]$ and $b\mapsto Ab$ is a linear map, the multilinearity of the determinant gives, for any column position $j$,If $b_i=b_j$ for $i\ne j$, the corresponding columns $A{b}_i,A{b}_j$ also agree, so $D(B)=0$. Thus $D$ is an alternating multilinear map of the columns of $B$, even when $A$ is not invertible.