= Solution
{c}
Write $B=[b_1,\ldots,b_n]$. Since $AB=[Ab_1,\ldots,Ab_n]$ and $b\mapsto Ab$ is a <linear map>, the <multilinearity> of the <determinant> gives, for any column position $j$,
$$
D(b_1,\ldots,\alpha u+\beta w,\ldots,b_n)
=\alpha D(b_1,\ldots,u,\ldots,b_n)
+\beta D(b_1,\ldots,w,\ldots,b_n).
$$
If $b_i=b_j$ for $i\ne j$, the corresponding columns $Ab_i,Ab_j$ also agree, so $D(B)=0$. Thus $D$ is an <alternating multilinear map> of the columns of $B$, even when $A$ is not invertible.
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