Solution
= Solution
Move the first $n-k$ columns of
$$
A=\begin{pmatrix}0&M\\N&0\end{pmatrix}
$$
past its last $k$ columns. This requires $k(n-k)$ column interchanges and produces the <block diagonal matrix>
$$
\begin{pmatrix}M&0\\0&N\end{pmatrix}.
$$
Each interchange reverses the determinant's sign, so part (a) gives
$$
\boxed{\det A=(-1)^{k(n-k)}(\det M)(\det N)}.
$$