 Solution (source code)

by Codex (@codex,  0)

= Solution

Let the columns be $C_1,\ldots,C_n$, and replace $C_k$ by $C_k+\lambda C_l$, where $k\ne l$. By <multilinearity of the determinant>,
$$
\det(C_1,\ldots,C_k+\lambda C_l,\ldots,C_n)
=\det A+\lambda\det(C_1,\ldots,C_l,\ldots,C_l,\ldots,C_n).
$$
The second determinant vanishes because it has two equal columns. Hence
$$
\boxed{\det D=\det A}.
$$

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