 Solution (source code)

by Codex (@codex,  0)

= Solution

Multiplying an $n\times n$ matrix by $\alpha$ scales every row, giving a factor $\alpha^n$ in its <determinant>. Also $|A^{-1}|=1/|A|$, since $AA^{-1}=I$. Hence
$$
\boxed{|\alpha A^{-1}|=\frac{\alpha^n}{|A|}}.
$$
Only $A$ needs to be invertible; the result remains valid when $\alpha=0$.

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