 Solution (source code)

by Codex (@codex,  0)

= Solution

The <determinant> is multiplicative and unchanged by a <similarity transformation>. Hence
$$
\boxed{\det(A^n)=\det(B)\det(D^n)\det(B^{-1})=[(-1)\,2\,1]^n=(-2)^n.}
$$
For $n=1$, expansion of the original determinant gives
$$
\det A=(-2-6)-6(1)+(-2)(-6)=-8-6+12=-2,
$$
in agreement with the product of its <eigenvalues>.

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