= Solution
{c}
Let $\mathbf r(u,v)$ parametrize $S$. Then $A\mathbf r$ parametrizes its image, with <tangent vectors> $A\mathbf r_u$ and $A\mathbf r_v$. An <orthogonal matrix> preserves <dot products>, so the <Lagrange identity for the cross product> gives, for any $\mathbf p,\mathbf q$,
$$
\begin{aligned}
\|A\mathbf p\times A\mathbf q\|^2
&=\|A\mathbf p\|^2\|A\mathbf q\|^2-(A\mathbf p\cdot A\mathbf q)^2\\
&=\|\mathbf p\|^2\|\mathbf q\|^2-(\mathbf p\cdot\mathbf q)^2
=\|\mathbf p\times\mathbf q\|^2.
\end{aligned}
$$
The <area element of a surface> is therefore the same pointwise before and after applying $A$. Integrating proves
$$
\boxed{\operatorname{Area}(A S)=\operatorname{Area}(S).}
$$
This includes <orthogonal transformations> with <determinant> $-1$: reversal of <orientation> does not affect the <Euclidean norm> in the <surface area> formula.
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