Vary $g^{ac}g_{cb}=\delta^a_b$. The <product rule> gives
$$
(\delta g^{ac})g_{cb}+g^{ac}\delta g_{cb}=0,
$$
and multiplication by $g^{bd}$ yields the <metric variation>
$$
\boxed{\delta g^{ab}=-g^{ac}g^{bd}\delta g_{cd}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 309 / 2 / a / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-309/2/a/i/solution