The <inverse metric> condition is
$$
(\eta^{\alpha\rho}+k^{\alpha\rho})
(\eta_{\rho\beta}+h_{\rho\beta})
=\delta^\alpha{}_\beta.
$$
Keeping only terms linear in the <metric perturbation> gives
$$
k^\alpha{}_\beta+h^\alpha{}_\beta=0.
$$
Thus the <linearized inverse metric> is
$$
\boxed{k^{\alpha\beta}=-h^{\alpha\beta}},
\qquad
\boxed{g^{\alpha\beta}
=\eta^{\alpha\beta}-h^{\alpha\beta}+O(\epsilon^2)},
$$
where indices on $h_{\alpha\beta}$ are raised with the Minkowski metric.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 357 / 1 / a / Solution

If $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$, then $g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}+O(h^2)$, with indices on $h$ raised using the background metric.


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