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Consider the disturbed metric $\widetilde{g}_{\mu\nu}=g_{\mu\nu}+h_{\mu\nu}$ with $|h_{\mu\nu}|<1$, meaning the pertubation is small and we can use $g$ to raise and lower indices, then $\widetilde{g}^{\mu\nu}=g^{\mu\nu}-h^{\mu\nu}$ with $|h^{\mu\nu}|<1$. The sign change arises since we want to have $\widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu}=\delta_\mu^\nu$ as well as $g_{\lambda\mu}g^{\mu\nu}=\delta_\mu^\nu$. Neglecting the pertubation in second order, we have: \begin{equation} \widetilde{g}_{\lambda\mu}\widetilde{g}^{\mu\nu} =(g_{\lambda\mu}+h_{\lambda\mu})(g^{\mu\nu}-h^{\mu\nu}) =\delta_\lambda^\nu -\underbrace{g_{\lambda\mu}h^{\mu\nu}}_{=-h_\lambda^\nu} +\underbrace{h_{\lambda\mu}g^{\mu\nu}}_{=h_\lambda^\nu} +\mathcal{O}(h^2). \end{equation} Therefore we need to put $\delta g^{\mu\nu}=-h^{\mu\nu}$. Furthermore, we have: \begin{equation} A_{\mu\nu}B^{\mu\nu} =A^{\kappa\lambda}g_{\kappa\mu}g_{\lambda\nu}B^{\mu\nu} =A^{\kappa\lambda}B_{\kappa\lambda}. \end{equation}