Perturbation metric problem I know this is an already answered question, but I couldn't make head or tail of it, and it's bugging me. I know I'm probably asking a silly question, but please bear with me as I'm 14 and this is my first post. 
I saw this perturbation while studying up a bit of linearised gravity:
$$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} \tag{1}$$
and then the next line says:
" you raise the indices to get":
$$g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu}. \tag{2}$$
I have done a bit of tensor calculus, but I still couldn't understand how the obvious fact $g_{\mu\nu}\, g^{\mu\rho} = \delta^{\rho}_{\nu}$ leads to the second equation. I tried everything I could, but all I get is the first equation with raised indices different from the given ones, and a persistent + sign. Please explain and derive the second equation step by step, as if to an idiot.
 A: If you have $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$, with $|h_{\mu\nu}|\ll 1$ a perturbation, $\eta_{\mu\nu} = \text{diag}(-1,1,1,1)$, then you can simply perform a Taylor expansion to obtain the inverse, $$g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu}+ \mathcal{O}(h^2),$$ where the indices of $h^{\mu\nu}$ are raised by $\eta^{\mu\nu}$. This is easy to see if you take the diagonal elements first, the off-diagonal ones are of second order, negligible. You don't need to solve a matrix equation for this, because the inverse is linearly dependent on $\eta_{\mu\nu}=\eta^{\mu\nu}$, $h_{\mu\nu}$ and their products, hence it must be of the form $g^{\mu\nu} = a\eta^{\mu\nu} + b h^{\mu\nu} + \ldots$, where the ellipses imply linear dependencies of higher order and the coefficients are constants. Also, note that $h_{\mu\nu} \eta^{\mu\mu} \eta^{\nu\nu} = h^{\mu\nu}$, no summation implied, namely $h^{\mu\nu} \propto h_{\mu\nu}$.
Also, a straightforward way to calculate $\delta g^{\mu\nu}$ is by use of $\delta g^{\mu\nu} = - g^{\mu\rho} g^{\nu\sigma} \delta g_{\rho\sigma}$ with $\delta g_{\mu\nu} = h_{\mu\nu}$.
A: Note that:
$$
h^{\mu \nu} = \eta^{\mu \rho}\eta^{\nu \lambda} h_{\rho \lambda}
$$
Therefore, up to first order, we have:
\begin{equation}
\begin{aligned}
g^{\mu \nu}g_{\nu \sigma} & = (\eta^{\mu \nu} - h^{\mu \nu})(\eta_{\nu \sigma} + h_{\nu \sigma}) \\&
=\eta^{\mu \nu}\eta_{\nu \sigma} + \eta^{\mu \nu}h_{\nu \sigma} - \eta_{\nu \sigma} h^{\mu \nu} + \mathcal{O}(h^2) \\&
= \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} - \eta_{\nu \sigma} \eta^{\mu \rho}\eta^{\nu \lambda} h_{\rho \lambda} + \mathcal{O}(h^2) \\&
= \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} - \delta_\sigma^\lambda \eta^{\mu \rho} h_{\rho \lambda} + \mathcal{O}(h^2) \\&
= \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} -  \eta^{\mu \rho} h_{\rho \sigma} + \mathcal{O}(h^2) \\&
= \delta^\mu_\sigma + \eta^{\mu \nu}h_{\nu \sigma} -  \eta^{\mu \nu} h_{\nu \sigma} + \mathcal{O}(h^2) \\&
= \delta^\mu_\sigma + \mathcal{O}(h^2)
\end{aligned}
\end{equation}
which is what we require.
Edit in response to comments. The inverse of $g_{\nu \sigma}$, which is denoted by $g^{\mu \nu}$, is defined to satisfy the following equation:
$$
g^{\mu \nu}g_{\nu \sigma} = \delta^\mu_\sigma
$$
In other words, we need to find an expression for $g^{\mu \nu}$ such that the following equations is satisfied:
$$
g^{\mu \nu}(\eta_{\nu \sigma} + h_{\nu \sigma}) = \delta^\mu_\sigma
$$
As I have shown above, the function that obeys the above equation up to first order in $h_{\mu \nu}$ is:
\begin{equation}
g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}
\end{equation}
That is really all that is going on. 
A: We consider a matrix approach:
$$
\begin{aligned}
g^{\mu\nu}\equiv(g_{\mu\nu})^{-1}&=(\eta_{\mu\nu}+h_{\mu\nu})^{-1}\\
&=(\eta_{\mu\sigma}(\delta^{\sigma}_{\nu}+h^{\sigma}_{\nu}))^{-1}\\
&=(\delta^{\sigma}_{\nu}+h^{\sigma}_{\nu})^{-1}(\eta_{\sigma\mu})^{-1}\\
&=(\delta^{\nu}_{\sigma}-h^{\nu}_{\sigma}+h^{\nu}_{\rho}h^{\rho}_{\sigma}-\dots)\eta^{\mu\sigma}\\
&=\eta^{\mu\nu}-h^{\mu\nu}+\mathcal{O}(h^2).
\end{aligned}$$
