Inverse metric in linearised gravity From what I've read, in the framework of linearised gravity, one perturbs the metric around a Minkowski background, $\eta_{\mu\nu}$, such that 
$$g_{\mu\nu}(x)=\eta_{\mu\nu}+h_{\mu\nu}(x)\tag{1}$$ 
where $h_{\mu\nu}(x)$ is a small perturbation, i.e. $$\big\lvert h_{\mu\nu}\big\rvert<<1.\tag{2}$$
The inverse metric is then found by assuming the following ansatz: 
$$g^{\mu\nu}(x)=\eta^{\mu\nu}+\tilde{h}^{\mu\nu}\tag{3}$$ 
where $\tilde{h}^{\mu\nu}$ is also small (i.e. $\big\lvert \tilde{h}_{\mu\nu}\big\rvert<<1$). 
Using this, it is easy to find that 
$$g^{\mu\nu}(x)=\eta^{\mu\nu}-h^{\mu\nu} \tag{4}$$ 
to first order. 
My question is, what is the justification for this ansatz? Is it simply that one expects the inverse metric to have a similar form to the metric in order to satisfy $$g^{\mu\alpha}g_{\alpha\nu}=\delta^{\mu}_{\;\nu}~?\tag{5}$$
 A: Hint: Ansatz (3) is not necessary. To derive eq. (4) from eqs. (1), (2) & (5), use instead that for an infinitesimal variation $$\delta (g^{-1})~=~- g^{-1}(\delta g)g^{-1},\tag{A}$$
where $g$ is an invertible matrix (with lower indices), and $g^{-1}$ is the inverse matrix (with upper indices). Can you see how eq. (A) is derived?
A: I am late, but I'll answer anyway. In linearized gravity, $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$. We wish to find the expression for $g^{\mu \nu}$ up to linear order, which in general, must be some linear combination of $\eta^{\mu \nu}$ and $h^{\mu \nu}$. To that end, we write:
$g^{\mu \nu} = a\eta^{\mu \nu} + b h^{\mu \nu}$ where  $a,b$ are coefficients to be determined
$g_{\mu \nu} g^{\nu \lambda} = (\eta_{\mu \nu} + h_{\mu \nu})(a\eta^{\nu \lambda} + b h^{\nu \lambda}) = \delta_\mu^\lambda$
$\Rightarrow a \delta_\mu^\lambda + (a+b)h_\mu^{\ \lambda} = \delta_\mu^\lambda \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ [$O(h^2)$ term neglected in the previous step]
$\Rightarrow a=1, b=-1$
So the inverse metric tensor in linearized gravity turns out to be $g^{\mu \nu} = \eta^{\mu \nu} - h^{\mu \nu}$.
