general relativity - 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 $$g^{\mu\alpha}g_{\alpha\nu}=\delta^{\mu}_{\;\nu}~?\tag{5}$$

Answer

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?