I'm having difficulty understanding some of this derivation. The relevant information, as I understand it is:

We've let the lorentzian metric be a perturbation of the minkowski metric:
$$g\_{\mu\nu} = \eta\_{\mu\nu}+h\_{\mu\nu},$$
where apparently if $h\_{\mu\nu}$ is small enough, we can approximate the inverse as:
$$g^{\mu\nu} = \eta^{\mu\nu}-h^{\mu\nu}.$$

By some assumptions, the stated results are:

$$ \Delta \bar{h}\_{00} = -2\kappa T\_{00}$$
$$ \Delta \bar{h}\_{0i} = 0 $$
$$ \Delta \bar{h}\_{ij} = 0 $$

where $\bar{h}\_{\mu \nu} = h\_{\mu \nu} - \frac{1}{2}h\_\beta^\beta\eta\_{\mu\nu}$.

Which we can substitute into the Newtonian gravitation equation to yield:

$$ \bar{h}\_{00} = \frac{-\kappa c^2}{2\pi G} \Phi $$
$$ \bar{h}\_{0i} = 0 $$
$$ \bar{h}\_{ij} = 0 $$

We then 'invert' the perturbation to: $ h\_{\mu \nu} = \bar{h}\_{\mu \nu} - \frac{1}{2}\bar{h}\_\beta^\beta\eta\_{\mu\nu}$.

and obtain from this:

$$h\_{00} = \frac{1}{2} \bar{h}\_{00} = \frac{-\kappa c^2}{2\pi G} \Phi $$
$$h\_{0i} = 0$$
$$h\_{ij} = \frac{1}{2} \delta\_{ij} \bar{h}\_{00} = -\delta\_{ij}\frac{\kappa c^2}{2\pi G} \Phi$$

I can follow most of it alright, I'm just struggling to figure out the working behind the last step if anyone could explain this.