Computing inverse Lorentz transformations in matrix form

Question

I have a question about finding inverse Lorentz transformations explicitly, in matrix form:

Suppose I have a Lorentz transformation $\Lambda^\mu_{\;\nu}$, with matrix representation $\underline{\underline{\Lambda}}$. The inverse Lorentz transformation $(\Lambda^{-1})^\mu_{\;\nu}=\Lambda_\nu^{\;\mu}$ can be found as $\Lambda_\nu^{\;\mu}=g_{\nu\rho}g^{\mu\sigma}\Lambda^\rho_{\;\sigma}$.

My question concerns how I should write the latter in matrix form, in order to compute it explicitly. If I write $g_{\mu\rho}\Lambda^\rho_{\;\sigma}g^{\nu\sigma}$ then, to me, that indicates the matrix expression is $\underline{\underline{\mathbb{g}}}\underline{\underline{\Lambda}}\underline{\underline{\mathbb{g}}}$. To me, this seems like the correct order as the indices of the neighbouring objects "match".

But if I look in a textbook, the inverse transformation is listed as $g^{\nu\sigma}\Lambda^\rho_{\;\sigma}g_{\mu\rho}$ instead, which gives $\underline{\underline{\mathbb{g}}}\underline{\underline{\Lambda}}^\mathrm{T}\underline{\underline{\mathbb{g}}}$. Which form is correct, and why?

The latter. Recall that $\Lambda^T\eta\Lambda=\eta$ (cf. here), and therefore $\eta\Lambda^T\eta\Lambda=1$. — AccidentalFourierTransform, May 26 '18 at 14:45
The latter is that correct. Lorenrz group is made of the matrices $A$ such that $A^tg A = g$ which implies $(gA^tg)A= gg=I$... — Valter Moretti, May 26 '18 at 14:47
Thanks for the replies! Am I correct in saying that if I wanted to compute, for the Faraday tensor, for example $F^{\mu\nu}$ from $F^\mu_{;\nu} = \underline{\underline{F}}$, I would use $F^{\mu\nu} = F^\mu_{;\rho}g^{\rho \nu} = \underline{\underline{\mathrm{F}}} \underline{\underline{\mathbb{g}}}$? — Mr. Bump, May 26 '18 at 15:01

Computing inverse Lorentz transformations in matrix form

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