Usually the notation is as follows. The Lorentz transformation matrix describes the transformation of contravariant tensor components. It's sufficient to consider the space-time four-vector:
$$x^{\prime \mu} = {\Lambda^{\mu}}_{\rho} x^{\rho}.$$
In order to be a Lorentz transformation you must have
$$\eta_{\mu \nu} x^{\prime \mu} x^{\prime \nu}=\eta_{\rho \sigma} x^{\rho} x^{\sigma},$$
where ##\eta_{\rho \sigma}## are the (purely covariant!) tensor components of the Minkowski fundamental form, i.e., in Minkowski orthonormal frames ##(\eta_{\rho \sigma})=\mathrm{diag}(1,-1,-1,-1)##. Since this should hold for any vector ##x##, the Lorentz-transformation matrix must fulfill
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma}=\eta_{\rho \sigma}.$$
Sometimes it's convenient to work in matrix-vector notation (to be a bit evil you can also say in order to utmostly confuse the student you switch between the self-explaining Ricci calculus and the notationally less precise matrix-vector notaion all the time ;-)). Then one defines matrices
$$\hat{\Lambda}=({\Lambda^{\mu}}_{\rho}), \quad \hat{\eta}=(\eta_{\rho \sigma}).$$
As you see, then the Ricci calculus's most important property to clearly distinguish between co- and contravariant components of tensors and transformation matrices, is lost, but that's no problem as long as you keep in mind what kind of components you are dealing with. Just write down the matrices and column and row vectors and then translate the Ricci-calculus formulae to the matrix-vector multiplication calculus.
Obviously, using column vectors ##\overline{x}=(x^{\mu})## for the contravariant vector components, then the Lorentz transformations read
$$\overline{x}'=\hat{\Lambda} \overline{x}.$$
The Lorentz-transformation property of the matrix ##\hat{\Lambda}## translates to
$$\hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta},$$
or since ##\hat{\eta}^2=1## after some simple matrix manipulation
$$\hat{\Lambda}^{-1} = \hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta}.$$
In the Ricci calculus this translates into
$${(\Lambda^{-1})^{\mu}}_{\nu}=\eta^{\mu \rho} \eta_{\nu \sigma} {\Lambda^{\sigma}}_{\rho}={\Lambda_{\nu}}^{\mu}.$$
This can be verified by the Ricci calculus easily, using the original Lorentz-transformation property:
$${\Lambda_{\nu}}^{\mu} {\Lambda^{\nu}}_{\rho}=g_{\nu \sigma} g^{\mu \alpha} {\Lambda^{\sigma}}_{\alpha} {\Lambda^{\nu}}_{\rho} = g^{\mu \alpha} g_{\alpha \rho} = {\delta^{\mu}}_{\rho}.$$