Pure Lorentz boost; transpose $\neq$ inverse? By definition a matrix representing a Lorentz transformation is orthogonal, so that its inverse is equal to its transpose. 
Consider a pure boost in the t-x plane;
$$\Lambda_x=\begin{pmatrix}
\cosh(\gamma) && \sinh(\gamma) && 0 && 0\\
\sinh(\gamma) && \cosh(\gamma) && 0 && 0\\
0 && 0 && 1 && 0\\
0&&0&&0&&1
\end{pmatrix}.$$
$\Lambda_x$ has inverse 
$$\Lambda_x^{-1}=\begin{pmatrix}
\cosh(\gamma) && -\sinh(\gamma) && 0&&0\\
-\sinh(\gamma) && \cosh(\gamma) && 0&&0\\
0 && 0 && 1 &&0\\
0&&0&&0&&1
\end{pmatrix}$$
but tranpose 
$$\Lambda_x^T=\begin{pmatrix}
\cosh(\gamma) && \sinh(\gamma) && 0&&0\\
\sinh(\gamma) && \cosh(\gamma) && 0&&0\\
0 && 0 && 1&&0\\
0&&0&&0&&1
\end{pmatrix}.$$
These are not equal. Where have I gone wrong?
 A: This is quite a common problem, and I feel it's usually made a bit clearer by using indices.
The Lorentz Group is $SO(1,3)$, not $SO(4)$, since (using the $(-+++)$ convention) the line element has an extra negative sign:
$$\text{d}s^2 = -c^2\text{d}t^2 + \text{d}x^2 + \text{d}y^2 + \text{d}z^2 $$
which is why Lorentz boosts are different from pure rotations in 4-dimensional space.
Such a boost $\Lambda$ takes $$x^\mu \longrightarrow x'^\mu = \Lambda^\mu_{\;\sigma} x^\sigma$$
Requiring that the element $\text{d}s^2$ be invariant, we have that
$$x^\mu x_\mu = x'^\alpha x'_\alpha$$
i.e.
$$x^\mu x^\nu \eta_{\mu\nu} = \left(\Lambda^\alpha_{\;\mu} x^\mu\right) \, \left(\Lambda^\beta_{\;\nu} x^\nu\right) \,\eta_{\alpha\beta} $$
Or in other words,
$$\eta_{\mu\nu} = \Lambda^\alpha_{\;\mu}\Lambda^\beta_{\;\nu}\eta_{\alpha\beta}$$
In terms of matrix multiplication, this is simply
$$\eta = \Lambda^\text{T} \, \eta \,\, \Lambda$$
which means, as you've shown yourself, $\Lambda^\text{T}\neq \Lambda^{-1}$, but rather
$$\Lambda^{-1} = \eta \, \Lambda^\text{T} \eta$$
A: The matrix representing a Lorentz boost is orthogonal with respect to the Minkowski metric $\eta = \mathrm{diag}(-1,1,1,1)$ (or reversed signs), which means
$$ \Lambda \eta \Lambda^T = \eta \text{   or    } \Lambda^{-1} = \eta \Lambda^T\eta.$$
A: There is a little more that can be added to the previous answers. 


*

*To see that your $\Lambda_x$ is not orthogonal, remember that orthogonal matrices preserve the square of the length $\vec a\cdot \vec a$ of a vector $\vec a$.  Using your $\Lambda_x$ and $\vec a=(1,0)$ one find, with $\vec a'=\Lambda_x \vec a$, that $\vec a'\cdot\vec a'=\cosh(2\gamma)\ne 1$, showing clearly  that $\Lambda_x$ is NOT orthogonal.

*The representation you have for $\Lambda_x$ is not only not orthogonal it is not unitary either.  If $\Lambda_x$ were unitary, then $\Lambda_x^{-1}=\Lambda_x^\dagger=\Lambda_x^T$ since $\Lambda_x$ is real.  The non-unitarity of $\Lambda_x$ is rooted in the fact that the Lorentz group is non-compact, and that there are no non-trivial finite dimensional representations of non-compact groups that are unitary.  

