special relativity - 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?

Answer

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 $$ \Lambda \eta \Lambda^T = \eta \text{ or } \Lambda^{-1} = \eta \Lambda^T\eta.$$