= Solution
Use the lower antisymmetric block shown in the original PDF. Write $M=\operatorname{diag}(aK,bJ)$ with
$$
K=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad J=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad K^2=I,\quad J^2=-I.
$$
Powers preserve the block structure. Splitting the <matrix exponential> series into even and odd powers gives $e^{aK}=I\cosh a+K\sinh a$. For the other block, the <matrix exponential when the square is minus the identity> gives $e^{bJ}=I\cos b+J\sin b$. Therefore
$$
\boxed{e^M=\begin{pmatrix}\cosh a&\sinh a&0&0\\\sinh a&\cosh a&0&0\\0&0&\cos b&\sin b\\0&0&-\sin b&\cos b\end{pmatrix}.}
$$
The upper block is hyperbolic and the lower block is a rotation. The transcription's altered last row and hyperbolic functions in the lower block would describe a different problem.
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