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For a constant matrix, the matrix exponential series and its differentiated series converge uniformly on bounded time intervals, as follows by comparison with the scalar exponential in a operator norm. Termwise derivatives therefore giveHence $\mathbf{x}(t)=e^{tM}{\mathbf{x}}_0$ solves the initial value problem. For this particular matrix, direct multiplication gives $M^2=-I$. Splitting the matrix exponential into its even and odd powers provesThis is the matrix exponential when the square is minus the identity. If ${\mathbf{x}}_0=(x_0,y_0{)}^{\mathsf{T}}$, the general solution isThe constants in part (a) are $\alpha =x_0/5$ and $\beta =y_0+2x_0/5$, which makes the two forms identical.

Use the lower antisymmetric block shown in the original PDF. Write $M=\mathrm{diag}(aK,bJ)$ withPowers 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 $e^{bJ}=I\cos b+J\sin b$. ThereforeThe 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.