Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 18 / c / ii / Solution

Let $J=A/a$, so $J^2=-I$. The even and odd powers in the matrix exponential giveHere the two scalar sums are the Taylor series for cosine and sine. ConsequentlyWith the usual convention that positive angles turn anticlockwise, this rotation matrix rotates by $-a$; the parameter in the form printed in the paper is $\theta =a$.