 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/5/e/solution

Here

A^{2} = - I

. Splitting the matrix exponential into even and odd powers therefore gives

U = I cos 1 + A sin 1 = (cos 1 - sin 1 sin 1 cos 1) .

(I + A) (I - A) = I - A^{2} = 2 I

, so

(I + A)^{- 1} = (I - A) /2

. Hence

V = \frac{1}{2} (I - A)^{2} = \frac{1}{2} (I - 2 A + A^{2}) = - A = (01 - 1 0) .

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