 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2/section-b/7/b/solution

The permutation matrix cycles the three coordinate vectors, so three applications give

C^{3} = I

. Consequently

C^{2} C = C C^{2} = I

, proving

C^{- 1} = C^{2}

. Since

16 = 5 \cdot 3 + 1

C^{16} = C = 010001100 .

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