 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/18/a/ii/solution

For either

A

from part (i), choose

B = - \frac{I + A}{2} .

Indeed, using

A^{2} = - I

and

A^{T} = - A

B^{2} = \frac{I + 2 A + A ^{2}}{2} = A, B^{T} B = \frac{( I - A ) ( I + A )}{2} = I .

- 2/ 2 = - 2 < 0

. Explicitly, for the two

A

matrices in the order displayed above, the corresponding orthogonal matrices are

B = \frac{1}{2} (- 1 - 1 1 - 1), B = \frac{1}{2} (- 1 1 - 1 - 1) .

Their negatives have positive matrix trace and would not meet the additional sign condition.

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