 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1/6/a/solution

A^{2} = (13121213), tr A = 6, det A = 5.

Consequently

A^{2} - 6 A + 5 I = 0

t^{2} - 6 t + 5

. Rearranging gives

A (6 I - A) = 5 I

, so

A^{- 1} = \frac{6 I - A}{5} = \frac{1}{5} (3 - 2 - 2 3) .

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