 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2/7/c/solution

Here the vertical bars denote determinants. Since

B = B^{- 1}

, taking determinants gives

(det B)^{2} = 1

. The condition

det B > 0

selects

det B = 1

. By the multiplicative property of the determinant,

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

The hypotheses determine this determinant, while allowing many possible matrices

A

and

B

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