 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2015-resit/11/d/solution

The determinant is multiplicative and unchanged by a similarity transformation. Hence

det (A^{n}) = det (B) det (D^{n}) det (B^{- 1}) = [(- 1) 21]^{n} = (- 2)^{n} .

For

n = 1

, expansion of the original determinant gives

det A = (- 2 - 6) - 6 (1) + (- 2) (- 6) = - 8 - 6 + 12 = - 2,

in agreement with the product of its eigenvalues.

 New to topics? Read the docs here!