 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/7/solution

A - λ B

precisely when its determinant vanishes:

0 = det (1 2 - 2 λ 2 - 2 λ 1) = 1 - 4 (1 - λ)^{2} .

Hence

λ = \frac{1}{2} or λ = \frac{3}{2} .

The corresponding eigenvectors can be taken as

(1, - 1)^{T}

and

(1, 1)^{T}

, respectively, which also checks that both values give nontrivial solutions.

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