 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/18/a/solution

Adding the diagonal entries gives

tr A = 4 + 4 + 2 = 10

. Expansion along the first row gives

det A = 4 (8 - 4) + 2 (0 - 8) = 0

. For the remaining trace, direct matrix multiplication gives

A^{2} = 2041242012121212 .

Thus

tr (A^{2}) = 20 + 20 + 12 = 52

det (A^{2}) = (det A)^{2} = 0

. Altogether,

tr A = 10, det A = 0, tr (A^{2}) = 52, det (A^{2}) = 0.

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