 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/18/a/solution

The middle row is

- 1/2

times the sum of the first and third rows, so the rows are linearly dependent and the determinant is zero. The matrix trace is the sum of diagonal entries,

1 + 4 + 1 = 6

. A transpose preserves the determinant, while the determinant of a product is the product of its determinants. Consequently

det A = 0, Tr A = 6, det (A^{T}) = 0, det (A^{2}) = 0.

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