 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/11/a/solution

Expanding the determinant along the first row gives

det A = 1 (4 - 16) + 4 (- 4 + 28) + 7 (16 - 28) = - 12 + 96 - 84 = 0 .

The matrix trace is the sum of the diagonal entries, $tr A = 1 + 4 + 1 = 6$ . Since the determinant is the product of the eigenvalues, at least one eigenvalue is zero. Thus

A

is singular; the determinant alone does not specify the two other eigenvalues.

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