 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/14/a/i/solution

Let

N = 8 A

be the unscaled matrix. Expansion along its third row gives

det N = 2 (2 + 18) + 2 (18 + 2) = 162 .

Scaling a three-by-three matrix by

1/ 8

scales its determinant by

(8)^{- 3}

. Therefore

det A = \frac{16 2}{( 8 ) ^{3}} = 1 .

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