 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/17/b/ii/solution

Multiplying an

n \times n

α

scales every row, giving a factor

α^{n}

in its determinant. Also

∣ A^{- 1} ∣ = 1/∣ A ∣

, since

A A^{- 1} = I

. Hence

∣ α A^{- 1} ∣ = \frac{α ^{n}}{∣ A ∣} .

Only

A

needs to be invertible; the result remains valid when

α = 0

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