 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/17/b/solution

R_{3} \leftarrow R_{3} + 3 R_{2}

. The new third row is

(0, 0, 9 - 3 α)

, so expansion of the determinant along it gives

det A = (9 - 3 α) det (11 2 α - 1) = 3 (α - 3) (2 α + 1) .

Therefore

det A = 0 ⟺ α = 3 or α = - \frac{1}{2}, β arbitrary .

β

does not affect the determinant.

 New to topics? Read the docs here!