 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/17/c/ii/solution

Subtract twice the second row from the third row when computing the determinant. The last row becomes

(0, 0, μ - 6)

, so

det A = (μ - 6) det (11 μ - 1) = - (μ - 6) (μ + 1) .

Its zeros give exactly the parameter values with a nontrivial null space:

μ = - 1 or μ = 6.

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