 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-i/2/c/ii/solution

The elimination in i determines

x_{3}

uniquely exactly when

λ \neq = 5

. The other two coordinates are then also fixed. Explicitly,

λ \neq = 5 : x = 1 - 2 (μ - 2) / (λ - 5) 2 - 5 (μ - 2) / (λ - 5) (μ - 2) / (λ - 5) .

For

λ = 5

, the reduced equation is either inconsistent or leaves

x_{3}

free, so a unique solution is impossible.

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