 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/17/b/iii/solution

For

λ \neq = \pm 2

, the reduced equation gives

y = (3 - λ) / (4 - λ^{2})

x = 1 - λ y

and

z = 1 - 4 y

gives all solutions:

r = \frac{1}{4 - λ ^{2}} 4 - 3 λ 3 - λ - λ^{2} + 4 λ - 8, λ \neq = \pm 2.

The nonzero determinant proves that there are no other solutions at these parameter values.

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