 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-2/26i/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Since one eigenvalue is one, the other is the determinant. Direct calculation gives

ϑ = \frac{μ ( λ + α )}{λ ( μ + β )}, v_{ϑ} = (b 1), b = \frac{α ( μ + β )}{β ( λ + α )} > 0.

Indeed the first row requires

(1 + α / λ - ϑ) b = α / λ

, and the second row then follows from the characteristic equation. At

ϑ = 1

b = 1

and the eigenvectors coincide; the nonidentity matrix has a repeated eigenvalue and a Jordan block.

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