 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-1/2/a/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

∥ v_{\pm} ∥ = 2

and

∥ v_{0} ∥ = 2

. With normalized eigenvectors as rows,

R = 1/2 1/ 2 1/2 1/ 2 0 - 1/ 2 1/2 - 1/ 2 1/2, RM R^{T} = diag (2, 0, - 2) .

The rows form an orthonormal basis, so

R R^{T} = I

and

R

M v_{0} = 0

for a nonzero vector; thus $M$ is singular and has no matrix inverse.

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