 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-1/7b/ii/solution

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

λ = 0

, then

A v = 0

for a nonzero eigenvector

v

, contradicting the existence of a matrix inverse. Hence

λ \neq = 0

. Multiplying

A v = λ v

A^{- 1}

gives

v = λ A^{- 1} v

, so

A^{- 1} v = λ^{- 1} v, λ^{- 1} is an eigenvalue of A^{- 1} .

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