 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-2/10f/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

A

B A = A^{- 1} (A B) A .

Thus $A B$ and $B A$ are similar matrices, and similar matrices have the same characteristic polynomial.

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