 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-2/7c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

x_{n} = A^{n} a + \sum_{j = 0}^{n - 1} A^{j} b = A^{n} a + (I - A^{n}) (I - A)^{- 1} b,

where the inverse exists by the hypothesis.

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