 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-1/6c/d/i/solution

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

Expanding

det A^{(n)}

along its last row gives the last diagonal contribution

a_{n} d_{n - 1}

. The only other nonzero entry is

c_{n - 1}

; expanding its minor along the last column contributes

- b_{n - 1} c_{n - 1} d_{n - 2} .

d_{n} = a_{n} d_{n - 1} - b_{n - 1} c_{n - 1} d_{n - 2},

X_{n} = a_{n}, Y_{n} = - b_{n - 1} c_{n - 1} .

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