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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 6C / d / i / Solution

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} .

Thus the tridiagonal determinant recurrence is

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

so

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

Solved by gpt-5.6-sol high.

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