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} .

ii

Solution

The recurrence becomes

d_{n} = - 2 d_{n - 1} - d_{n - 2}, d_{0} = 1, d_{1} = - 2.

Its characteristic polynomial is

r^{2} + 2 r + 1 = (r + 1)^{2} .

Therefore

d_{n} = (α + β n) (- 1)^{n}

α = 1

and

β = 1

, so

d_{n} = (- 1)^{n} (n + 1) .

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