Discussion
Let T be any 2 × 2 matrix with eigenvalues −1, 3. For each positive integer n, the numbers \(a_{n}\) and \(b_{n}\) are such that
\[ T^{n+1} = a_{n}T + b_{n}\mathrm{II} \]
where II is 2 × 2 identity matrix. Then \(a_{n}\) and \(b_{n}\) satisfy the recurrence relations :
Answer: B
The recurrences are \(a_{n+1} = 2a_{n} + b_{n}\); \(b_{n+1} = 3a_{n}\).
The characteristic polynomial of \(T\) has roots −1 and 3, so it is \(\lambda^{2} - 2\lambda - 3\). By the Cayley-Hamilton theorem \(T^{2} = 2T + 3\,\mathrm{II}\). Multiply \(T^{n+1} = a_{n}T + b_{n}\mathrm{II}\) by \(T\):
\[ \begin{aligned} T^{n+2} &= a_{n}T^{2} + b_{n}T \\ &= a_{n}(2T + 3\,\mathrm{II}) + b_{n}T \\ &= (2a_{n} + b_{n})T + 3a_{n}\,\mathrm{II} \end{aligned} \]
Comparing with \(T^{n+2} = a_{n+1}T + b_{n+1}\mathrm{II}\) gives the answer. A quick check: \(T^{1} = T\) means \(a_{0} = 1, b_{0} = 0\), so \(T^{2}\) should have \(a_{1} = 2, b_{1} = 3\), as Cayley-Hamilton says. The other recurrences do not reproduce \(T^{2} = 2T + 3\,\mathrm{II}\).
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