$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ then $A^{2014} = \lambda A^{2013} + \mu A^{2012}$, $\lambda + \mu = $_____.
Step-by-Step Solution
Key Concept: Use the Cayley-Hamilton theorem: since A satisfies its characteristic equation λ² - 5λ - 2 = 0, we have A² = 5A + 2I. This recurrence relation A^n = 5A^(n-1) + 2A^(n-2) holds for all n ≥ 2, giving A^2014 = 5A^2013 + 2A^2012, so λ = 5 and μ = 2.
Given $A^2 - 5A - 2I = 0$, we have $A^2 = 5A + 2I$. Then $A^3 = 5A^2 + 2A = 5(5A+2I) + 2A = 27A + 10I$ and $A^{2014} = 5A^{2013} + 2A^{2012}$. By finding the characteristic equation $\lambda^2 - 5\lambda - 2 = 0$, the eigenvalues satisfy $\lambda + 7 = 7$, giving $\lambda = 0, 5$ or using the recurrence relation yields $\lambda = 7$.
Correct Answer: 7