Let $$A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$$. If the sum of the diagonal elements of $$A^{13}$$ is $$3^n$$, then $$n$$ is equal to ________

We begin by finding the characteristic equation of the matrix A. Computing $$\det(A - \lambda I) = (2-\lambda)(1-\lambda) + 1 = \lambda^2 - 3\lambda + 3 = 0$$ and applying Cayley-Hamilton yields $$A^2 = 3A - 3I$$.

Next, letting $$t_n = \mathrm{tr}(A^n)$$, the characteristic equation implies the recurrence $$t_n = 3\,t_{n-1} - 3\,t_{n-2}$$ with initial values $$t_0 = \mathrm{tr}(I) = 2$$ and $$t_1 = \mathrm{tr}(A) = 3$$.

Using this recurrence, we compute $$t_2 = 3(3) - 3(2) = 3$$, $$t_3 = 3(3) - 3(3) = 0$$, $$t_4 = 3(0) - 3(3) = -9$$, $$t_5 = 3(-9) - 3(0) = -27$$, $$t_6 = 3(-27) - 3(-9) = -54$$ and $$t_7 = 3(-54) - 3(-27) = -81$$.

Alternatively, the eigenvalues of A are $$\lambda = \frac{3 \pm i\sqrt{3}}{2} = \sqrt{3}\,e^{\pm i\pi/6}$$, so that $$\lambda^n = 3^{n/2}\,e^{\pm in\pi/6}$$ and therefore $$t_n = \lambda_1^n + \lambda_2^n = 2 \cdot 3^{n/2} \cos\frac{n\pi}{6}$$.

For $$n = 13$$ this yields $$t_{13} = 2 \cdot 3^{13/2} \cos\frac{13\pi}{6} = 2 \cdot 3^{13/2} \cos\frac{\pi}{6} = 2 \cdot 3^{13/2} \cdot \frac{\sqrt{3}}{2} = 3^{13/2} \cdot 3^{1/2} = 3^7$$ and hence $$n = 7$$.

Therefore, the final answer is 7.