Let A = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mn>2</mn></mtd><mtd><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math>. If the sum of the diagonal elements of A<sup>13</sup> is 3<sup>n</sup>, then n is equal to ____.
Step-by-Step Solution
Key Concept: The sum of the diagonal elements of a matrix is its trace. For a 2x2 matrix A, the trace of A^n can be found using the characteristic equation A^2 - tr(A)A + det(A)I = 0. Here, tr(A) = 3 and det(A) = 3. Thus, A^2 - 3A + 3I = 0. This leads to a recurrence relation for the trace of A^n.
Let T_n = tr(A^n). The characteristic equation of A is det(A - \lambda I) = 0, which is (2-\lambda)(1-\lambda) - (-1)(1) = 0, or \lambda^2 - 3\lambda + 3 = 0. By Cayley-Hamilton theorem, A^2 - 3A + 3I = 0. Multiplying by A^(n-2), we get A^n - 3A^(n-1) + 3A^(n-2) = 0. Taking the trace, T_n - 3T_(n-1) + 3T_(n-2) = 0. We have T_1 = tr(A) = 3, and T_2 = tr(A^2) = tr(3A - 3I) = 3(3) - 3(2) = 9 - 6 = 3. Using the recurrence, T_3 = 3T_2 - 3T_1 = 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) = -81 + 27 = -54. T_7 = 3(-54) - 3(-27) = -162 + 81 = -81. Wait, let's re-evaluate. The eigenvalues are \lambda = (3 \pm i\sqrt{3})/2 = \sqrt{3}(cos(\pi/6) \pm i sin(\pi/6)) = \sqrt{3} e^(\pmi\pi/6). Then \lambda^n = (\sqrt{3})^n e^(\pmin\pi/6) = 3^(n/2) (cos(n\pi/6) \pm i sin(n\pi/6)). The trace T_n = \lambda1^n + \lambda2^n = 2 * 3^(n/2) cos(n\pi/6). For n=13, T_13 = 2 * 3^(13/2) cos(13\pi/6) = 2 * 3^6 * \sqrt{3} * cos(\pi/6) = 2 * 3^6 * \sqrt{3} * (\sqrt{3}/2) = 3^6 * 3 = 3^7. Thus n=7.
Correct Answer: 7