Let P = <table><tr><td><img src="https://latex.codecogs.com/svg.image?\frac{\sqrt{3}}{2}" /></td><td><img src="https://latex.codecogs.com/svg.image?\frac{1}{2}" /></td></tr><tr><td><img src="https://latex.codecogs.com/svg.image?-\frac{1}{2}" /></td><td><img src="https://latex.codecogs.com/svg.image?\frac{\sqrt{3}}{2}" /></td></tr></table>, A = <table><tr><td>1</td><td>1</td></tr><tr><td>0</td><td>1</td></tr></table> and Q = PAP<sup>T</sup>. If P<sup>T</sup>Q<sup>2007</sup>P = <table><tr><td>a</td><td>b</td></tr><tr><td>c</td><td>d</td></tr></table>, then 2a + b - 3c - 4d equal to
Step-by-Step Solution
Key Concept: The matrix P is a rotation matrix. Q = PAP^T implies Q^n = PA^nP^T. Thus P^T Q^n P = A^n. For A = [[1, 1], [0, 1]], A^n = [[1, n], [0, 1]].
Given P = [[cos(pi/6), sin(pi/6)], [-sin(pi/6), cos(pi/6)]], which is a rotation matrix. Q = PAP^T. Then Q^2007 = (PAP^T)(PAP^T)...(PAP^T) = PA^2007P^T. Therefore, P^T Q^2007 P = P^T (PA^2007P^T) P = (P^T P) A^2007 (P^T P) = I A^2007 I = A^2007. Since A = [[1, 1], [0, 1]], A^n = [[1, n], [0, 1]]. Thus A^2007 = [[1, 2007], [0, 1]]. Comparing with [[a, b], [c, d]], we get a=1, b=2007, c=0, d=1. The expression 2a + b - 3c - 4d = 2(1) + 2007 - 3(0) - 4(1) = 2 + 2007 - 4 = 2005.
Correct Answer: 2