Let $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ & $P = \begin{bmatrix} \cos \frac{\pi}{12} & \sin \frac{\pi}{12} \\ -\sin \frac{\pi}{12} & \cos \frac{\pi}{12} \end{bmatrix}$ and $Q = P^T AP$, then if $PQ^{2014}P^T = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ then sum of digits of $b$ is _____.
Step-by-Step Solution
Key Concept: Recognize that P is an orthogonal matrix (rotation matrix) satisfying P^T P = I, and use the property that similarity transformations preserve matrix structure: PQ^{2014}P^T = (P^T AP)^{2014} when expanded telescopically. Since A is the identity matrix, Q = I, making the calculation reduce to finding A^{2014}.
Given $P^T P = I = PP^T$, we compute $PQ^{2014}P^T = P(P^TAP)(P^TAP)\cdots(P^TAP)P^T = A^{2014}$. The matrix $A^{2014} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^{2014} = \begin{bmatrix} 1 & 2014 \\ 0 & 1 \end{bmatrix}$, so $b = 2014$ and the sum of digits is $2 + 0 + 1 + 4 = 7$.
Correct Answer: 7