Question

If \(P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}\), \(A = \begin{bmatrix} 1 & 1 0 & 1 \end{bmatrix}\), and \(Q = PAP'\), then \(P'Q^{2005}P\) is

• A. \(\begin{bmatrix} 1 & 1 2005 & 1 \end{bmatrix}\)
• B. \(\begin{bmatrix} 1 & 2005 0 & 1 \end{bmatrix}\)
• C. \(\begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix}\)
• D. \(\begin{bmatrix} 1 & 2005 2005 & 1 \end{bmatrix}\)

Hint:

For orthogonal matrix \(P\), \(P' = P^{-1}\). Also, \(\begin{bmatrix} 1 & 1 0 & 1 \end{bmatrix}^n = \begin{bmatrix} 1 & n 0 & 1 \end{bmatrix}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
\(P\) is orthogonal rotation matrix. \(P' = P^{-1}\).

Step 2: Detailed Explanation:
\(Q = PAP'\). Then \(Q^{2005} = PA^{2005}P'\).
\(P'Q^{2005}P = P'(PA^{2005}P')P = (P'P)A^{2005}(P'P) = I \cdot A^{2005} \cdot I = A^{2005}\).
\(A = \begin{bmatrix} 1 & 1 0 & 1 \end{bmatrix}\) is a shear matrix. \(A^n = \begin{bmatrix} 1 & n 0 & 1 \end{bmatrix}\).
So \(A^{2005} = \begin{bmatrix} 1 & 2005 0 & 1 \end{bmatrix}\).

Step 3: Final Answer:
Option (B).