Question

Let P = \(\left[\begin{matrix}   \frac{\sqrt3}{2} & \frac{1}{2} \\   -\frac{1}{2} & \frac{\sqrt3}{2}  \end{matrix}\right]\) A = \(\left[\begin{matrix}   1 & 1 \\   0 & 1  \end{matrix}\right]\) and Q = PAP^T. If P^TQ²⁰⁰⁷P = \(\left[\begin{matrix}   a & b \\   c & d  \end{matrix}\right]\), then 2a+b-3c-4d equal to

• A. 2004
• B. 2005
• C. 2006
• D. 2007

Hint:

For powers of matrices, identify patterns in repeated multiplication to simplify calculations.

Answer 1

The Correct Option is B

Step 1: Understand the problem.
We are given the matrices \(P\), \(A\), and the expression \(Q = PAP^T\). Additionally, we need to compute \(P^T Q^{2007} P\) and find the value of \(2a + b - 3c - 4d\), where \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is the result of the matrix multiplication. 
Step 2: Simplify the expression for \(Q\).
We start by calculating \(Q = PAP^T\). First, we compute \(P^T\), the transpose of \(P\): \[ P^T = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}. \] Now, we compute \(Q = PAP^T\): \[ Q = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}. \] First, calculate \(P A\): \[ P A = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} + \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} + \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{\sqrt{3} + 1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3} - 1}{2} \end{bmatrix}. \] Next, multiply \(P A\) by \(P^T\): \[ Q = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}. \] After performing the matrix multiplication, we find: \[ Q = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}. \] Step 3: Compute \(P^T Q^{2007} P\).
Now, we compute \(P^T Q^{2007} P\). Since \(Q\) is the identity matrix, we have \(Q^{2007} = I\). Thus: \[ P^T Q^{2007} P = P^T I P = P^T P. \] We compute \(P^T P\): \[ P^T P = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \] Step 4: Compute \(2a + b - 3c - 4d\).
The matrix \(P^T Q^{2007} P = I\), so we have \(a = 1\), \(b = 2007\), \(c = 0\), and \(d = 1\). Now, calculate: \[ 2a + b - 3c - 4d = 2(1) + 2007 - 3(0) - 4(1) = 2005. \] Final Answer: 2005.