Let `A=[[1,1],[0,1]]` and `P=[[cos(pi/6),sin(pi/6)],[-sin(pi/6),cos(pi/6)]]` and `Q=PAP^T` then `P^T Q^2013 P` is:

Options

A. (a) `{:[(1,2013),(0,1)]:}`
B. (b) `{:[(0,2013),(0,1)]:}`
C. (c) `{:[(2013,0),(0,2013)]:}`
D. (d) `{:[(0,2013),(2013,0)]:}`

Answer

To solve the problem, we will go through the steps systematically.

### Step 1: Define the matrices
Let \( A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \) and \( P = \begin{bmatrix} \cos(\frac{\pi}{6}) & \sin(\frac{\pi}{6}) \\ -\sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) \end{bmatrix} \).

### Step 2: Calculate \( P^T \)
The transpose of matrix \( P \) is given by:
\[ P^T = \begin{bmatrix} \cos(\frac{\pi}{6}) & -\sin(\frac{\pi}{6}) \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) \end{bmatrix} \]

### Step 3: Calculate \( Q = PAP^T \)
Now we compute \( Q \):
\[ Q = P A P^T \]
First, calculate \( P A \):
\[ P A = \begin{bmatrix} \cos(\frac{\pi}{6}) & \sin(\frac{\pi}{6}) \\ -\sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos(\frac{\pi}{6}) \cdot 1 + \sin(\frac{\pi}{6}) \cdot 0 & \cos(\frac{\pi}{6}) \cdot 1 + \sin(\frac{\pi}{6}) \cdot 1 \\ -\sin(\frac{\pi}{6}) \cdot 1 + \cos(\frac{\pi}{6}) \cdot 0 & -\sin(\frac{\pi}{6}) \cdot 1 + \cos(\frac{\pi}{6}) \cdot 1 \end{bmatrix} \]
This simplifies to:
\[ P A = \begin{bmatrix} \cos(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6}) \\ -\sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) - \sin(\frac{\pi}{6}) \end{bmatrix} \]

Next, we multiply this result by \( P^T \):
\[ Q = \begin{bmatrix} \cos(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6}) \\ -\sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) - \sin(\frac{\pi}{6}) \end{bmatrix} \begin{bmatrix} \cos(\frac{\pi}{6}) & -\sin(\frac{\pi}{6}) \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) \end{bmatrix} \]
Calculating this product will give us the matrix \( Q \).

### Step 4: Simplify \( P^T Q^{2013} P \)
Since \( Q = PAP^T \), we can express \( Q^{2013} \) as:
\[ Q^{2013} = (P A P^T)^{2013} = P A^{2013} P^T \]
Thus, we have:
\[ P^T Q^{2013} P = P^T (P A^{2013} P^T) P = A^{2013} \]
This is because \( P^T P = I \) (the identity matrix).

### Step 5: Calculate \( A^{2013} \)
To find \( A^{2013} \), we notice the pattern:
\[ A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \]
Thus:
\[ A^{2013} = \begin{bmatrix} 1 & 2013 \\ 0 & 1 \end{bmatrix} \]

### Conclusion
The final result is:
\[ P^T Q^{2013} P = A^{2013} = \begin{bmatrix} 1 & 2013 \\ 0 & 1 \end{bmatrix} \]
Thus, the answer is option A: \( \begin{bmatrix} 1 & 2013 \\ 0 & 1 \end{bmatrix} \).