If `A=((1,2),(0,1)),P=((costheta, sintheta),(-sintheta, costheta)),Q=P^(T)AP`, find `PQ^(2014)P^(T)`: (a) `((1,2^(2014)),(0,1))` (b) `((1,4028),(0,1))` (c) `(P^(T))^(2013)A^(2014)P^(2013)` (d)`P^(T)A^(2014)P`

Options

A. `((1,2^(2014)),(0,1))`
B. `((1,4028),(0,1))`
C. `(P^(T))^(2013)A^(2014)P^(2013)`
D. `P^(T)A^(2014)P`

Answer

To solve the problem, we will follow the steps outlined in the video transcript and derive the solution step by step.

### Step 1: Define the matrices
We have:
- \( A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \)
- \( P = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \)

### Step 2: Compute \( Q \)
We need to compute \( Q = P^T A P \).

First, calculate \( P^T \):
\[ P^T = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

Now, compute \( P^T A \):
\[ P^T A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \cos \theta & 2\cos \theta - \sin \theta \\ \sin \theta & 2\sin \theta + \cos \theta \end{pmatrix} \]

Next, compute \( Q = P^T A P \):
\[ Q = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} \cos \theta & 2\cos \theta - \sin \theta \\ \sin \theta & 2\sin \theta + \cos \theta \end{pmatrix} \]

Calculating this product:
\[ Q = \begin{pmatrix} \cos^2 \theta + (-\sin \theta)(\sin \theta) & \cos \theta(2\cos \theta - \sin \theta) + (-\sin \theta)(2\sin \theta + \cos \theta) \\ \sin \theta(\cos \theta) + \cos \theta(-\sin \theta) & \sin \theta(2\cos \theta - \sin \theta) + \cos \theta(2\sin \theta + \cos \theta) \end{pmatrix} \]

This simplifies to:
\[ Q = \begin{pmatrix} \cos^2 \theta - \sin^2 \theta & 2\cos \theta \\ 0 & 1 \end{pmatrix} \]

### Step 3: Find \( PQ^{2014}P^T \)
We need to compute \( PQ^{2014}P^T \).

Using the property of matrix powers:
\[ PQ^nP^T = P(P^TAP)^nP^T = P^TAP^T = A^n \]

Thus:
\[ PQ^{2014}P^T = P(P^TAP)^{2014}P^T = PA^{2014}P^T \]

### Step 4: Calculate \( A^{2014} \)
To find \( A^{2014} \), we first find \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \]

Next, we find \( A^3 \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix} \]

From the pattern, we can conclude:
\[ A^n = \begin{pmatrix} 1 & 2n \\ 0 & 1 \end{pmatrix} \]

Thus:
\[ A^{2014} = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} \]

### Step 5: Final computation
Now substituting back:
\[ PQ^{2014}P^T = A^{2014} = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} \]

### Conclusion
The final answer is:
\[ PQ^{2014}P^T = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} \]