If `A=((1,2),(0,1)),P=((costheta, sintheta),(-sintheta, costheta)),Q=P^(T)AP`, find `PQ^(2014)P^(T)`:

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 step by step, we need to compute \( PQ^{2014}P^T \) where \( Q = P^T A P \).

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

### Step 2: Calculate \( P^T \)
The transpose of matrix \( P \) is:
\[ P^T = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

### Step 3: Compute \( Q = P^T A P \)
Now we compute \( Q \):
\[ Q = P^T A P \]
First, calculate \( A P \):
\[ A P = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} = \begin{pmatrix} 1 \cdot \cos \theta + 2 \cdot (-\sin \theta) & 1 \cdot \sin \theta + 2 \cdot \cos \theta \\ 0 \cdot \cos \theta + 1 \cdot (-\sin \theta) & 0 \cdot \sin \theta + 1 \cdot \cos \theta \end{pmatrix} \]
This simplifies to:
\[ A P = \begin{pmatrix} \cos \theta - 2 \sin \theta & \sin \theta + 2 \cos \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

Now compute \( P^T (A P) \):
\[ Q = P^T \begin{pmatrix} \cos \theta - 2 \sin \theta & \sin \theta + 2 \cos \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]
Calculating this gives:
\[ Q = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} \cos \theta - 2 \sin \theta & \sin \theta + 2 \cos \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]
After performing the multiplication, we find:
\[ Q = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = A \]

### Step 4: Compute \( PQ^{2014}P^T \)
Since \( Q = A \), we can write:
\[ PQ^{2014}P^T = PA^{2014}P^T \]

### Step 5: Calculate \( A^{2014} \)
To find \( A^{2014} \), we observe the pattern:
- \( A^1 = A \)
- \( A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \)
- \( A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix} \)

From this pattern, we can deduce:
\[ A^n = \begin{pmatrix} 1 & 2n \\ 0 & 1 \end{pmatrix} \]
Thus,
\[ A^{2014} = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} \]

### Step 6: Substitute Back
Now substituting back into our expression:
\[ PQ^{2014}P^T = PA^{2014}P^T = P \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} P^T \]

### Step 7: Final Calculation
Calculating \( PA^{2014}P^T \):
\[ PA^{2014} = P \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} \]
This results in:
\[ \begin{pmatrix} \cos \theta & 4028 \cos \theta + \sin \theta \\ -\sin \theta & 4028 \sin \theta + \cos \theta \end{pmatrix} \]
Now multiply by \( P^T \):
\[ = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} \]

### Final Result
After performing the multiplication, we find:
\[ PQ^{2014}P^T = \begin{pmatrix} 1 & 4028 \\ 0 & 1 \end{pmatrix} \]