Question

If \( A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \), \( P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \) and \( X = A P A^T \), then \( A^T X^{50} A = \) 
 

• A. \(\begin{pmatrix} 0 & 1 1 & 0 \end{pmatrix}\)
• B. \(\begin{pmatrix} 2 & 1 0 & -1 \end{pmatrix}\)
• C. \(\begin{pmatrix} 25 & 1 1 & -25 \end{pmatrix}\)
• D. \(\begin{pmatrix} 1 & 50 0 & 1 \end{pmatrix}\)

Hint:

For any matrix of the form \(\begin{pmatrix} 1 & a 0 & 1 \end{pmatrix}\), the \(n^{th}\) power is simply \(\begin{pmatrix} 1 & na 0 & 1 \end{pmatrix}\). This is a very common shortcut in matrix algebra!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept: 
We are dealing with matrix powers and transformations. Note that for matrix \( A \), \( A^T = A \) and \( A^2 = I \) (the identity matrix), because \( A \) is an involutory matrix. 

Step 2: Key Formula or Approach: 
If \( X = A P A^T \), then \( X^n = (A P A^T)(A P A^T)\dots(A P A^T) \). Since \( A^T A = I \), the inner terms cancel out. \[ X^n = A P^n A^T \] 

Step 3: Detailed Explanation: 
1. We need to find \( A^T X^{50} A \).
2. Substitute \( X^{50} = A P^{50} A^T \): \[ A^T (A P^{50} A^T) A = (A^T A) P^{50} (A^T A) \] 3. Since \( A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \), we have \[ A^T A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \] \[ I \cdot P^{50} \cdot I = P^{50} \] 4. For \( P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \), the power \( P^n \) follows the pattern: \[ P^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} \] 5. Therefore, \[ P^{50} = \begin{pmatrix} 1 & 50 \\ 0 & 1 \end{pmatrix} \] 

Step 4: Final Answer 
The result is \( \begin{pmatrix} 1 & 50 \\ 0 & 1 \end{pmatrix} \).