Question

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

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

Hint:

For a matrix of the form \( P = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \), its \( n \)th power is given by: \[ P^n = \begin{bmatrix} 1 & nk \\ 0 & 1 \end{bmatrix}. \]

Answer 1

The Correct Option is D

Step 1: {Show that \( A \) is an orthogonal matrix}
Since \( A A^T = I \), the matrix \( A \) is orthogonal.
Step 2: {Simplify \( A^T X^{50} A \)}
\[ A^T X^{50} A = A^T X^{49} (A P A^T) A \] \[ = A^T X^{49} A P (A^T A) = A^T X^{49} A P \] \[ = A^T X^{48} (A P A^T) A P = A^T X^{48} A P^2 \dots \] \[ = A^T A P^{50} = I P^{50} = P^{50}. \] Step 3: {Compute \( P^{50} \)}
\[ P = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] \[ P^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \] \[ P^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \] \[ \vdots \] \[ P^{50} = \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix} \] Step 4: {Conclusion}
\[ A^T X^{50} A = P^{50} = \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix}. \]

Answer 2

Step 1: Given Matrices
We are given the following matrices: \[ A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad P = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}. \] We are also given the matrix \( X \), defined as: \[ X = A P A^T. \] We need to find \( A^T X^{50} A \).

Step 2: Compute \( X \)
First, let's compute \( X = A P A^T \). We already know that: \[ A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad A^T = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \] \[ P = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}. \] We now calculate \( A P \): \[ A P = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix}. \] Next, we compute \( A P A^T \): \[ A P A^T = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}. \] Thus, we have: \[ X = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}. \]

Step 3: Compute \( X^{50} \)
Now, let's find \( X^{50} \). Notice that \( X \) is a simple matrix of the form: \[ X = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}. \] This is a shear matrix, and we can compute higher powers of \( X \) as follows: \[ X^n = \begin{bmatrix} 1 & -n \\ 0 & 1 \end{bmatrix}. \] Thus, \( X^{50} \) is: \[ X^{50} = \begin{bmatrix} 1 & -50 \\ 0 & 1 \end{bmatrix}. \]

Step 4: Compute \( A^T X^{50} A \)
We now need to compute \( A^T X^{50} A \). Using the values of \( A^T \) and \( A \) and the computed \( X^{50} \), we have: \[ A^T X^{50} A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & -50 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}. \] First, compute the product \( X^{50} A \): \[ X^{50} A = \begin{bmatrix} 1 & -50 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 50 \\ 0 & -1 \end{bmatrix}. \] Now compute \( A^T X^{50} A \): \[ A^T X^{50} A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 50 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix}. \]

Step 5: Conclusion
Thus, the value of \( A^T X^{50} A \) is: \[ \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix}. \]

The correct answer is: \( \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix} \)