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:

Show Hint

For a matrix of the form \( P = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \), its \( n \)th power is given by: \[ P^n = \begin{bmatrix} 1 & nk \\ 0 & 1 \end{bmatrix}. \]
Updated On: Mar 26, 2025
  • \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
  • \( \begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix} \)
  • \( \begin{bmatrix} 25 & 1 \\ 1 & -25 \end{bmatrix} \)
  • \( \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix} \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

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}
\[ A^T X^{50} A = P^{50} = \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix}. \]
Was this answer helpful?
0
0