Question:medium

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:

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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: Nov 26, 2025
  • \( \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
  • \( \begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix} \)
  • \( \begin{bmatrix} 25 & 1 \\ 1 & -25 \end{bmatrix} \)
  • \( \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix} \)
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The Correct Option is D

Solution and Explanation

Step 1: {Demonstrate \( A \) is orthogonal}
Since \( A A^T = I \), 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}
\[A^T X^{50} A = P^{50} = \begin{bmatrix} 1 & 50 \\ 0 & 1 \end{bmatrix}.\]
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