Let the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 1 \\0 & 1 & 0 \end{pmatrix} $ satisfy $ A^n = A^{n-2} + A^2 - I $ for $ n \geq 3 $. Then the sum of all the elements of $ A^{50} $ is:

Question

Let $ A = \begin{bmatrix} 1 & \sqrt{2} \\ -2 & 0 & 1 \end{bmatrix} $ and $ P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} $, $ \theta > 0 $. If $ B = P A P^T $, $ C = P^T B P $, and the sum of the diagonal elements of $ C $ is $ \frac{m}{n} $, where $ \gcd(m, n) = 1 $, then $ m + n $ is:

Answer 1

To determine higher powers of matrix $ A $, the recurrence relation $ A^n = A^{n-2} + A^2 - I $ for $ n \geq 3 $ is employed. Matrix algebra confirms that the sum of the elements in $ A^{50} $ equals 53.
The correct answer is therefore $ 53 $.

Let the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 1 \\0 & 1 & 0 \end{pmatrix} $ satisfy $ A^n = A^{n-2} + A^2 - I $ for $ n \geq 3 $. Then the sum of all the elements of $ A^{50} $ is:

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The Correct Option is A

Solution and Explanation

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