<p><strong>For Problems 9–11</strong><br>Let \(A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\) satisfies \(A^n = A^{n-2} + A^2 - I\) for \(n \geq 3\). And trace of a square matrix \(X\) is equal to the sum of elements in its principal diagonal.<br>Further consider a matrix \(U_{3\times 3}\) with its columns as \(U_1, U_2, U_3\) such that<br>\[A^{50}U_1 = \begin{bmatrix}1\\25\\25\end{bmatrix},\quad A^{50}U_2 = \begin{bmatrix}0\\1\\0\end{bmatrix},\quad A^{50}U_3 = \begin{bmatrix}0\\0\\1\end{bmatrix}\]<br>The value of \(|A^{50}|\) equals</p>
Step-by-Step Solution
Key Concept: Use the recurrence relation A^n = A^(n-2) + A^2 - I to find a pattern in determinants, then recognize that |A^50| = |A|^50 can be computed by finding the characteristic pattern of |A^n| through the recurrence, or directly compute |A| from the given matrix and raise to power 50.
<p><strong>Step 1:</strong> Calculate |A| from the given matrix A = [[1,0,0], [1,0,1], [0,1,0]].</p><p>Expanding along first row: |A| = 1·|[0,1], [1,0]| = 1·(0-1) = -1</p><p><strong>Step 2:</strong> Verify consistency with recurrence relation A^n = A^(n-2) + A^2 - I.</p><p>Taking determinants: |A^n| = |A^(n-2) + A^2 - I|. For a linear recurrence with constant terms, we can verify that |A^n| follows a pattern determined by |A|.</p><p><strong>Step 3:</strong> Use the fundamental property |A^n| = |A|^n.</p><p>Therefore: |A^50| = |A|^50 = (-1)^50 = 1</p><p><strong>Verification:</strong> The recurrence relation ensures consistency. The vectors U₁, U₂, U₃ define columns of U where A^50·U = [U₁', U₂', U₃'] with the given right-hand sides, which is independent of the determinant calculation.</p><p>∴ Answer: B (|A^50| = 1)</p>
Correct Answer: B