<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>Trace 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 powers of A, then compute trace(A^50) using the trace's linearity property: trace(A^n) = trace(A^(n-2)) + trace(A^2) - trace(I).
<p><strong>Step 1:</strong> Compute A², A³, and trace values.</p><p>A² = [[1,0,0],[2,0,1],[1,0,0]] (by matrix multiplication)</p><p>trace(A²) = 1 + 0 + 0 = 1</p><p>trace(I) = 3</p><p>trace(A) = 1 + 0 + 0 = 1</p><p><strong>Step 2:</strong> Apply linearity of trace to the recurrence relation.</p><p>Given: A^n = A^(n-2) + A^2 - I for n ≥ 3</p><p>Taking trace: trace(A^n) = trace(A^(n-2)) + trace(A^2) - trace(I)</p><p>trace(A^n) = trace(A^(n-2)) + 1 - 3</p><p>trace(A^n) = trace(A^(n-2)) - 2</p><p><strong>Step 3:</strong> Solve the recurrence relation.</p><p>This is a linear recurrence with period. For even n = 2k:</p><p>trace(A^(2k)) = trace(A^(2k-2)) - 2</p><p>Compute initial values: trace(A) = 1, trace(A²) = 1, trace(A³) = trace(A) - 2 = -1, trace(A⁴) = trace(A²) - 2 = -1</p><p>Pattern: trace(A^(2k)) = 1 - 2(k-1) = 3 - 2k</p><p><strong>Step 4:</strong> Calculate for n = 50.</p><p>50 = 2(25), so k = 25</p><p>trace(A^50) = 3 - 2(25) = 3 - 50 = -47</p><p>∴ Answer: D (trace of A^50 = -47)</p>
Correct Answer: D