<p>We have \[BC = \begin{bmatrix}3 & 4\\2 & 3\end{bmatrix}\begin{bmatrix}3 & -4\\-2 & 3\end{bmatrix} = I.\] Find \[\text{tr}(A) + \text{tr}\!\left(\frac{A(BC)}{2}\right) + \text{tr}\!\left(\frac{A(BC)^2}{4}\right) + \text{tr}\!\left(\frac{A(BC)^3}{8}\right) + \cdots\infty,\] given that \(A = \begin{bmatrix}2 & 1\\1 & 1\end{bmatrix}\) (so that \(\text{tr}(A)=3\) and the series converges).</p>
Step-by-Step Solution
Key Concept: Since BC = I, the series becomes tr(A) + tr(A/2) + tr(A/4) + tr(A/8) + ⋯ = tr(A)·(1 + 1/2 + 1/4 + 1/8 + ⋯). The trace is linear, so tr(kM) = k·tr(M), and the geometric series sums to 2.
<p><strong>Step 1:</strong> Verify BC = I by multiplication:</p><p>BC = [3·3 + 4·(-2), 3·(-4) + 4·3; 2·3 + 3·(-2), 2·(-4) + 3·3] = [9-8, -12+12; 6-6, -8+9] = [1,0; 0,1] = I ✓</p><p><strong>Step 2:</strong> Since (BC)ⁿ = Iⁿ = I for all n ≥ 1, rewrite the series:</p><p>S = tr(A) + tr(A·I/2) + tr(A·I/4) + tr(A·I/8) + ⋯</p><p><strong>Step 3:</strong> Use linearity of trace: tr(kM) = k·tr(M)</p><p>S = tr(A) + (1/2)tr(A) + (1/4)tr(A) + (1/8)tr(A) + ⋯</p><p>S = tr(A)·[1 + 1/2 + 1/4 + 1/8 + ⋯]</p><p><strong>Step 4:</strong> The bracketed term is a geometric series with first term a = 1 and common ratio r = 1/2:</p><p>Geometric sum = 1/(1 - 1/2) = 1/(1/2) = 2</p><p><strong>Step 5:</strong> Substitute tr(A) = 3:</p><p>S = 3 × 2 = 6</p><p>∴ <strong>Answer: 6</strong></p>
Correct Answer: 6