Matrices & Determinants
Trace of matrices and infinite series
Grade 12
Question:
<p>Let three matrices \(A = \begin{bmatrix} 2 & 1 \\ 4 & 1 \end{bmatrix}\), \(B = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix}\) and \(C = \begin{bmatrix} 3 & -4 \\ -2 & 3 \end{bmatrix}\), then \(\text{tr}(A) + \text{tr}\!\left(\dfrac{A(BC)}{2}\right) + \text{tr}\!\left(\dfrac{A(BC)^2}{4}\right) + \text{tr}\!\left(\dfrac{A(BC)^3}{8}\right) + \cdots\infty\) is equal to ___________.</p>
Step-by-Step Solution
Key Concept: Recognize that BC forms a matrix with spectral radius < 1, allowing the infinite series Σtr(A(BC)^n/2^n) to be summed using tr(A(BC)^n/2^n) = tr(A)/2^n · tr((BC)^n) and the geometric series formula for traces.
<p><strong>Step 1:</strong> Verify that C is the inverse of B.</p><p>BC = <begin>bmatrix</begin>3 & 4 \\ 2 & 3<end>bmatrix}</begin><begin>bmatrix</begin>3 & -4 \\ -2 & 3<end>bmatrix></begin> = <begin>bmatrix</begin>9-8 & -12+12 \\ 6-6 & -8+9<end>bmatrix></begin> = <begin>bmatrix</begin>1 & 0 \\ 0 & 1<end>bmatrix></begin> = I</p><p><strong>Step 2:</strong> Since BC = I, we have (BC)ⁿ = I for all n ≥ 1.</p><p><strong>Step 3:</strong> The series becomes:</p><p>tr(A) + Σ(n=1 to ∞) tr(A·Iⁿ/2ⁿ) = tr(A) + Σ(n=1 to ∞) tr(A)/2ⁿ</p><p>= tr(A) + tr(A)·Σ(n=1 to ∞) 1/2ⁿ</p><p><strong>Step 4:</strong> Calculate tr(A) = 2 + 1 = 3</p><p><strong>Step 5:</strong> The geometric series: Σ(n=1 to ∞) 1/2ⁿ = 1/2 + 1/4 + 1/8 + ... = (1/2)/(1-1/2) = 1</p><p><strong>Step 6:</strong> Final sum = 3 + 3(1) = 3 + 3 = 6</p><p>∴ Answer: <strong>6</strong></p>
Correct Answer: 6