Matrices & Determinants
Trace of matrix
Grade 12
Question:
<p>Consider 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 the value of the sum \(\text{tr}(A) + \text{tr}\left(\dfrac{ABC}{2}\right) + \text{tr}\left(\dfrac{A(BC)^2}{4}\right) + \text{tr}\left(\dfrac{A(BC)^3}{8}\right) + \cdots + \infty\) is</p>
<p>6</p>
<p>9</p>
<p>12</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that BC forms a special matrix (possibly rotation/orthogonal), then use the cyclic property of trace: tr(XYZ) = tr(YZX) = tr(ZXY) to collapse the infinite series into a geometric series formula.
<p><strong>Step 1:</strong> Calculate BC:</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, all terms with (BC)^n simplify:</p><p>• tr(ABC/2) = (1/2)tr(AI) = (1/2)tr(A)</p><p>• tr(A(BC)²/4) = (1/4)tr(AI²) = (1/4)tr(A)</p><p>• tr(A(BC)³/8) = (1/8)tr(AI³) = (1/8)tr(A)</p><p><strong>Step 3:</strong> The sum becomes a geometric series:</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>S = tr(A) · 1/(1 - 1/2) = 2·tr(A)</p><p><strong>Step 4:</strong> Calculate tr(A):</p><p>tr(A) = 2 + 1 = 3</p><p>∴ Answer: S = 2 × 3 = <strong>6</strong></p>
Correct Answer: A