Let three matrices be 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 <br> tr(A) + tr\left(\frac{ABC}{2}\right) + tr\left(\frac{A(BC)^2}{4}\right) + tr\left(\frac{A(BC)^3}{8}\right) + ....... + \infty is equal to
Step-by-Step Solution
Key Concept: The expression is a sum of traces of terms forming a geometric series of matrices. Specifically, it is tr(A(I + (BC/2) + (BC/2)^2 + ...)). Since BC = I, the series simplifies to tr(A(I + I/2 + I/4 + ...)) = tr(A(2I)) = 2tr(A).
First, calculate BC: <br> BC = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} 9-8 & -12+12 \\ 6-6 & -8+9 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I. <br> The given series is tr(A) + tr(A(BC)/2) + tr(A(BC)^2/4) + ... <br> Since BC = I, this becomes tr(A) + tr(A/2) + tr(A/4) + tr(A/8) + ... <br> = tr(A) (1 + 1/2 + 1/4 + 1/8 + ...) <br> = tr(A) * (1 / (1 - 1/2)) = tr(A) * 2. <br> Given A = \begin{bmatrix} 2 & 1 \\ 4 & 1 \end{bmatrix}, tr(A) = 2 + 1 = 3. <br> Therefore, the sum is 3 * 2 = 6.
Correct Answer: 1