Let three matrices A = begin{bmatrix} 2 & 1 | vedclass

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(A) First,calculate the product $BC$: $BC = \begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix} = \begin{bmatr

Vedclass · Accepted Answer

$6$

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(A) First,calculate the product $BC$: $BC = \begin{bmatrix} 3 & 4 \ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \ -2 & 3 \end{bmatrix} = \begin{bmatrix} 9-8 & -12+12 \ 6-6 & -8+9 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$. Since $BC = I$,then $(BC)^n = I^n = I$ for any positive integer $n$. The given series is $S = Tr(A) + Tr\left( \frac{AI}{2} ight) + Tr\left( \frac{AI^2}{4} ight) + Tr\left( \frac{AI^3}{8} ight) + \dots$ $S = Tr(A) + \frac{1}{2} Tr(A) + \frac{1}{4} Tr(A) + \frac{1}{8} Tr(A) + \dots$ This is an infinite geometric series with first term $a = Tr(A)$ and common ratio $r = \frac{1}{2}$. The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{Tr(A)}{1 - 1/2} = 2 Tr(A)$. Given $A = \begin{bmatrix} 2 & 1 \ 4 & 1 \end{bmatrix}$,$Tr(A) = 2 + 1 = 3$. Therefore,$S = 2(3) = 6$.

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