Let A = begin{bmatrix} 1 & frac{1}{51} 0 & 1 | vedclass

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(A) Let $C = \begin{bmatrix} 1 & 2 \ -1 & -1 \end{bmatrix}$ and $D = \begin{bmatrix} -1 & -2 \ 1 & 1 \end{bmatrix}$.Note that $CD = \begin{bmatrix}

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$100$

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(A) Let $C = \begin{bmatrix} 1 & 2 \ -1 & -1 \end{bmatrix}$ and $D = \begin{bmatrix} -1 & -2 \ 1 & 1 \end{bmatrix}$.Note that $CD = \begin{bmatrix} 1 & 2 \ -1 & -1 \end{bmatrix} \begin{bmatrix} -1 & -2 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$.Since $B = CAD$,we have $B^n = (CAD)(CAD)...(CAD) = CA^n D$.Given $A = \begin{bmatrix} 1 & \frac{1}{51} \ 0 & 1 \end{bmatrix}$,by induction $A^n = \begin{bmatrix} 1 & \frac{n}{51} \ 0 & 1 \end{bmatrix}$.Thus,$B^n = C A^n D = \begin{bmatrix} 1 & 2 \ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & \frac{n}{51} \ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -2 \ 1 & 1 \end{bmatrix}$.$B^n = \begin{bmatrix} 1 & \frac{n}{51} + 2 \ -1 & -\frac{n}{51} - 1 \end{bmatrix} \begin{bmatrix} -1 & -2 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{n}{51} + 1 & \frac{n}{51} \ -\frac{n}{51} & 1 - \frac{n}{51} \end{bmatrix}$.Summing from $n=1$ to $50$:$\sum_{n=1}^{50} B^n = \begin{bmatrix} \sum_{n=1}^{50} (\frac{n}{51} + 1) & \sum_{n=1}^{50} \frac{n}{51} \ \sum_{n=1}^{50} (-\frac{n}{51}) & \sum_{n=1}^{50} (1 - \frac{n}{51}) \end{bmatrix}$.Using $\sum_{n=1}^{50} n = \frac{50 	imes 51}{2} = 1275$,we get $\sum_{n=1}^{50} \frac{n}{51} = \frac{1275}{51} = 25$.$\sum_{n=1}^{50} B^n = \begin{bmatrix} 25 + 50 & 25 \ -25 & 50 - 25 \end{bmatrix} = \begin{bmatrix} 75 & 25 \ -25 & 25 \end{bmatrix}$.The sum of all elements is $75 + 25 - 25 + 25 = 100$.

Let $A = \begin{bmatrix} 1 & \frac{1}{51} \\ 0 & 1 \end{bmatrix}$. If $B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} A \begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix}$,then the sum of all the elements of the matrix $\sum_{n=1}^{50} B^n$ is equal to

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