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  • Let <img alt="A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]" src="https://entrancecorner.oncodecogs.com/gif.latex?A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%20%26%20%

Let A=\left[\begin{array}{cc}1 & \frac{1}{51} \\ 0 & 1\end{array}\right]. If B=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right] A\left[\begin{array}{cc}-1 & -2 \\ 1 & 1\end{array}\right], then the sum of all the elements of the matrix \sum_{n=1}^{50} B^a is equal to

Option: 1

50


Option: 2

75


Option: 3

125


Option: 4

100


Answers (1)

best_answer

\begin{aligned} & \text { Let } \mathrm{C}=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right], \mathrm{D}=\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\ & D C=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I \\ \end{aligned}\begin{aligned} & \mathrm{~B}=\mathrm{CAD} \\ & B^n=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots . . .(\mathrm{CAD})}_{\mathrm{a}-\text {-ims }} \\ & \Rightarrow B^n=C^n D \quad \ldots(1) \\ & \mathrm{A}^2=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right] \\ & A^3=\left[\begin{array}{cc} 1 & \frac{3}{51} \\ 0 & 1 \end{array}\right] \\ & \text { Similarly } A^{\mathrm{n}}=\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right] \\ & \mathrm{Bn}=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\ & \end{aligned}

$$ \begin{aligned} & =\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51}+2 \\ -1 & -\frac{\mathrm{n}}{51}-1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\ & =\left[\begin{array}{cc} \frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\ -\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51} \end{array}\right] \\ & \sum_{\mathrm{n}=1}^{50} \mathrm{~B}^{\mathrm{n}}=\left[\begin{array}{cc} 25-50 & 25 \\ -25 & -25-50 \end{array}\right]=\left[\begin{array}{cc} 75 & 25 \\ -25 & 25 \end{array}\right] \\ & \text { Sum of the elements }=100 \end{aligned}

 

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Ritika Harsh

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