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

$$ \begin{aligned} & \text { Let } C=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right], \mathrm{D}=\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & \mathrm{DC}=\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\mathrm{I} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \mathrm{B}=\mathrm{CAD} \\\\ & \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots(\mathrm{CAD})}_{\text {n-times }} \end{aligned} $$ <br/><br/>$\Rightarrow \mathrm{B}^{\mathrm{n}}=\mathrm{CA}^{\mathrm{n}} \mathrm{D}$ <br/><br/>$$ \mathrm{A}^2=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right] $$ <br/><br/>$$ \mathrm{A}^3=\left[\begin{array}{cc} 1 & \frac{1}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{2}{51} \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 1 & \frac{3}{51} \\ 0 & 1 \end{array}\right] $$ <br/><br/>$$ \text { Similarly } A^{\mathrm{n}}=\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right] $$ <br/><br/>$$ \begin{aligned} \mathrm{B}^{\mathrm{n}} & =\left[\begin{array}{cc} 1 & 2 \\ -1 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51}+2 \\ -1 & -\frac{\mathrm{n}}{51}-1 \end{array}\right]\left[\begin{array}{cc} -1 & -2 \\ 1 & 1 \end{array}\right] \\\\ & =\left[\begin{array}{cc} \frac{\mathrm{n}}{51}+1 & \frac{\mathrm{n}}{51} \\ -\frac{\mathrm{n}}{51} & 1-\frac{\mathrm{n}}{51} \end{array}\right] \end{aligned} $$ <br/><br/>$$ \sum\limits_{n=1}^{50} B^n=\left[\begin{array}{cc} 50+\frac{50 \cdot 51}{2 \cdot 51} & \frac{50 \cdot 51}{2 \cdot 51} \\\\ \frac{-50 \cdot 51}{2 \cdot 51} & 50-\frac{50 \cdot 51}{2 \cdot 51} \end{array}\right]=\left[\begin{array}{cc} 75 & 25 \\ -25 & 25 \end{array}\right] $$ <br/><br/>$\therefore$ Sum of the elements = 100