Let A=[array|c|c| 1 & (1)/(51) 0 & 1 array].

Question

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

Accepted Answer

$$
\begin{aligned}
& \text { Let } C=\left[\[\begin{array}{cc} 1 & 2 \ -1 & -1 \end{array}\end{aligned}$$\right], \mathrm{D}=\left[$$\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}

\]

$$
\begin{aligned}
& \mathrm{B}=\mathrm{CAD} \\
& \mathrm{B}^{\mathrm{n}}=\underbrace{(\mathrm{CAD})(\mathrm{CAD})(\mathrm{CAD}) \ldots(\mathrm{CAD})}_{\text {n-times }}
\end{aligned}
$$

$$
\Rightarrow \mathrm{B}^{\mathrm{n}}=\mathrm{CA}^{\mathrm{n}} \mathrm{D}
$$

$$
\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]
\]

$$
\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]
\]

$$
\text { Similarly } A^{\mathrm{n}}=\left[\[\begin{array}{cc} 1 & \frac{\mathrm{n}}{51} \ 0 & 1 \end{array}$$\right]
\]

$$
\begin{aligned}
\mathrm{B}^{\mathrm{n}} & =\left[\[\begin{array}{cc} 1 & 2 \ -1 & -1 \end{array}\end{aligned}$$\right]\left[$$\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]

\]

$$
\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]
\]

$$ \therefore $$ Sum of the elements = 100

Solutions · 1