JEE Main 2023 (Online) 12th April Morning ShiftQuestion 1
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
Matrices
Solution
With $P=\begin{bmatrix}1&2\\-1&-1\end{bmatrix}$, $P^{-1}=\begin{bmatrix}-1&-2\\1&1\end{bmatrix}$, so $B=PAP^{-1}$ and $B^n=PA^nP^{-1}$, $A^n=\begin{bmatrix}1&n/51\\0&1\end{bmatrix}$. $\sum_{n=1}^{50}B^n=P\begin{bmatrix}50&\frac{1275}{51}\\0&50\end{bmatrix}P^{-1}=P\begin{bmatrix}50&25\\0&50\end{bmatrix}P^{-1}$. $P\begin{bmatrix}50&25\\0&50\end{bmatrix}=\begin{bmatrix}50&125\\-50&-75\end{bmatrix}$; times $P^{-1}$: $\begin{bmatrix}-50+125&-100+125\\50-75&100-75\end{bmatrix}=\begin{bmatrix}75&25\\-25&25\end{bmatrix}$. Sum of elements $=75+25-25+25=100$. Option (C).