• # question_answer If$\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right].\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right].\left[ \begin{matrix} 1 & 3 \\ 0 & 1 \\ \end{matrix} \right]......\left[ \begin{matrix} 1 & n-1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 78 \\ 0 & 1 \\ \end{matrix} \right],$then the inverse of$\left[ \begin{matrix} 1 & n \\ 0 & 1 \\ \end{matrix} \right]$is [JEE Main 9-4-2019 Morning] A) $\left[ \begin{matrix} 1 & -13 \\ 0 & 1 \\ \end{matrix} \right]$              B) $\left[ \begin{matrix} 1 & 0 \\ 12 & 1 \\ \end{matrix} \right]$C) $\left[ \begin{matrix} 1 & -12 \\ 0 & 1 \\ \end{matrix} \right]$              D) $\left[ \begin{matrix} 1 & 0 \\ 13 & 1 \\ \end{matrix} \right]$

$\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 3 \\ 0 & 1 \\ \end{matrix} \right].....\left[ \begin{matrix} 1 & n-1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 78 \\ 0 & 1 \\ \end{matrix} \right]$ $\Rightarrow \left[ \begin{matrix} 1 & 1+2+3+.....+n-1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 78 \\ 0 & 1 \\ \end{matrix} \right]$ $\Rightarrow \frac{n(n-2)}{2}=78\Rightarrow n=13,-12$(reject) $\therefore$We have to find inverse of$\left[ \begin{matrix} 1 & 13 \\ 0 & 1 \\ \end{matrix} \right]$ $\therefore$$\left[ \begin{matrix} 1 & -13 \\ 0 & 1 \\ \end{matrix} \right]$