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) \[\left[ \begin{matrix} 1 & 0 \\ 13 & 1 \\ \end{matrix} \right]\]
Correct Answer: A
Solution :
\[\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 3 \\ 0 & 1 \\ \end{matrix} \right].....\left[ \begin{matrix} 1 & n-1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 78 \\ 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \left[ \begin{matrix} 1 & 1+2+3+.....+n-1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 78 \\ 0 & 1 \\ \end{matrix} \right]\] \[\Rightarrow \frac{n(n-2)}{2}=78\Rightarrow n=13,-12\](reject) \[\therefore \]We have to find inverse of\[\left[ \begin{matrix} 1 & 13 \\ 0 & 1 \\ \end{matrix} \right]\] \[\therefore \]\[\left[ \begin{matrix} 1 & -13 \\ 0 & 1 \\ \end{matrix} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
