Question

If $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]..\left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$, then the inverse of $\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$ is?

• A. $\left[\begin{array}{cc}1& -13\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 0\\ 12& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& -12\\ 0& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 0\\ 13& 1\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& -13\\ 0& 1\end{array}\right]$

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