Question

If
$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\cdots \left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$,
then the inverse of $\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$ will be:

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

Answer 1

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

$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\cdots \left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}1& 1+2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\cdots \left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}1& 1+2+3\\ 0& 1\end{array}\right]\cdots \left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}1& 1+2+3+\cdots n-1\\ 0& 1\end{array}\right]\cdots \left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$
On comparing, we get
$\frac{n(n-1)}{2}=78\Rightarrow n^2-n-136=0\Rightarrow n^2-13n+12n-136=0\Rightarrow n=13,n=-12\left(Reject\right)$
$\therefore$ Inverse of $$\left[\begin{array}{cc}1& 13\\ 0& 1\end{array}\right]$$ is
$$\left[\begin{array}{cc}1& -13\\ 0& 1\end{array}\right]$$