Mathematics - Matrix Question with Solution | TestHub

Given $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]....\left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$

And, we have $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 1+2\\ 0& 1\end{array}\right]$

Also, $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 1+2+3\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]....\left[\begin{array}{cc}1& n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 1+2+3+....+n-1\\ 0& 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& 1+2+3+....+n-1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 78\\ 0& 1\end{array}\right]$

Using the sum of first $n$ natural numbers i.e. $1+2+3+...+n=\frac{n\left(n+1\right)}{2},$ we get

$\frac{n(n-1)}{2}=78$

$\Rightarrow n^2-n-156=0$

$\Rightarrow \left(n-13\right)\left(n+12\right)=0$

$\Rightarrow n=13$ or $n=-12$ (reject as $n$ is a natural number)
$\therefore$ We have to find inverse of $\left[\begin{array}{cc}1& 13\\ 0& 1\end{array}\right]$

Which can be find by using $A^{-1}=\frac{adjA}{\left|A\right|}$

Now, $\left|A\right|=\left|\begin{array}{cc}1& 13\\ 0& 1\end{array}\right|=1-0=1$

$A_{11}=1, A_{12}=-13, A_{21}=0, A_{22}=1$

Hence, $adjA=\left[\begin{array}{cc}1& -13\\ 0& 1\end{array}\right]$

$\therefore$$$$A^{-1}=\left[\begin{array}{cc}1& -13\\ 0& 1\end{array}\right].$