If [ cases 1 & 1 0 & 1 cases ][ cases 1 & 2 0 & 1 cases ] [ cases 1...

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

And, we have \left[$\begin{cases}1 & 1 \\ 0 & 1\end{cases}$\right]\left[$\begin{cases}1 & 2 \\ 0 & 1\end{cases}$\right] = \left[$\begin{cases}1 & 3 \\ 0 & 1\end{cases}$\right] = \left[$\begin{cases}1 & 1 + 2 \\ 0 & 1\end{cases}$\right]</p><p>Also, \left[\begin{cases}1 & 1 \\ 0 & 1\end{cases}\right]\left[\begin{cases}1 & 2 \\ 0 & 1\end{cases}\right]\left[\begin{cases}1 & 3 \\ 0 & 1\end{cases}\right] = \left[\begin{cases}1 & 3 \\ 0 & 1\end{cases}\right]\left[\begin{cases}1 & 3 \\ 0 & 1\end{cases}\right]

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

⇒ \left[\begin{cases}1 & 1 \\ 0 & 1\end{cases}\right]\left[\begin{cases}1 & 2 \\ 0 & 1\end{cases}\right]\left[\begin{cases}1 & 3 \\ 0 & 1\end{cases}\right] . . . . \left[\begin{cases}1 & n - 1 \\ 0 & 1\end{cases}\right] = \left[\begin{cases}1 & 1 + 2 + 3 + . . . . + n - 1 \\ 0 & 1\end{cases}\right]

⇒ \left[\begin{cases}1 & 1 + 2 + 3 + . . . . + n - 1 \\ 0 & 1\end{cases}\right] = \left[\begin{cases}1 & 78 \\ 0 & 1\end{cases}\right]

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

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

$⇒ {n}^{2} - n - 156 = 0$

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

$⇒ n = 13$ or $n = - 12$ (reject as $n$ is a natural number)
$∴$ We have to find inverse of \left[$\begin{cases}1 & 13 \\ 0 & 1\end{cases}$\right]$</p><p>Which can be find by using${A}^{- 1} = \dfrac{adjA}{\left|A\right|}</p><p>Now, \left|A\right| = \left|\begin{cases}1 & 13 \\ 0 & 1\end{cases}\right| = 1 - 0 = 1

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

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

$∴${A}^{- 1} = \left[\begin{cases}1 & - 13 \\ 0 & 1\end{cases}\right] .