જો $\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]\,........\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$ તો $\left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right]$ નો વ્યસ્ત શ્રેણિક મેળવો.
1&1\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]\,........\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right]\, = \,\left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$ તો $\left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right]$ નો વ્યસ્ત શ્રેણિક મેળવો.
JEE MAIN 2019, Difficult
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$\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]........\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$
1&1\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]........\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right]$ $ = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
1&{1 + 2 + 3 + ... + n - 1}\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]$
$ \Rightarrow \frac{{n\left( {n - 1} \right)}}{2} = 78 \Rightarrow n = 13, - 12$ (reject)
$\therefore $ we have to find inverse of $\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]$
$\therefore \left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]$
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