Question
If \[\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right]......\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right],\] then the inverse of \[\left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right]\] is:
A.
\[\left[ {\begin{array}{*{20}{c}}
1&0\\
{12}&1
\end{array}} \right]\]
B.
\[\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]\]
C.
\[\left[ {\begin{array}{*{20}{c}}
1&{ - 12}\\
0&1
\end{array}} \right]\]
D.
\[\left[ {\begin{array}{*{20}{c}}
1&0\\
{13}&1
\end{array}} \right]\]
Answer :
\[\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]\]
Solution :
\[\begin{array}{l}
\left[ {\begin{array}{*{20}{c}}
1&1\\
0&1
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
1&2\\
0&1
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
1&3\\
0&1
\end{array}} \right].\left[ {\begin{array}{*{20}{c}}
1&4\\
0&1
\end{array}} \right]......\left[ {\begin{array}{*{20}{c}}
1&{n - 1}\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]\\
\Rightarrow \,\,\left[ {\begin{array}{*{20}{c}}
1&{1 + 2 + 3 + ...... + \left( {n - 1} \right)}\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{78}\\
0&1
\end{array}} \right]\\
\Rightarrow \,\,\frac{{\left( {n - 1} \right)n}}{2} = 78\\
\Rightarrow \,\,{n^2} - n - 15 = 0\\
\Rightarrow \,\,n = 13\\
{\rm{Now, \,the \,matrix }}\left[ {\begin{array}{*{20}{c}}
1&n\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{13}\\
0&1
\end{array}} \right]
\end{array}\]
Then, the required inverse of
\[\left[ {\begin{array}{*{20}{c}}
1&{13}\\
0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]\]