JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(\left[ {\begin{array}{*{20}{c}}
1&{ - 12}\\
0&1
\end{array}} \right]\) - B \(\left[ {\begin{array}{*{20}{c}}
1&0\\
{13}&1
\end{array}} \right]\) - C \(\left[ {\begin{array}{*{20}{c}}
1&0\\
{12}&1
\end{array}} \right]\) - D \(\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]\)
Answer & Solution
Correct Answer
(D) \(\left[ {\begin{array}{*{20}{c}}
1&{ - 13}\\
0&1
\end{array}} \right]\)
Step-by-step Solution
Detailed explanation
\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[…
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