JEE Main2019Math
Q.
If \left[ {\begin{array}{*{20}{c}}11 \\ 01 \end{array}} \right]\,.\,\left[ {\begin{array}{*{20}{c}}12 \\ 01 \end{array}} \right]\,\,\,\,\left[ {\begin{array}{*{20}{c}}13 \\ 01 \end{array}} \right]\,\,......\left[ {\begin{array}{*{20}{c}}1{n - 1} \\ 01 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1{78} \\ 01 \end{array}} \right],then the inverse of \left[ {\begin{array}{*{20}{c}} 1n \\ 01 \end{array}} \right] is
- A
\left[ {\begin{array}{*{20}{c}}10 \\ {13}1 \end{array}} \right] - B
\left[{\begin{array}{*{20}{c}}1{ - 13} \\ 01 \end{array}} \right] - C
\left[ {\begin{array}{*{20}{c}}1{ - 12} \\ 01 \end{array}} \right] - D
\left[{\begin{array}{*{20}{c}}10 \\ {12}1 \end{array}} \right]
Solution
