JEE Main 20199 Apr, Morning shiftMathematicsHardMatrices and Determinants245× since 2002If [1101][1201]\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right][1011][1021][1301]\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right][1031]....[1n−101]=[17801]\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right][10n−11]=[10781], then the inverse of [1n01]\left[ {\matrix{ 1 & n \cr 0 & 1 \cr } } \right][10n1] isAOption A. [10121]\left[ {\matrix{ 1 & { 0} \cr {12} & 1 \cr } } \right][11201]BOption B. [10131]\left[ {\matrix{ 1 & { 0} \cr {13} & 1 \cr } } \right][11301]COption C. [1−1301]\left[ {\matrix{ 1 & { - 13} \cr 0 & 1 \cr } } \right][10−131]DOption D. [1−1201]\left[ {\matrix{ 1 & { - 12} \cr 0 & 1 \cr } } \right][10−121]CheckSkip · show solution
