JEE Main 202325 Jan, Evening shiftMathematicsHardMatrices and Determinants245× since 2002Let A=[110310−310110]A = \left[ {\matrix{ {{1 \over {\sqrt {10} }}} & {{3 \over {\sqrt {10} }}} \cr {{{ - 3} \over {\sqrt {10} }}} & {{1 \over {\sqrt {10} }}} \cr } } \right]A=[10110−3103101] and B=[1−i01]B = \left[ {\matrix{ 1 & { - i} \cr 0 & 1 \cr } } \right]B=[10−i1], where i=−1i = \sqrt { - 1}i=−1. If M=ATBA\mathrm{M=A^T B A}M=ATBA, then the inverse of the matrix AM2023AT\mathrm{AM^{2023}A^T}AM2023AT isAOption A. [1−2023i01]\left[ {\matrix{ 1 & { - 2023i} \cr 0 & 1 \cr } } \right][10−2023i1]BOption B. [102023i1]\left[ {\matrix{ 1 & 0 \cr {2023i} & 1 \cr } } \right][12023i01]COption C. [12023i01]\left[ {\matrix{ 1 & {2023i} \cr 0 & 1 \cr } } \right][102023i1]DOption D. [10−2023i1]\left[ {\matrix{ 1 & 0 \cr { - 2023i} & 1 \cr } } \right][1−2023i01]CheckSkip · show solution
