JEE Main 2025April, Unspecified (session-level dataset)MathematicsMediumMatrices and Determinants245× since 2002Let A=(α−1−16β)A=\begin{pmatrix}\alpha-1 & -1\\6 & \beta\end{pmatrix}A=(α−16−1β), α>0\alpha>0α>0, such that det(A)=0\det(A)=0det(A)=0 and α+β=1\alpha+\beta=1α+β=1. If III denotes the 2×22\times22×2 identity matrix, then the matrix (I+A)8(I+A)^8(I+A)8 is:AOption A. (4−16−1)\begin{pmatrix}4 & -1\\6 & -1\end{pmatrix}(46−1−1)BOption B. (257−64514−127)\begin{pmatrix}257 & -64\\514 & -127\end{pmatrix}(257514−64−127)COption C. (1025−5112024−1024)\begin{pmatrix}1025 & -511\\2024 & -1024\end{pmatrix}(10252024−511−1024)DOption D. (766−2551530−509)\begin{pmatrix}766 & -255\\1530 & -509\end{pmatrix}(7661530−255−509)CheckStep-by-step hintSkip · show solution
