1.Let A=(α−16β)A = \begin{pmatrix} \alpha & -1 \\ 6 & \beta \end{pmatrix}A=(α6−1β), 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 \times 22×2 identity matrix, then the matrix (I+A)8(I + A)^{8}(I+A)8 is:a.(4−16−1)\begin{pmatrix} 4 & -1 \\ 6 & -1 \end{pmatrix}(46−1−1)b.(257−64514−127)\begin{pmatrix} 257 & -64 \\ 514 & -127 \end{pmatrix}(257514−64−127)c.(1025−5112024−1024)\begin{pmatrix} 1025 & -511 \\ 2024 & -1024 \end{pmatrix}(10252024−511−1024)d.(766−2551530−509)\begin{pmatrix} 766 & -255 \\ 1530 & -509 \end{pmatrix}(7661530−255−509)Login to continueOnly logged in users canattempt or see the solution.
