1.Let AAA be a 2×22 \times 22×2 matrix with real entries such that AT=αA+IA^T = \alpha A + IAT=αA+I, where α∈R∖{−1,1}\alpha \in \mathbb{R} \setminus \{-1, 1\}α∈R∖{−1,1}. If det(A2−A)=4\det(A^2 - A) = 4det(A2−A)=4, the sum of all possible values of α\alphaα is equal to:a.000b.12\dfrac{1}{2}21c.222d.52\dfrac{5}{2}25Login to continueOnly logged in users canattempt or see the solution.
