Let A be a 2 × 2 symmetric matrix such that A [ … ] and the determinant of A be 1 . If A^-1=α A+β I, where I is an identity matrix of order 2 × 2,…

$$\begin{aligned} & \text { Let } A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right] \\ & {\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right], a d-b^2=1} \\ & a+b=3, b+d=7,(3-b)(7-b)-b^2=1 \\ & 21-10 b=1 \rightarrow b=2, a=1, d=5 \\ & A=\left[\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right], A^{-1}=\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right] \\ & A^{-1}=\alpha A+\beta I \\ & {\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \alpha \\ 2 \alpha & 5 \alpha+\beta\end{array}\right]} \\ & \alpha=-1, \beta=6 \rightarrow \alpha+\beta=5\end{aligned}$$​ Let A=[ab​bd​][ab​bd​][11​]=[37​],ad−b2=1a+b=3,b+d=7,(3−b)(7−b)−b2=121−10b=1→b=2,a=1,d=5A=[12​25​],A−1=[5−2​−21​]A−1=αA+βI[5−2​−21​]=[α+β2α​2α5α+β​]α=−1,β=6→α+β=5​