The correct option is (A): \(24\)
If \( A = \begin{bmatrix} 1 & 5 \\ \lambda & 10 \end{bmatrix} \), \( A^{-1} = \alpha A + \beta I \) and \( \alpha + \beta = -2 \), then \( 4\alpha^2 + \beta^2 + \lambda^2 \) is equal to:
Let P = \(\left[\begin{matrix} \frac{\sqrt3}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt3}{2} \end{matrix}\right]\) A = \(\left[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right]\) and Q = PAPT. If PTQ2007P = \(\left[\begin{matrix} a & b \\ c & d \end{matrix}\right]\), then 2a+b-3c-4d equal to