1.If A=(15λ10)A = \begin{pmatrix} 1 & 5 \\ \lambda & 10 \end{pmatrix}A=(1λ510), A−1=αA+βIA^{-1} = \alpha A + \beta IA−1=αA+βI and α+β=−2\alpha + \beta = -2α+β=−2, then 4α2+β2+λ24\alpha^2 + \beta^2 + \lambda^24α2+β2+λ2 is equal to:a.121212b.191919c.141414d.101010Login to continueOnly logged in users canattempt or see the solution.
