JEE Main 20238 Apr, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002If A=[15λ10],A−1=αA+βIA=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], \mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}A=[1λ510],A−1=αA+βI and α+β=−2\alpha+\beta=-2α+β=−2, then 4α2+β2+λ24 \alpha^{2}+\beta^{2}+\lambda^{2}4α2+β2+λ2 is equal to :AOption A. 12BOption B. 10COption C. 19DOption D. 14CheckSkip · show solution
