JEE Main 20246 Apr, Morning shiftMathematicsMediumMatrices and Determinants245× since 2002For α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R and a natural number nnn, let Ar=∣r1n22+α2r2n2−β3r−23n(3n−1)2∣A_r=\left|\begin{array}{ccc}r & 1 & \frac{n^2}{2}+\alpha \\ 2 r & 2 & n^2-\beta \\ 3 r-2 & 3 & \frac{n(3 n-1)}{2}\end{array}\right|Ar=r2r3r−21232n2+αn2−β2n(3n−1). Then 2A10−A82 A_{10}-A_82A10−A8 isAOption A. 4α+2β4 \alpha+2 \beta4α+2βBOption B. 0COption C. 2n2 n2nDOption D. 2α+4β2 \alpha+4 \beta2α+4βCheckSkip · show solution
