JEE Main 2023 (Online) 11th April Evening ShiftQuestion 1
If $\begin{vmatrix}x+1&x&x\\x&x+\lambda&x\\x&x&x+\lambda^{2}\end{vmatrix}=\frac{9}{8}(103x+81)$, then $\lambda,\frac{\lambda}{3}$ are the roots of the equation
Determinant
Solution
Expand: $D=\lambda^3(x+1)+\lambda^2x+\lambda x$ (i.e. $\lambda^3+x(\lambda^3+\lambda^2+\lambda)$). Comparing with $\frac98(103x+81)$: constant $\lambda^3=\frac{729}{8}\Rightarrow\lambda=\frac92$; check $x$-coefficient $\lambda^3+\lambda^2+\lambda=\frac{927}{8}$ ✓. Roots $\frac92,\frac32$: sum $=6$, product $=\frac{27}{4}$, so $4x^2-24x+27=0$. Option D.