JEE Main 2023 — Mathematics Algebra

Given that, $|\begin{matrix}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+{\lambda }^{2}\end{matrix}|=\frac{9}{8}(103x+81)$∣x+1xx​xx+λx​xxx+λ2​∣=89​(103x+81)

Put $x=0$x=0 as $x\in R$x∈R

$$\Rightarrow |\begin{matrix}1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & {\lambda }^{2}\end{matrix}|=\frac{9}{8}(103(0)+81)$$⇒∣100​0λ0​00λ2​∣=89​(103(0)+81)

On expanding the determinant we get,

$$\Rightarrow {\lambda }^{3}=\frac{9}{8}\times 81$$⇒λ3=89​×81

$$\Rightarrow {\lambda }^{3}=\frac{{9}^{3}}{{2}^{3}}$$⇒λ3=2393​

$\Rightarrow \lambda =\frac{9}{2}$⇒λ=29​ and $\frac{\lambda }{3}=\frac{9}{6}$3λ​=69​

Now the required quadratic equation is ${x}^{2}-(\frac{9}{2}+\frac{9}{6})x+\frac{9}{2}\times \frac{9}{6}=0$x2−(29​+69​)x+29​×69​=0

$$\Rightarrow {x}^{2}-(6)x+\frac{27}{4}=0$$⇒x2−(6)x+427​=0

$$\Rightarrow 4{x}^{2}-24x+27=0$$⇒4x2−24x+27=0

Hence, the required quadratic equation is $4{x}^{2}-24x+27=0$4x2−24x+27=0