JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 8)
$$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$$, then $$\lambda, \frac{\lambda}{3}$$ are the roots of the equation :
$$4 x^{2}+24 x-27=0$$
$$4 x^{2}-24 x+27=0$$
$$4 x^{2}-24 x-27=0$$
$$4 x^{2}+24 x+27=0$$
Explanation
$$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$$
Put $x=0$
$$ \begin{aligned} & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{array}\right|=\frac{9}{8} \times 81 \\\\ & \lambda^3=\frac{3^6}{2^3} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \lambda=\frac{9}{2} \\\\ & \Rightarrow \frac{\lambda}{3}=\frac{3}{2} \end{aligned} $$
$$ \begin{aligned} & \text { Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\\\ & \Rightarrow 4 x^2-24 x+27=0 \end{aligned} $$
Put $x=0$
$$ \begin{aligned} & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{array}\right|=\frac{9}{8} \times 81 \\\\ & \lambda^3=\frac{3^6}{2^3} \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \lambda=\frac{9}{2} \\\\ & \Rightarrow \frac{\lambda}{3}=\frac{3}{2} \end{aligned} $$
$$ \begin{aligned} & \text { Equation: } x^2-\left(\frac{9}{2}+\frac{3}{2}\right) x+\frac{9}{2} \times \frac{3}{2}=0 \\\\ & \Rightarrow 4 x^2-24 x+27=0 \end{aligned} $$
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