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.

Question

Aicharya · Accepted Answer

Given that, $\begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix} = \frac{9}{8}(103x + 81)$ Put $x=0$ as $x \in \mathbb{R}$ $\Rightarrow \begin{vmatrix} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^2 \end{vmatrix} = \frac{9}{8}(103 \cdot 0 + 81)$

On expanding the determinant we get,

$\Rightarrow λ^{3} = \frac{9}{8} \times 81$

$\Rightarrow λ^{3} = \frac{9^{3}}{2^{3}}$

$\Rightarrow λ = \frac{9}{2}$ and $\frac{λ}{3} = \frac{9}{6}$

Now the required quadratic equation is $x^{2} - (\frac{9}{2} + \frac{9}{6}) x + \frac{9}{2} \times \frac{9}{6} = 0$

$\Rightarrow x^{2} - (6) x + \frac{27}{4} = 0$

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

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

If $\begin{vmatrix} x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2 \end{vmatrix} =…

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