If left|begin{array}{ccc}x+1 & x & x | vedclass

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Question

Put $x =0$

$\begin{aligned}& \left|\begin{array}{ccc}1 & 0 & 0 \0 & \lambda & 0 \0 & 0 & \lambda^2\end{array}ight|\

Vedclass · Accepted Answer

$4 x ^2-24 x +27=0$

Vedclass · Answer

Put $x =0$

$\begin{aligned}& \left|\begin{array}{ccc}1 & 0 & 0 \0 & \lambda & 0 \0 & 0 & \lambda^2\end{array}ight|\end{aligned}=\frac{9}{8} 	imes 81$

$\lambda^3=\frac{9^3}{8} 	herefore \lambda=\frac{9}{2}$

$	herefore \frac{\lambda}{3}=\frac{3}{2}$

$	herefore$ Required equation is : $x^2-x\left(\frac{9}{2}+\frac{3}{2}ight) x +\frac{27}{4}=0$ $4 x^2-24 x+27=0$

If $\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^2\end{array}\right|=\frac{9}{8}(103 x+81)$, then $\lambda$, $\frac{\lambda}{3}$ are the roots of the equation

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