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3 and 4 .Determinants and Matrices
hard
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
A
$4 x ^2+24 x -27=0$
B
$4 x ^2-24 x +27=0$
C
$4 x ^2+24 x +27=0$
D
$4 x ^2-24 x -27=0$
(JEE MAIN-2023)
Solution
Put $x =0$
$\begin{aligned}& \left|\begin{array}{ccc}1 & 0 & 0 \\0 & \lambda & 0 \\0 & 0 & \lambda^2\end{array}\right|\end{aligned}=\frac{9}{8} \times 81$
$\lambda^3=\frac{9^3}{8} \therefore \lambda=\frac{9}{2}$
$\therefore \frac{\lambda}{3}=\frac{3}{2}$
$\therefore$ Required equation is : $x^2-x\left(\frac{9}{2}+\frac{3}{2}\right) x +\frac{27}{4}=0$ $4 x^2-24 x+27=0$
Standard 12
Mathematics