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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
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
    $4 x ^2+24 x -27=0$
  • ✓
    $4 x ^2-24 x +27=0$
  • C
    $4 x ^2+24 x +27=0$
  • D
    $4 x ^2-24 x -27=0$
✓

Answer

Correct option: B.
$4 x ^2-24 x +27=0$
b
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 : $x^2-x\left(\frac{9}{2}+\frac{3}{2}\right) x +\frac{27}{4}=0$ $4 x^2-24 x+27=0$

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