JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(4 x ^2-24 x -27=0\)
Answer & Solution
Correct Answer
(B) \(4 x ^2-24 x +27=0\)
Step-by-step Solution
Detailed explanation
Put \(x =0\) \(\begin{aligned}& \left|\begin{array}{ccc}1 & 0 & 0 \\0 & \lambda & 0 \\0 & 0 & \lambda^2\end{array}\right|\end{aligned}=\frac{9}{8} \times 81\) \(\lambda^3=\frac{9^3}{8} \therefore \lambda=\frac{9}{2}\) \(\therefore \frac{\lambda}{3}=\frac{3}{2}\) \(\therefore\)…
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