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  • If <img alt="\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)" src="https://entrancecorner.oncodecog

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

Option: 1

4 x^{2}-24 x-27=0


Option: 2

4 x^{2}+24 x+27=0


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\left|\begin{array}{ccc}x+1 & x & x \\ x & x+d & x \\ x & x & x+d^{2}\end{array}\right|=\frac{9}{8}(103 x+81)

Put \: \mathrm{x}=0
\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda^{2} \end{array}\right|=\frac{9}{8} \times 81
\lambda^{3}=\frac{9^{3}}{8}
\lambda=\frac{9}{2}

\frac{\lambda}{3}=\frac{9}{2 \times 3} \Rightarrow \frac{3}{2}
\frac{\lambda}{3}=\frac{3}{2}

Option (C) \: 4 x^{2}-24 x+27=0
has \: Root\: \frac{3}{2}, \frac{9}{2}

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shivangi.shekhar

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