AP EAMCET · Maths · Determinants
If \(\omega\) is a root of the equation \(x+\frac{1}{x}+1=0\), then \(\left|\begin{array}{ccc}1 & 1+\omega & 1+\omega+\omega^2 \\ 3 & 4+3 \omega & 5+4 \omega+3 \omega^2 \\ 6 & 9+6 \omega & 11+9 \omega+6 \omega^2\end{array}\right|\) is equal to
- A 1
- B -1
- C 0
- D \(1+\infty\)
Answer & Solution
Correct Answer
(B) -1
Step-by-step Solution
Detailed explanation
\(\omega\) is root of \(x+\frac{1}{x}+1=0\), i.e. \(x^2+x+1=0\) This gives roots \(\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm i \sqrt{3}}{2}\) \(\therefore \omega=\frac{-1+\sqrt{-3}}{2}\) and \(\omega^2=\frac{-1-\sqrt{-3}}{2}\) are roots of \(x^2+x+1=0\) Also,…
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