JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
Let \(\alpha\) be a solution of \(x^2+x+1=0\), and for some \(a\) and \(b\) in
\(\mathbb{R},\left[\begin{array}{lll}4 & \mathrm{a} & \mathrm{b}\end{array}\right]\left[\begin{array}{ccc}1 & 16 & 13 \\ -1 & -1 & 2 \\ -2 & -14 & -8\end{array}\right]=\left[\begin{array}{ccc}0 & 0 & 0\end{array}\right]\). If \(\frac{4}{\alpha^4}\) \(+\frac{\mathrm{m}}{\alpha^{\mathrm{a}}}+\frac{\mathrm{n}}{\alpha^{\mathrm{b}}}=3\), then \(\mathrm{m}+\mathrm{n}\) is equal to
- A 3
- B 11
- C 7
- D 8
Answer & Solution
Correct Answer
(B) 11
Step-by-step Solution
Detailed explanation
\(x^2+x+1=0\) \(\alpha\) is root \(\begin{aligned} & \therefore \alpha^2+\alpha+1=0 \\ & \Rightarrow \alpha=\omega \text { as } \omega^2 \text { [cube root of unity] } \end{aligned}\) also…
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