Let be a solution of , and for some a and b in R, . If , then m + n is equal to __________ [2025]

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Solution

Q.
Let $α$ be a solution of $x^{2} + x + 1 = 0$ , and for some a and b in R, $[\begin{matrix} 4 & a & b \end{matrix}] [\begin{matrix} 1 & 16 & 13 \\ - 1 & - 1 & 2 \\ - 2 & - 14 & - 8 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \end{matrix}]$ . If $\frac{4}{α^{4}} + \frac{m}{α^{a}} + \frac{n}{α^{b}} = 3$ , then m + n is equal to __________ [2025]

1 8

2 3

3 7

4 11

Ans.
(4)

Given, $α$ is a solution of $x^{2} + x + 1 = 0$

$\Rightarrow α^{2} + α + 1 = 0$

$\Rightarrow α = ω or ω^{2}$ (cube root of unity)

Now, $[\begin{matrix} 4 & a & b \end{matrix}] [\begin{matrix} 1 & 16 & 13 \\ - 1 & - 1 & 2 \\ - 2 & - 14 & - 8 \end{matrix}] [\begin{matrix} 0 & 0 & 0 \end{matrix}]$

$\Rightarrow [\begin{matrix} 4 - a - 2 b & 64 - a - 14 b & 52 + 2 a - 8 b \end{matrix}] = [\begin{matrix} 0 & 0 & 0 \end{matrix}]$

$\Rightarrow a + 2 b = 4; a + 14 b = 64 and a - 4 b = - 26$

On solving above equations we get b = 5 and a = –6

Now, $\frac{4}{α^{4}} + \frac{m}{α^{a}} + \frac{n}{α^{b}} = 3 \Rightarrow \frac{4}{ω^{4}} + \frac{m}{ω^{- 6}} + \frac{n}{ω^{5}} = 3$

$\Rightarrow \frac{4}{ω} + \frac{m}{1} + \frac{n}{ω^{2}} = 3$ [ $∵ ω^{3} = 1$ ]

$\Rightarrow 4 ω^{2} + m + n ω = 3$

$\Rightarrow 4 [\frac{- 1 - \sqrt{3} i}{2}] + m + n [\frac{- 1 + \sqrt{3} i}{2}] = 3$

$\Rightarrow - 2 - 2 \sqrt{3} i + m - \frac{n}{2} + \frac{n \sqrt{3} i}{2} = 3$

$\Rightarrow - 2 + m - \frac{n}{2} = 3 and \frac{n \sqrt{3}}{2} - 2 \sqrt{3} = 0$ [Comparing real and imaginary part]

$\Rightarrow$ m = 7, n = 4 $∴$ m + n = 11.

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