If omega is cube root of unity, then root of equation $left| {begin{array}{*{20}{c}} {x + 2}&

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3 and 4 .Determinants and Matrices

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If $\omega $ is cube root of unity, then root of the equation $\left| {\begin{array}{*{20}{c}}
{x + 2}&\omega &{{\omega ^2}} \\
\omega &{x + 1 + {\omega ^2}}&1 \\
{{\omega ^2}}&1&{x + 1 + \omega }
\end{array}} \right| = 0$ is

A

$0$

B

$\omega $

C

$\omega ^2$

D

$-1$

Solution

$\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3}$

$\Rightarrow(x+1)\left|\begin{array}{ccc}{1} & {1} & {1} \\ {\omega} & {x+1+\omega^{2}} & {1} \\ {\omega^{2}} & {1} & {x+1+\omega}\end{array}\right|$

$ = 0 \Rightarrow {({\text{x}} + 1)^3} = 0 \Rightarrow \boxed{{\text{x}} = – 1}$

Standard 12

Mathematics

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