If $ω$ is a complex cube root of unity, then a root of the equation $\begin{vmatrix}x+1&ω&ω^2\\ω&x+ω^2&1\\ω^2&1&x+ω\end{vmatrix}=0$, is
Answer & explanation
Correct answer: option 4
We have,
$\begin{vmatrix}x+1&ω&ω^2\\ω&x+ω^2&1\\ω^2&1&x+ω\end{vmatrix}=0$
$⇒\begin{vmatrix}x+1+ω+ω^2&ω&ω^2\\x+1+ω+ω^2&x+ω^2&1\\x+1+ω+ω^2&1&x+ω\end{vmatrix}=0$ [Applying $C_1→C_1+C_2+C_3$]
$⇒(x+1+ω+ω^2)\begin{vmatrix}1&ω&ω^2\\1&x+ω^2&1\\1&1&x+ω\end{vmatrix}=0$
$⇒x\begin{vmatrix}1&ω&ω^2\\0&x+ω^2-ω&1-ω^2\\0&1-ω&x+ω-ω^2\end{vmatrix}=0$ [Using $R_2 → R_2-R_1, R_3→R_3-R_1$]
$⇒x\{(x+ω-ω^2) (x + ω^2 - ω) − (1 − ω) (1 - ω^2)\}=0$
$⇒x=0$ is a root of the given equation.