If ω is a non-real cube root of unity, then what is a root of the following equation?\( \begin{vmatrix} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{vmatrix} = 0 \)

 

Calculation:

Given,

Let ω be a non-real cube root of unity, so \( \omega^{3}=1 \) and \( 1+\omega+\omega^{2}=0 \).

Consider the determinant

\( \Delta(x)= \begin{vmatrix} x+1 & \omega & \omega^{2}\\ \omega & x+\omega^{2} & 1\\ \omega^{2} & 1 & x+\omega \end{vmatrix}=0. \)

Step 1 — Column operation:  Replace the first column by \(C_{1}-C_{2}\):

\( \Delta(x)= \begin{vmatrix} x^{2}-1 & \,2\!k\!+\!1\, & 1\\ k-1 & k+2 & 1\\ 0 & 3 & 1 \end{vmatrix}. \)

Step 2 — Expansion along the third row:

\( \Delta(x)= -3\! \begin{vmatrix} k^{2}-1 & 1\\ k-1 & 1 \end{vmatrix} + \begin{vmatrix} k^{2}-1 & 2k+1\\ k-1 & k+2 \end{vmatrix}, \)

which simplifies to

\( \Delta(x)=x(x^{2}-1)-x\bigl(\omega+\omega^{2}\bigr) =x(x^{2}-1)+x =x^{3}. \)

Step 3 — Equate to zero:

\( \Delta(x)=0 \;\Longrightarrow\; x^{3}=0 \;\Longrightarrow\; x=0. \)

∴ The root of the equation is  \( x = 0 \).

Hence, the correct answer is Option 1.