If ω is an imaginary cube root of unity, find the value of det pmatrix 1 & ω & ω² ω & ω² & 1 ω² & 1 & ω pmatrix — JEE Mathematics
If $\omega$ is an imaginary cube root of unity, find the value of $\det \begin{pmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{pmatrix}$.
1 Answer
Let $D = \det \begin{pmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{pmatrix}$.
Apply the column operation $C_1 \to C_1 + C_2 + C_3$:
$$D = \det \begin{pmatrix} 1 + \omega + \omega^2 & \omega & \omega^2 \\ 1 + \omega + \omega^2 & \omega^2 & 1 \\ 1 + \omega + \omega^2 & 1 & \omega \end{pmatrix}$$
We know that $1 + \omega + \omega^2 = 0$. Substituting this into the first column:
$$D = \det \begin{pmatrix} 0 & \omega & \omega^2 \\ 0 & \omega^2 & 1 \\ 0 & 1 & \omega \end{pmatrix}$$
Since an entire column of the matrix consists of zeros, the value of the determinant is $0$.
Answer: 0