If $\omega$ is cube root of unity, then what is the value of the determinant $\begin{vmatrix}\omega^{27}&\omega^{105}\\omega^{426}&\omega^{21}\end{vmatrix}$?

(i) $-1$
(ii) $0$
(iii) $1$
(iv) $2$

Question

If $\omega$ is cube root of unity, then what is the value of the determinant $\begin{vmatrix}\omega^{27}&\omega^{105}\\omega^{426}&\omega^{21}\end{vmatrix}$?

(i) $-1$
(ii) $0$
(iii) $1$
(iv) $2$

Accepted Answer

>

> [!TLDR]
> Reduce each exponent modulo 3 using $\omega^3=1$; all entries become 1, making the determinant zero.
>
> Since $\omega$ is a cube root of unity, $\omega^3=1$, so $\omega^{n}=\omega^{n \bmod 3}$.
>
> - $27 = 3	imes9 \implies \omega^{27}=(\omega^3)^9=1$
> - $105 = 3	imes35 \implies \omega^{105}=1$
> - $426 = 3	imes142 \implies \omega^{426}=1$ …

Q.If $\omega$ is cube root of unity, then what is the value of the determinant $\begin{vmatrix}\omega^{27}&\omega^{105}\\\omega^{426}&\omega^{21}\end{vmatrix}$ ?

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

Why determinants change under row operations

Step-by-step evaluation

Q.If ω\omegaω is cube root of unity, then what is the value of the determinant ∣ω27ω105ω426ω21∣\begin{vmatrix}\omega^{27}&\omega^{105}\\\omega^{426}&\omega^{21}\end{vmatrix}​ω27ω426​ω105ω21​​?

🎓Unlock everything free for 14 days

Q.If $\omega$ is cube root of unity, then what is the value of the determinant $\begin{vmatrix}\omega^{27}&\omega^{105}\\\omega^{426}&\omega^{21}\end{vmatrix}$ ?

Unlock everything free for 14 days