If $\omega
e 1, \omega^3 = 1$ and $\begin{vmatrix} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$ then $x =$

(a) $0$
(b) $\omega$
(c) $\omega^2$
(d) none of these

Question

If $\omega 
e 1, \omega^3 = 1$ and $\begin{vmatrix} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$ then $x =$

(a) $0$
(b) $\omega$
(c) $\omega^2$
(d) none of these

Accepted Answer

>

> [!TLDR]
> Adding all three columns makes every row sum to $x$, so the determinant is proportional to $x$; setting it to $0$ gives $x=0$.
>
> We use the cube-root-of-unity identities: $\omega^3=1$ and $1+\omega+\omega^2=0$.
>
> Apply the column operation $C_1 	o C_1+C_2+C_3$. The entries of the new first column become:
>
> Row 1: $(x+1)+\omega+\omega^2 = x+(1+\omega+\omega^2) = x+0 = x$.
>
> Row 2: $\omega+(x+\omega^2)+1 = x+(1+\omega+\omega^2) = x$.
>
> Row 3: $\omega^2+1+(x+\omega) = x+(1+\omega+\omega^2) = x$.
>
> So the first column is $\begin{bmatrix} x \ x \ x \end{bmatrix} = x\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$, and $x$ can be pulled out:
> …

Q.If $\omega \ne 1, \omega^3 = 1$ and $\begin{vmatrix} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$ then $x =$

Concept understanding — Determinant Equality Equation

Determinant Equality Equation

The basic move

The important special case: equals zero

A classic application: three points on one line

Solution

Q.If ω≠1,ω3=1\omega \ne 1, \omega^3 = 1ω=1,ω3=1 and ∣x+1ωω2ωx+ω21ω21x+ω∣=0\begin{vmatrix} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{vmatrix} = 0​x+1ωω2​ωx+ω21​ω21x+ω​​=0 then x=x =x=

🎓Unlock everything free for 14 days

Q.If $\omega \ne 1, \omega^3 = 1$ and $\begin{vmatrix} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$ then $x =$

Unlock everything free for 14 days