Solve: $\begin{vmatrix} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$.

Question

Accepted Answer

>

> [!TLDR]
> Adding all three rows collapses the determinant to a common factor of $x$, and simplifying the remaining $3	imes3$ determinant (using $1+\omega+\omega^2=0$) shows the equation reduces to $x^3=0$.
>
> $\begin{vmatrix} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$, where $\omega$ is a complex cube root of unity, so $\omega^3=1$ and $1+\omega+\omega^2=0$.
>
> **Step 1: $R_1 	o R_1+R_2+R_3$.**
>
> Each entry of the new $R_1$:
> - Col 1: $(x+1)+\omega+\omega^2 = x+(1+\omega+\omega^2) = x+0 = x$
> - Col 2: $\omega+(x+\omega^2)+1 = x+(1+\omega+\omega^2) = x$
> - Col 3: $\omega^2+1+(x+\omega) = x+(1+\omega+\omega^2) = x$
>
> So $R_1$ becomes $(x,\ x,\ x)$, and the determinant is
>
> $\begin{vmatrix} x & x & x \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix} = x\begin{vmatrix} 1 & 1 & 1 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{vmatrix}$ (factoring $x$ from $R_1$).
>
> **Step 2:** $C_2	o C_2-C_1$, $C_3	o C_3-C_1$:
>
> $\begin{vmatrix} 1 & 0 & 0 \ \omega & x+\omega^2-\omega & 1-\omega \ \omega^2 & 1-\omega^2 & x+\omega-\omega^2 \end{vmatrix}$
>
> Expanding along $R_1$:
>
> $= (x+\omega^2-\omega)(x+\omega-\omega^2) - (1-\omega)(1-\omega^2)$
> …

Q.Solve: $\begin{vmatrix} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$ .

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.Solve: ∣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.

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Q.Solve: $\begin{vmatrix} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{vmatrix} = 0$ .

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