Show that the following matrix is equal to 0, where $\omega$ is a complex cube root of unity.

Question

Show that $$ \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & 0 \\ \end{vmatrix} =0 $$ Where $\omega$ is a complex cube root of unity.

I am not sure if I could get the proof right. I tried to refer some online, but so far I couldn't. Really appreciate any help by you.

Thanks

Edit: This was my attempt

$1(0-1)-\omega(0-\omega^2)+\omega^2(\omega-\omega^4)=0$

$0-1-0+\omega^3+\omega^3-\omega^2=0$

$-1+2\omega^3-\omega^2=0$

Since $\omega^3=1$

$-1+2-\omega^2=0$

$1-\omega^2=0$

$1=\omega^2$ >>> Obviously this is not true.

Answer 1

Assuming you want to show that the determinant of the matrix is $0$, we can just apply the Laplace expansion respect to the first row to get :

$$1 \cdot det \begin{pmatrix}\omega^{2} & 1 \\ 1 & 0\end{pmatrix} - \omega \cdot det\begin{pmatrix}\omega & 1 \\ \omega^{2} & 0\end{pmatrix} + \omega^{2}det \begin{pmatrix}\omega & \omega^{2} \\ \omega^{2} & 1\end{pmatrix}$$

Expanding the $2x2$ determinants we get

$$-1 + \omega^{3} + \omega^{2}(\omega - \omega^{4})$$

Since $\omega$ is a third root of unity has (as suggested in the comments) the following property $\omega^{3} = 1$

We do get further semplification noticing that if $\omega^{3}=1$, then $\omega^{4}=\omega$

In the end we just have $-1 +\omega^{3} + \omega^{2}(\omega-\omega) = -1 + 1 + 0 = 0 $

Answer 2

Use the property: $1+ \omega + \omega^2 = 0$

Answer 3

Having made the connection that $\omega$ is a cube root of unity is equivalent to the fact that $\omega^3=1$ or $\omega^3-1$, all we need to do is pick our favourite method to expand the determinant.

For example, using the Rule of Sarrus, we get $$\begin{vmatrix}1&\omega&\omega^2\\\omega&\omega^2&1\\\omega^2&1&0\end{vmatrix}=\omega^3+\omega^3-\omega^6-1=-\omega^6+2\omega^3-1=-(\omega^6-2\omega^3+1)=-(\omega^3-1)^2,$$ which is obviously zero if $\omega$ is a cube root of unity. One can also simply replace $\omega^3$ with $1$ in the expression to see that it vanishes: $$2\omega^3-\omega^6-1=2\omega^3-(\omega^3)^2-1=2\cdot 1-1-1=0.$$

We can also expand the determinant using Laplace expansion. This is perhaps easiest on the rightmost column (or the bottom row) since it contains a $0$. This gives: \begin{align*} \begin{vmatrix}1&\omega&\omega^2\\\omega&\omega^2&1\\\omega^2&1&0\end{vmatrix}&=\omega^2\begin{vmatrix}\omega&\omega^2\\\omega^2&1\end{vmatrix}-1\cdot\begin{vmatrix}1&\omega\\\omega^2&1\end{vmatrix}=\omega^2(\omega-\omega^4)-(1-\omega^3) \\ &=\omega^3(1-\omega^3)-(1-\omega^3)=(\omega^3-1)(1-\omega^3), \end{align*} which is once again obviously zero if $\omega$ is a cube root of unity.

Answer 4

There’s no compelling reason to expand this determinant in order to evaluate it. Comparing the first and second rows, we can see that $(\omega,\omega^2,1)=\omega(1,\omega,\omega^2)$. This means that the two rows are linearly dependent, which in turn means that the determinant vanishes.

One can grind this out by subtracting $\omega$ times the first row from the second, which doesn’t change the value of the determinant. You now have a row of zeros, and expanding the determinant along this row obviously produces a zero value for the determinant.