Solving $D=\begin{vmatrix}a&\omega b&\omega^2c\\\omega^2b&c&\omega a\\\omega c&\omega^2a&b\end{vmatrix}$

complex numbersdeterminantmatrices

If $D=\begin{vmatrix}a&\omega b&\omega^2c\\\omega^2b&c&\omega a\\\omega c&\omega^2a&b\end{vmatrix}$ and $D'=\begin{vmatrix}a& b&c\\b&c& a\\\ c&a&b\end{vmatrix}$, where $\omega$ is the non real cube root of unity, then which of the following does not hold:

  • $1)\ D=0$ if $a+b+c=0$ and $a,b,c$ all distinct.
  • $2)\ D'=0$ if $a=b=c$ and $a+b+c\ne0$
  • $3)\ D=-D'$
  • $4)\ D=D'$

My Attempt:

I am getting $D'=-\frac12(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$. So, $2)$ holds.

For $D$, I tried the row operation $R_1\to R_1+R_2+R_3$ but couldn't conclude.

I also tried multiplying each row with $\omega$ twice but in vain.

I tried writing the value of $\omega$ in $i$ but couldn't finish again.

Any help?

Best Answer

We know that multiplying a row of a determinant by a constant changes the determinant by that constant. Therefore, $$ \begin{vmatrix} a&\omega b&\omega^2c\\\omega^2b & c & \omega a\\\omega c & \omega^2 a & b \end{vmatrix} = \frac{1}{\omega^3}\begin{vmatrix} a&\omega b&\omega^2c\\b & \omega c & \omega^2 a\\ c & \omega a & \omega^2b \end{vmatrix} $$ where we multiplied the second row by $\omega$ and third row by $\omega^2$. But actually $\omega^3=1$ so the constant factor in front is just $1$. Now, multiplying a column by a constant also has the same effect, so we multiply the second column by $\omega^2$ and third column by $\omega$ to conclude that, in fact,

$$D = \begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix}.$$

So really $D=D'$. It is also easier to now compute the determinant by directly expanding it to get $D = 3abc-a^3-b^3-c^3 = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Note that the final form of the determinant is a circulant matrix, for which there is a standard formula to compute the determinant. So, this problem can be generalised to $n\times n$ matrices and you can still write out an algebraic expression for the determinant explicitly.

Also, it is generally a horrible idea to write $\omega$ out in its real and imaginary part (i.e. write $\omega = (-1+i\sqrt{3})/2$) and try to simplify from there. What you need for things to simplify is just the property that $\omega^3=1$, and you are just making your life harder by writing out the real and imaginary parts. It really won't help at all.

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