Determine the determinant

Question

$$ \begin{vmatrix} 1 & -\omega & \omega^2 \\ -\omega & \omega^2 & 1 \\ \omega^2 & 1 & -\omega \\ \end{vmatrix} $$ I can solve the question by doing $c_1'=c_1+c_3\omega~~$.this simply gives the answer $-4$.but when I am trying to go on another procedure like ,if I perform $r_1'=r_1\omega+r_2~~$ then I am getting the answer $-4\omega$ which was not same as the previous procedure.please let me know my mistake. Here,\omega is a root of $\sqrt[3]{1}\text{ and is} =\dfrac{1}{2}(-1+i \sqrt{3})$ .And $c_n,r_n$ represent $n_{th}$ row or column. Thank you.

Answer 1

$$ \begin{vmatrix} 1 & -\omega & \omega^2 \\ -\omega & \omega^2 & 1 \\ \omega^2 & 1 & -\omega \\ \end{vmatrix} $$ $$ =\dfrac{1}{\omega}\begin{vmatrix} 0 & 0 & \omega^3+1 \\ -\omega & \omega^2 & 1 \\ \omega^2 & 1 & -\omega \\ \end{vmatrix}[r_1'=\omega r_{1}+r_2] $$ $$=\dfrac{(\omega^3+1)}{\omega} \begin{vmatrix} \ -\omega & \omega^2 \\ \omega^2 & 1 \\ \end{vmatrix} $$ $$=\dfrac{(1+1)}{\omega}(-\omega (1+\omega^3))$$ $$=\dfrac{-4\omega}{\omega}=-4$$

Answer 2

$r_1' = r_1 \omega + r_2$ does not preserve the determinant. What you're actually doing are two elementary operations : $r_1'' = \omega r_1$ which multiplies the determinant by $\omega$ and then $r_1' = r_1'' + r_2$ which preserve the determinant.

So you when you do $r_1' = r_1 \omega + r_2$, you must divide by $\omega$ to counterbalance the fact that this operation scaled the determinant by $\omega$.

Answer 3

I'd use a more systematic way: \begin{align} \begin{bmatrix} 1 & -\omega & \omega^2 \\ -\omega & \omega^2 & 1 \\ \omega^2 & 1 & -\omega \end{bmatrix} &\to \begin{bmatrix} 1 & -\omega & \omega^2 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{bmatrix} &&\begin{aligned}R_2&\gets R_2+\omega R_1\\ R_3 &\gets R_3-\omega^2R_1\end{aligned} \\[6px] \color{red}{(-1)}\qquad&\to \begin{bmatrix} 1 & -\omega & \omega^2 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} &&R_2\leftrightarrow R_3 \end{align} (assuming $\omega^3=1$). Then the determinant is $(-1)\cdot 1\cdot 2\cdot 2$. The factor $-1$ is due to the row swapping operation.

Note that an operation such as $$ R_1\gets \omega R_1+R_2 $$ (in your notation $r_1'=\omega r_1+r_2$) does change the determinant, multiplying it by $\omega$. And indeed, you find $-4\omega$, which should be divided by $\omega$.

An operation such as $$ R_i\gets aR_i+bR_j $$ (with $a\ne0$ and $i\ne j$) is admissible in this context, but remember it multiplies the determinant by $a$. Similarly for columns.

In my elimination I always used $a=1$.