TITLE: determinant diagonal zero symmetric matrix
QUESTION [5 upvotes]: Let
$$M=\begin{pmatrix}      
    0 & 1    &  1   &     1\\
    1 & 0    &  \alpha + \beta &  \alpha + \gamma\\
    1 & \beta + \alpha & 0    &  \beta + \gamma\\
    1& \gamma + \alpha  & \gamma + \beta  &   0
\end{pmatrix},$$
then it holds $$\det(M) = −4(\alpha\beta + \beta\gamma + \gamma\alpha).$$

What is the value of this when $\alpha,\beta,\gamma$ are the three roots of the equation $x^3 − 1 = 0$?

Can anyone help me to do it by elementary row operation?
My idea is just solving $\alpha,\beta,\gamma$ and then plug in $−4(\alpha\beta + \beta\gamma + \gamma\alpha)$
and when I try to find the determinant of the LHS I got $-4\beta\gamma$.

REPLY [8 votes]: By the fundamental theorem of algebra / factor theorem we have
\begin{align}
x^3 - 1 
& =
(x - \alpha) (x - \beta) (x - \gamma) 
\\
& = x^3 - x^2 (\alpha + \beta + \gamma) + x (\alpha \beta + \beta \gamma + \alpha \gamma) - \alpha \beta \gamma
.
\end{align}
From this, you can read off, by comparing the coefficients, that $\alpha \beta + \beta \gamma + \alpha \gamma = 0$.
Of course, you can also compute the roots explicitly and do the algebra. The roots of $x^3 - 1$ are $\{ e^{i 2 \pi k/3} : k = 0, 1, 2 \}$. Here is a funny observation: the products $\{ e^{i 2 \pi (k + \ell) / 3} : k \neq \ell \}$ are still the roots of the same polynomial, so their sum vanishes by the above argument.