# determinant diagonal zero symmetric matrix

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 $-4\beta\gamma$.

• I got -4βγ before considering x^3-1=0 May 5, 2017 at 18:05

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.
• @stedmoaoa Yes only in the condition that $\alpha, \beta, \gamma$ are the cube roots of unity, the determinant is zero. May 5, 2017 at 19:01