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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 $-4\beta\gamma$.

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  • $\begingroup$ I got -4βγ before considering x^3-1=0 $\endgroup$
    – stedmoaoa
    May 5, 2017 at 18:05

1 Answer 1

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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.

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  • $\begingroup$ do you mean that the determinant is 0 $\endgroup$
    – stedmoaoa
    May 5, 2017 at 18:02
  • $\begingroup$ with the α+β+γ satisfies x^3-1 but what if the general condition? $\endgroup$
    – stedmoaoa
    May 5, 2017 at 18:03
  • $\begingroup$ @stedmoaoa Yes only in the condition that $\alpha, \beta, \gamma$ are the cube roots of unity, the determinant is zero. $\endgroup$ May 5, 2017 at 19:01

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