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The identity comes from expanding the determinant $$\begin{vmatrix} a & b & c \\ c & a & b \\ b & c & a \\ \end{vmatrix}$$ in two ways.

The LHS comes from expanding the determinant by Sarrus' rule. The RHS comes from adding up all columns to the first, factoring (a+b+c), and then expanding the remaining determinant.

The derivation of the RHS is what I don't understand. I'm guessing that there is some elementary determinant operation at work that I'm unfamiliar with.

Anonymous196
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Sid
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  • Well, once you have factored out the $a+b+c$ you can apply Sarrus' rule again. I don't think there's more to it than that ... – Ewan Delanoy Oct 17 '17 at 05:05

3 Answers3

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In more detail, the thing that's being suggested to get the RHS is the following reasoning: \begin{align} \begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} &= \begin{vmatrix}a+c & a+b & b+c \\ c & a & b \\ b & c & a\end{vmatrix} & \text{(Add row 2 to row 1)} \\ &= \begin{vmatrix}a+b+c & a+b+c & a+b+c \\ c& a & b \\ b & c & a\end{vmatrix} & \text{(Add row 3 to row 1)} \\ &= (a+b+c) \begin{vmatrix}1 & 1 & 1\\c & a & b\\ b & c & a\end{vmatrix} & \text{(Factor out $a+b+c$)} \\ &= (a+b+c) \left(\begin{vmatrix}a & b \\ c & a\end{vmatrix} - \begin{vmatrix}c & b \\ b & a\end{vmatrix} + \begin{vmatrix}c & a \\ b & c\end{vmatrix}\right) & \text{(Expand on $1^{\text{st}}$ row)} \\ &= (a+b+c)(a^2-bc - (ca-b^2) + c^2-ab) \\ &= (a+b+c)(a^2+b^2+c^2 - ab - bc - ca). \end{align}

Misha Lavrov
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  • ok thanks for making the reasoning explicit, i forgot about the column operation. – Sid Oct 17 '17 at 05:17
  • I guess I did row operations instead of the column operations that "adding up all columns to the first" suggests. (Unintentional; I'm just more used to row operations.) But those are equivalent and if you take the transpose of my answer, you get an answer with column operations. – Misha Lavrov Oct 17 '17 at 05:20
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The matrix is a circulant matrix. Its eigenvectors are $(1,\zeta,\zeta^2)^t$ where $\zeta$ runs through the cube roots of unity. The corresponding eigenvalues are $a+b\zeta+c\zeta^2$. The determinant is thus $$(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$ where $\omega=\exp(2\pi i/3)$. Pair off the last two factors to get your form.

As a follow-up problem, consider the determinant of $$\pmatrix{a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a}.$$

Angina Seng
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  • I'm familiar with roots of unity but not eigenvectors. Thanks for the food for thought though. – Sid Oct 17 '17 at 05:18
  • I've learned about eigenvectors and that the determinant of an n x n circulant matrix is given by $(-1)^{n-1}\prod_{j=0}^{n-1}\left(\sum_{k=1}^{n}\zeta^{jk}x_{k}\right)$, although it isn't clear to me why the determinant should be written as a product of eigenvalues. I guess I need to learn more linear algebra. – Sid Oct 17 '17 at 22:09
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Add $\omega$ times the second row and $\omega^2$ times the third row to the first row to get $\begin{align} \begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} &= \begin{vmatrix}a+c\omega+b\omega^2 & b+a\omega+c\omega^2 & c+b\omega + a\omega^2 \\ c & a & b \\ b & c & a\end{vmatrix} \\ &=\begin{vmatrix}a+c\omega+b\omega^2 & \omega(a+b\omega^2+c\omega) & \omega^2(c\omega+b\omega^2 + a) \\ c & a & b \\ b & c & a\end{vmatrix} \end{align}$

and hence $a+c\omega+b\omega^2$ is a factor. Similar arguments can be used to show the other two factors in the proof given by Lord Shark the Unkniwn.