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Let $a,b,c,d,e,f \in \mathbb{R}$. Prove that $$ \begin{vmatrix} (a+b)de-(d+e)ab & de-ab & a+b-d-e \\ (b+c)ef-(e+f)bc & ef-bc & b+c-e-f \\ (c+d)fa-(f+a)cd & fa-cd & c+d-f-a \end{vmatrix}=0 $$

I tried to write the determinant as $$ \begin{vmatrix} x_1x_3-x_2x_4 & x_3-x_4 & x_1-x_2 \\ y_1y_3-y_2y_4 & y_3-y_4 & y_1-y_2 \\ z_1z_3-z_2z_4 & z_3-z_4 & z_1-z_2 \end{vmatrix} $$ where $x_1=a+b, x_2=d+e,x_3=de, x_4=ab$ and the other cyclical ones are defined similarly. But now I have $12$ variables instead of $6$, so they depend on one another somehow. I tried to expand it, but it doesn't look too good.

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  • $\begingroup$ Equivalently, you can show that the rows or columns are linearly dependent. Or show that there is at least one zero eigenvalue. $\endgroup$ – AlexanderJ93 Feb 13 '18 at 21:03
  • $\begingroup$ You could make use of Sarrus' Rule and actually calculate the six summands which yield the determinant. $\endgroup$ – Axel Kemper Feb 13 '18 at 23:05
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The combinations $a+b$ and $ab$ make one think of quadratic polynomials with root $a,b$. The way I shall use it is by saying that the points $(a,a^2)$ and $(b,b^2)$ belong to the straight line given by the equation $y=(a+b)x-ab$.

Consider the line passing through $(a,a^2)$ and $(b,b^2)$. Let us determine where it meets the line passing through $(d,d^2)$ and $(e,e^2)$. For that, we need to solve the system of linear equations $$ \left\{ \begin{aligned} -(a+b)x+y&=-ab,\\ -(d+e)x+y&=-de. \end{aligned} \right. $$ By Cramer's rule, we get the coordinates of the intersection point $$(x_{ab,de},y_{ab,de})=\left(\frac{-ab+de}{-a-b+d+e},\frac{(a+b)de-(d+e)ab}{-a-b+d+e}\right).$$

Similar formulas hold for $(x_{bc,ef},y_{bc,ef})$ and $(x_{cd,af},y_{cd,af})$. Your determinant identity, if you multiply the last column by $(-1)$ and then divide every row by its last element, can be written as $$ \det\begin{pmatrix}x_{ab,de}&y_{ab,de}&1\\ x_{bc,ef}&y_{bc,ef}&1\\ x_{cd,af}&y_{cd,af}&1\end{pmatrix}=0, $$ which holds iff the points $(x_{ab,de},y_{ab,de})$, $(x_{bc,ef},y_{bc,ef})$, and $(x_{cd,af},y_{cd,af})$ are collinear. Since $(a,a^2)$,...,$(f,f^2)$ are one the same quadric (the parabola $y=x^2$), this follows from Pascal's theorem (https://en.wikipedia.org/wiki/Pascal%27s_theorem).

Remark: you may say that sometimes we cannot divide by the last element of the row, as it may vanish. Basically, the best way to think about it is to say that we just proved that the determinant vanishes for $a,\ldots,f$ generic enough (where those last row elements do not vanish), and for the rest it follows by continuity (the determinant is a polynomial in these variables, so a continuous function).

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