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My assignment requires me to show for the following Vandermonde matrix:

\begin{bmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{bmatrix}

the determinant is $(x-y)(y-z)(z-x)$.

But when I calculate, I got $(y-x)(z-x)(z-y)$. How do I change the sign so that I get $(x-y)(y-z)(z-x)$ ?

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$$(y-x)(z-x)(z-y)=[-(x-y)](z-x)[-(y-z)]=(-1)(-1)(x-y)(z-x)(y-z)=(x-y)(z-x)(y-z)$$ Because: $-(a-b)=-a-(-b)=-a+b=(b-a)$

Note: The multiplication of the (real/complex) numbers is commutative, unlike the multiplication of matrices.

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  • $\begingroup$ I can just simply put a negative sign for this? $\endgroup$ – dembrownies Dec 4 '17 at 18:09
  • $\begingroup$ @dembrownies what do you mean? Your answer is the same as the given answer. $\endgroup$ – Botond Dec 4 '17 at 18:10
  • $\begingroup$ I thought I can only have the negative sign if I interchange the rows $\endgroup$ – dembrownies Dec 4 '17 at 18:11
  • $\begingroup$ There is no interchange of rows. There are no negative signs to account for. Your answer and the book answer are the same $(y-x)(z-x)(z-y) = (x-y)(y-z)(z-x)$ $\endgroup$ – Doug M Dec 4 '17 at 18:12
  • $\begingroup$ I guess I got confused with some other stuff. Thank you! $\endgroup$ – dembrownies Dec 4 '17 at 18:12

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