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Using determinant properties (without expanding), prove that

$$ \begin{vmatrix}yz & z^2 & y^2 \\ z^2 & xz & x^2 \\ y^2 & x^2 & xy \end{vmatrix} = xyz\begin{vmatrix}x & z & y \\ z & y & x \\ y & x &z\end{vmatrix}$$

I'm completely lost and I don't know what to try. Any hints?

Thanks in advance.

Mahmoud Ahmed
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2 Answers2

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Start from

$$\begin{vmatrix}x & z & y \\ z & y & x \\ y & x &z\end{vmatrix}.$$

Multiply the colums by $y, x, \frac{xy}z$; multiply the rows by $\frac{z}{x},\frac z y, 1$. You have now multiplied the determinant with $xyz$.

Edit: More symmetric: multiply the colums $yz, zx, xy$; multiply the rows by $\frac1x,\frac1y, \frac1z$.

Myself
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    Ingenious. Any tip on how you discovered that this sequence of operations would work? – hmakholm left over Monica Jan 10 '15 at 23:22
  • This uses that $x,y,z$ must each be invertible, which is against the spirit of determinants. More precisely makes the proof invalid over general commutative rings. And in any case it obliges you to do some special cases. – Marc van Leeuwen Jan 11 '15 at 06:02
  • @Henning If you guess $y$ and $z/x$ then there is at most one possibility for the other options. There is further freedom because you can multiply all rows by the same number and divide all colums by the same number so there is really only one degree of freedom. – Myself Jan 11 '15 at 10:17
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    @MarcvanLeeuwen Actually the above argument proves it for the 'universal' ring $\Bbb Z[x,y,z]$ by considering its fraction field where $x,y,z$ are invertible and then, because the it is just a polynomial, it holds in any ring $R$ for any variables $a,b,c$ by using the evaluation morphism $\Bbb Z[x,y,z] \mapsto R: f(x,y,z)\mapsto \bar f(a,b,c)$. – Myself Jan 11 '15 at 10:19
  • @Myself: Yes I know. But (just to be obnoxious) how do you prove the determinant is a polynomial in the matrix entries without using expansion? – Marc van Leeuwen Jan 11 '15 at 11:30
  • @MarcvanLeeuwen I guess that depends on how you wish to define the determinant... Some people define it as a polynomial (and then you must prove the multilinear properties). Others as the unique multilinear map with some properties, or the induced map ${\wedge^n} V\to {\wedge^n} V : x\mapsto \mathrm{det}(\alpha)x$ etc. But inevitably at some point someone, like an author of a text on linear algebra, has to do a bit of work to tie everything together :-) – Myself Jan 11 '15 at 17:53
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HINT: Sarrus' scheme gives it immediately.

Kola B.
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