Determinant

Likon · New Member Offline Joined: 12 Aug 2009 Posts: 13

determinant $\left[ \begin{array}{ccc} x+y+z & y & z \\ x+y+z & z & x \\ x+y+z & x & y \end{array} \right]$

How can i get in the value $-x^{3} - y^{3} - z^{3} + 3xyz$ ?

i wait
thank you.

**Posted:** Sat Nov 07, 2009 2:16 pm

isabella2296

Can you please clarify your question? I may be missing something but I don't understand what you want to find.

**Posted:** Sat Nov 07, 2009 3:05 pm

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Likon · New Member Offline Joined: 12 Aug 2009 Posts: 13

I want the value of this determinant , the answer is $-x^{3}-y^{3}-z^{3}+3xyz$ , but do not know how to get in it

**Posted:** Sat Nov 07, 2009 3:11 pm

isabella2296

Oh, okay.

Do you know how to find the determinants of 3 x 3 matrices in general? If so, just apply the same method here, evaluating in terms of the variables.

**Posted:** Sat Nov 07, 2009 3:16 pm

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Likon · New Member Offline Joined: 12 Aug 2009 Posts: 13

Yes , i know.
Acctualy , i want some easy way to solv it.

using $(x+y+z).z.y + y.z(x+y+z....$ is so Large and confusing.
Seeing the way the determinant, I think there is another way to solve

**Posted:** Sat Nov 07, 2009 3:42 pm

azjps

This determinant is a quick way to verify the well-known identity involving the elementary symmetric polynomial decomposition of $x^3 + y^3 + z^3$ . Hint

**Posted:** Sat Nov 07, 2009 3:53 pm

Tomekk

Notice that

$-x^{3}-y^{3}-z^{3}+3xyz=(x+y+z)(zy+xy+zx-z^2-x^2-y^2)$

And using what azjps said, subtracting second and third columns from the first you easily get the equality I wrote.

**Posted:** Sun Nov 08, 2009 1:20 pm