# Determinant

• Jul 11th 2009, 06:28 AM
m3th0dman
Determinant
$x^{3}+mx+n=0$ is given with m and n real numbers and $x_1, x_2, x_3$ it's solutions.
It's asked for the value of the determinant:
$det = \begin{vmatrix}
1 & 1 & 1\\
x_1 & x_2 & x_3 \\
x_1^{2} & x_2^{2} & x_3^{2}
\end{vmatrix}$

I have tried to solve it in the Vandermonde way
$det=(x_1-x_2)(x_1-x_3)(x_2-x_3)$

Or to sum up columns 2 and 3 to the first (to obtain the sum, and the sum of squares of the solutions - which are found easily with Viete).
$det=(x_3-x_2)(3x_3x_2+2m)$

I have managed to find out that:
$(x_1^{2}+x_1x_2+x_2^{2})(x_1^{2}+x_1x_3+x_3^{2})(x _2^{2}+x_2x_3+x_3^{2})=m$
the value of the determinant beeing $\pm \sqrt{-4m^{3}-27n^{2}}$ thinking that I should sqare the first identety and make some conection with this one.

I'm sorry for posting this problem to pre-algebra too. This is a linear algebra problem, but it's for highschool.
• Jul 11th 2009, 08:59 AM
halbard
Let the roots of $x^3+mx+n=0$ be $a$, $b$ and $c$. Then $a+b+c=0$, $bc+ca+ab=m$ and $abc=-n$.

Here's one way to find $(a-b)(b-c)(c-a)$, probably not the best.

Start with $a^2+b^2+c^2=(a+b+c)^2-2(bc+ca+ab)=-2m$, and so $a^2+b^2=-2m-c^2$.

Since $c$ is a root, $c^3=-mc-n=-mc+abc$, so $c^2=ab-m$. So now $a^2+b^2=-m-ab$.

Therefore $(a-b)^2=a^2+b^2-2ab=-m-3ab$. Similarly $(b-c)^2=-m-3bc$ and $(c-a)^2=-m-3ca$.

Thus $(a-b)^2(b-c)^2(c-a)^2=-(m+3ab)(m+3bc)(m+3ca)$ $=-(27a^2b^2c^2+9m(a^2bc+b^2ca+c^2ab)+3m^2(ab+bc+ca)+ m^3)$.

Using the root relations, this last quantity is $-(27n^2+4m^3)$. Hope this helps.
• Jul 11th 2009, 09:23 AM
m3th0dman
It was very helpful. I have found out, in other place, antoher method.

Thank you very much, this one I have understood it.
• Jul 11th 2009, 09:51 AM
m3th0dman
I will post the other solution, it's not so complicated.

Let it be
$A= \begin{pmatrix}
1 & 1 & 1\\
x_1 & x_2 & x_3\\
x_1^{2} & x_2^{2} & x_3^{2}
\end{pmatrix}, A^{t}=\begin{pmatrix}
1 & x_1 & x_1^{2}\\
1 & x_2 & x_2^{2}\\
1 & x_3 & x_3^{2}
\end{pmatrix}, A^{tr}-transposed \, matrix$

hope transposed is the name in English.

$det(AA^{tr})=det(A)det(A^{tr})=det(A)^{2}=\begin{v matrix}
1 & x_1+x_2+x_3 & x_1^{2}+x_2^{2}+x_3^{2}\\
x_1+x_2+x_3 &x_1^{2}+x_2^{2}+x_3^{2} &x_1^{3}+x_2^{3}+x_3^{3} \\
x_1^{2}+x_2^{2}+x_3^{2} & x_1^{3}+x_2^{3}+x_3^{3} & x_1^{4}+x_2^{4}+x_3^{4}
\end{vmatrix}$

From Viete's relations:
$x_1+x_2+x_3=0$

$x_1^{2}+x_2^{2}+x_3^{2}=(x_1+x_2+x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=0-2m$

$x^{3}=-mx-n$
$x^{4}=-mx^2-nx$

$x_1^{3}+x_2^{3}+x_3^{3}=-m(x_1+x_2+x_3)-3n$

$x_1^{4}+x_2^{4}+x_3^{4}=-m(x_1^{2}+x_2^{2}+x_3^{2})-n(x_1+x_2+x_3)$

I didn't solve it. Someone from another forum (from my country) did. Hope it helps others too.