1
$\begingroup$

I have to prove that by rotation of axes the expression $F(x,y,z) = ax^2 + by^2 + cz^2 +2fyz + 2gzx + 2hxy $ transforms to $ λ_1x^2 + λ_2y^2 + λ_3z^2 =0 $ where $λ_1, λ_2,λ_3 $ are roots of cubic

$ \begin{vmatrix} a-λ&h&g\\ h&b-λ&f\\ g&f&c-λ\\ \end{vmatrix}=0 $

Solution in book says that $ax^2 + by^2 + cz^2 +2fyz + 2gzx + 2hxy - λ(x^2 +y^2+z^2)$ should reduce to $ λ_1x^2 + λ_2y^2 + λ_3z^2-λ(x^2 +y^2+z^2)$ . Hence, book says, both these expressions will be product of linear factors for same value of λ. I don't understand why this is necessary.

Further, it is said that if $(a-λ)x^2 + (b-λ)y^2 + (c-λ)z^2 +2fyz + 2gzx + 2hxy $ is a product of two linear factors then $ \begin{vmatrix} a-λ&h&g\\ h&b-λ&f\\ g&f&c-λ\\ \end{vmatrix}=0 $

I also don't get how this condition is arrived at. Any help in clearing these two doubts will be helpful.

$\endgroup$
  • $\begingroup$ Do you mean maybe three degree equation? $\endgroup$ – Michael Rozenberg Sep 23 '17 at 8:19
  • $\begingroup$ No, condition is derived for general equation of second degree in x,y, and z $\endgroup$ – LoneCuriousWolf Sep 23 '17 at 8:21
  • $\begingroup$ @MichaelRozenberg are you able to understand the question? $\endgroup$ – LoneCuriousWolf Sep 23 '17 at 10:29
  • $\begingroup$ Yes of course! It should be $(a_1x+b_1y+c_1z+d_1)(a_2x+b_2y+c_2z+d_2)=0$, but I don't see something interesting for this, otherwise, just to write $f(x,y,z)=(a_1x+b_1y+c_1z+d_1)(a_2x+b_2y+c_2z+d_2)$ and to solve the system. $\endgroup$ – Michael Rozenberg Sep 23 '17 at 10:33
  • $\begingroup$ @MichaelRozenberg I have edited and added some background to the question. Let me know if you could find anything interesting now. $\endgroup$ – LoneCuriousWolf Sep 23 '17 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.