Question

1. Reduce the following quadratic form into canonical form by an orthogonal transformation, 6x2 +3y2 + 3z2 – 4yz + 4zx – 2xy Ap CO1 15

Answer 1

Answer:

Diagonalization of matrices – Reduction of a quadratic form to canonical form by orthogonal transformation – Nature of quadratic forms. UNIT II ... AP, find the Matrix A. K where k is a.

Answer 2

Answer:

Quadratic form into canonical form by an orthogonal transformation is given by $8 \tilde{x}^{2}+2 \tilde{y}^{2}+2 \tilde{z}^{2}=0$

Step-by-step explanation:

Given equation of the surface of 2 -order:

$6 x^{2}-4 x y+4 x z+3 y^{2}-2 y z+3 z^{2}=0$

This equation looks like:

$a_{11} x^{2}+2 a_{12} x y+2 a_{13} x z+2 a_{14} x+a_{22} y^{2}+2 a_{23} y z+2 a_{24} y+a_{33} z^{2}+2 a_{34} z+a_{44}=0$

where

$a_{11}=6\\ a_{12}=-2\\ a_{13}=2\\ a_{14}=0\\ a_{22}=3$

$a_{23}=-1\\ a_{24}=0\\ a_{33}=3\\ a_{34}=0\\ a_{44}=0$

The invariants of the equation when converting coordinates are determinants:

$I_{1}=a_{11}+a_{22}+a_{33}$

$I_{3}=\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\a_{12} & a_{22} & a_{23} \\a_{13} & a_{23} & a_{33}\end{array}\right|$

$&I_{3}=\left|\begin{array}{ccc}6 & -2 & 2 \\-2 & 3 & -1 \\2 & -1 & 3\end{array}\right|$

$&I_{4}=\left|\begin{array}{cccc}6 & -2 & 2 & 0 \\-2 & 3 & -1 & 0 \\2 & -1 & 3 & 0 \\0 & 0 & 0 & 0\end{array}\right|$

$&I 2=\left|\begin{array}{ll}6 & -2\end{array}\right|+\left|\begin{array}{ll}3 & -1\end{array}\right|+\left|\begin{array}{rr}6 & 2\end{array}\right|$

$&I_{3}=\left|\begin{array}{ccc}6 & -2 & 2 \\-2 & 3 & -1 \\2 & -1 & 3\end{array}\right|$

$&I_{4}=\left|\begin{array}{cccc}6 & -2 & 2 & 0 \\-2 & 3 & -1 & 0 \\2 & -1 & 3 & 0 \\0 & 0 & 0 & 0\end{array}\right|$

$&I(\lambda)=\left|\begin{array}{ccc}6-\lambda & -2 & 2 \\-2 & 3-\lambda & -1 \\2 & -1 & 3-\lambda\end{array}\right|\\&K 2=\left|\begin{array}{ll}6 & 0 \\0 & 0\end{array}\right|+\left|\begin{array}{ll}3 & 0\end{array}\right|+\left|\begin{array}{ll}3 & 0 \\\mid 0 & 0\end{array}\right|+\left|\begin{array}{ll}\mid 0 & 0\end{array}\right|\end{aligned}$

$\begin{aligned}&K 2=\left|\begin{array}{ll}6 & 0\end{array}\right|+\left|\begin{array}{ll}3 & 0\end{array}\right|+\left|\begin{array}{ll}3 & 0\end{array}\right|\\&I_{1}=12\\&I_{2}=36\\&I_{3}=32\\&I_{4}=0\\&I(\lambda)=-\lambda^{3}+12 \lambda^{2}-36 \lambda+32\\&K_{2}=0\end{aligned}$

$K_{3}=0$

Because

$I_3 !=0$

then by type of surface:

you need to

Make the characteristic equation for the surface:

$-I_{1} \lambda^{2}+I_{2} \lambda-I_{3}+\lambda^{3}=0$

or

$\lambda^{3}-12 \lambda^{2}+36 \lambda-32=0\\ \lambda_{1}=8\\ \lambda_{2}=2\\ \lambda_{3}=2$

then the canonical form of the equation will be

$\left(\tilde{z}^{2} \lambda_{3}+\left(\tilde{x}^{2} \lambda_{1}+\tilde{y}^{2} \lambda_{2}\right)\right)+\frac{I_{4}}{I_{3}}=0$

$8 \tilde{x}^{2}+2 \tilde{y}^{2}+2 \tilde{z}^{2}=0$

$\frac{\tilde{z}^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}}+\left(\frac{\tilde{x}^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}}+\frac{\tilde{y}^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}}\right)=0$

this equation is for a type imaginary cone

- reduced to canonical form