3y^2+6xz-2x^2-2xy-4yz-z^2 forma canónica

Se da la ecuación de superficie de 2 grado:
$$- 2 x^{2} - 2 x y + 6 x z + 3 y^{2} - 4 y z - z^{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
donde
$$a_{11} = -2$$
$$a_{12} = -1$$
$$a_{13} = 3$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = -2$$
$$a_{24} = 0$$
$$a_{33} = -1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22} + a_{33}$$

     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$

     |a11  a14|   |a22  a24|   |a33  a34|
K2 = |        | + |        | + |        |
     |a14  a44|   |a24  a44|   |a34  a44|

     |a11  a12  a14|   |a22  a23  a24|   |a11  a13  a14|
     |             |   |             |   |             |
K3 = |a12  a22  a24| + |a23  a33  a34| + |a13  a33  a34|
     |             |   |             |   |             |
     |a14  a24  a44|   |a24  a34  a44|   |a14  a34  a44|

sustituimos coeficientes
$$I_{1} = 0$$

     |-2  -1|   |3   -2|   |-2  3 |
I2 = |      | + |      | + |      |
     |-1  3 |   |-2  -1|   |3   -1|

$$I_{3} = \left|\begin{matrix}-2 & -1 & 3\\-1 & 3 & -2\\3 & -2 & -1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-2 & -1 & 3 & 0\\-1 & 3 & -2 & 0\\3 & -2 & -1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & -1 & 3\\-1 & 3 - \lambda & -2\\3 & -2 & - \lambda - 1\end{matrix}\right|$$

     |-2  0|   |3  0|   |-1  0|
K2 = |     | + |    | + |     |
     |0   0|   |0  0|   |0   0|

     |-2  -1  0|   |3   -2  0|   |-2  3   0|
     |         |   |         |   |         |
K3 = |-1  3   0| + |-2  -1  0| + |3   -1  0|
     |         |   |         |   |         |
     |0   0   0|   |0   0   0|   |0   0   0|

$$I_{1} = 0$$
$$I_{2} = -21$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 21 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
entonces por razón de tipos de rectas:
hay que
Formulamos la ecuación característica para nuestra superficie:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
o
$$\lambda^{3} - 21 \lambda = 0$$
$$\lambda_{1} = - \sqrt{21}$$
$$\lambda_{2} = \sqrt{21}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \sqrt{21} \tilde x^{2} + \sqrt{21} \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{21^{\frac{3}{4}}}{21}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{21^{\frac{3}{4}}}{21}\right)^{2}} = 0$$
es la ecuación para el tipo dos planos intersectantes
- está reducida a la forma canónica