6x+5y+7z-4xy+4zx-108=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$- 4 x y + 4 x z + 6 x + 5 y + 7 z - 108 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = -2$$
$$a_{13} = 2$$
$$a_{14} = 3$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = \frac{5}{2}$$
$$a_{33} = 0$$
$$a_{34} = \frac{7}{2}$$
$$a_{44} = -108$$
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
$$I_{1} = 0$$

     |0   -2|   |0  0|   |0  2|
I2 = |      | + |    | + |    |
     |-2  0 |   |0  0|   |2  0|

$$I_{3} = \left|\begin{matrix}0 & -2 & 2\\-2 & 0 & 0\\2 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & -2 & 2 & 3\\-2 & 0 & 0 & \frac{5}{2}\\2 & 0 & 0 & \frac{7}{2}\\3 & \frac{5}{2} & \frac{7}{2} & -108\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & -2 & 2\\-2 & - \lambda & 0\\2 & 0 & - \lambda\end{matrix}\right|$$

     |0   3  |   | 0   5/2 |   | 0   7/2 |
K2 = |       | + |         | + |         |
     |3  -108|   |5/2  -108|   |7/2  -108|

     |0   -2    3  |   | 0    0   5/2 |   |0   2    3  |
     |             |   |              |   |            |
K3 = |-2   0   5/2 | + | 0    0   7/2 | + |2   0   7/2 |
     |             |   |              |   |            |
     |3   5/2  -108|   |5/2  7/2  -108|   |3  7/2  -108|

$$I_{1} = 0$$
$$I_{2} = -8$$
$$I_{3} = 0$$
$$I_{4} = 144$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 8 \lambda$$
$$K_{2} = - \frac{55}{2}$$
$$K_{3} = 876$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - 8 \lambda = 0$$
$$\lambda_{1} = - 2 \sqrt{2}$$
$$\lambda_{2} = 2 \sqrt{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$- 2 \sqrt{2} \tilde x^{2} + 2 \sqrt{2} \tilde y^{2} + 6 \sqrt{2} \tilde z = 0$$
y
$$- 2 \sqrt{2} \tilde x^{2} + 2 \sqrt{2} \tilde y^{2} - 6 \sqrt{2} \tilde z = 0$$
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{2}} - \frac{\tilde y^{2}}{\frac{3}{2}}\right) = 0$$
y
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{2}} - \frac{\tilde y^{2}}{\frac{3}{2}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica