Se da la ecuación de superficie de 2 grado:
$$x^{2} - x y + x z + 3 x + y^{2} + y z + 3 y + z^{2} - 3 z = 0$$
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
$$a_{11} = 1$$
$$a_{12} = - \frac{1}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = \frac{3}{2}$$
$$a_{22} = 1$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 1$$
$$a_{34} = - \frac{3}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 3$$
| 1 -1/2| | 1 1/2| | 1 1/2|
I2 = | | + | | + | |
|-1/2 1 | |1/2 1 | |1/2 1 |
$$I_{3} = \left|\begin{matrix}1 & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & 1 & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & - \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\- \frac{1}{2} & 1 & \frac{1}{2} & \frac{3}{2}\\\frac{1}{2} & \frac{1}{2} & 1 & - \frac{3}{2}\\\frac{3}{2} & \frac{3}{2} & - \frac{3}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & 1 - \lambda & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & 1 - \lambda\end{matrix}\right|$$
| 1 3/2| | 1 3/2| | 1 -3/2|
K2 = | | + | | + | |
|3/2 0 | |3/2 0 | |-3/2 0 |
| 1 -1/2 3/2| | 1 1/2 3/2 | | 1 1/2 3/2 |
| | | | | |
K3 = |-1/2 1 3/2| + |1/2 1 -3/2| + |1/2 1 -3/2|
| | | | | |
|3/2 3/2 0 | |3/2 -3/2 0 | |3/2 -3/2 0 |
$$I_{1} = 3$$
$$I_{2} = \frac{9}{4}$$
$$I_{3} = 0$$
$$I_{4} = - \frac{243}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - \frac{9 \lambda}{4}$$
$$K_{2} = - \frac{27}{4}$$
$$K_{3} = - \frac{81}{4}$$
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
$$\lambda^{3} - 3 \lambda^{2} + \frac{9 \lambda}{4} = 0$$
$$\lambda_{1} = \frac{3}{2}$$
$$\lambda_{2} = \frac{3}{2}$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} + 3 \sqrt{3} \tilde z = 0$$
y
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} - 3 \sqrt{3} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\sqrt{3}} + \frac{\tilde y^{2}}{\sqrt{3}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\sqrt{3}} + \frac{\tilde y^{2}}{\sqrt{3}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica