Se da la ecuación de superficie de 2 grado:
$$7 x^{2} + 4 x y + 10 x z - 2 x + 4 y^{2} - 4 y z - 4 y + 7 z^{2} + 2 z - 5 = 0$$
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
$$a_{11} = 7$$
$$a_{12} = 2$$
$$a_{13} = 5$$
$$a_{14} = -1$$
$$a_{22} = 4$$
$$a_{23} = -2$$
$$a_{24} = -2$$
$$a_{33} = 7$$
$$a_{34} = 1$$
$$a_{44} = -5$$
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
$$I_{1} = 18$$
|7 2| |4 -2| |7 5|
I2 = | | + | | + | |
|2 4| |-2 7 | |5 7|
$$I_{3} = \left|\begin{matrix}7 & 2 & 5\\2 & 4 & -2\\5 & -2 & 7\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}7 & 2 & 5 & -1\\2 & 4 & -2 & -2\\5 & -2 & 7 & 1\\-1 & -2 & 1 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}7 - \lambda & 2 & 5\\2 & 4 - \lambda & -2\\5 & -2 & 7 - \lambda\end{matrix}\right|$$
|7 -1| |4 -2| |7 1 |
K2 = | | + | | + | |
|-1 -5| |-2 -5| |1 -5|
|7 2 -1| |4 -2 -2| |7 5 -1|
| | | | | |
K3 = |2 4 -2| + |-2 7 1 | + |5 7 1 |
| | | | | |
|-1 -2 -5| |-2 1 -5| |-1 1 -5|
$$I_{1} = 18$$
$$I_{2} = 72$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 18 \lambda^{2} - 72 \lambda$$
$$K_{2} = -96$$
$$K_{3} = -432$$
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
$$\lambda^{3} - 18 \lambda^{2} + 72 \lambda = 0$$
$$\lambda_{1} = 12$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$12 \tilde x^{2} + 6 \tilde y^{2} - 6 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{6} \sqrt{3}}{\frac{1}{6} \sqrt{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{6} \sqrt{6}}\right)^{2}} = 1$$
es la ecuación para el tipo cilindro elíptico
- está reducida a la forma canónica