Se da la ecuación de superficie de 2 grado:
$$- 4 x - 4 y - 8 z + \left(x + y + 2 z\right)^{2} + 3 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 1$$
$$a_{13} = 2$$
$$a_{14} = -2$$
$$a_{22} = 1$$
$$a_{23} = 2$$
$$a_{24} = -2$$
$$a_{33} = 4$$
$$a_{34} = -4$$
$$a_{44} = 3$$
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
$$I_{1} = 6$$
|1 1| |1 2| |1 2|
I2 = | | + | | + | |
|1 1| |2 4| |2 4|
$$I_{3} = \left|\begin{matrix}1 & 1 & 2\\1 & 1 & 2\\2 & 2 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 1 & 2 & -2\\1 & 1 & 2 & -2\\2 & 2 & 4 & -4\\-2 & -2 & -4 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 1 & 2\\1 & 1 - \lambda & 2\\2 & 2 & 4 - \lambda\end{matrix}\right|$$
|1 -2| |1 -2| |4 -4|
K2 = | | + | | + | |
|-2 3 | |-2 3 | |-4 3 |
|1 1 -2| |1 2 -2| |1 2 -2|
| | | | | |
K3 = |1 1 -2| + |2 4 -4| + |2 4 -4|
| | | | | |
|-2 -2 3 | |-2 -4 3 | |-2 -4 3 |
$$I_{1} = 6$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2}$$
$$K_{2} = -6$$
$$K_{3} = 0$$
Como
$$I_{2} = 0 \wedge I_{3} = 0 \wedge I_{4} = 0 \wedge K_{3} = 0 \wedge I_{1} \neq 0$$
entonces por razón de tipos de rectas:
hay que
entonces la forma canónica de la ecuación será
$$I_{1} \tilde x^{2} + \frac{K_{2}}{I_{1}} = 0$$
$$6 \tilde x^{2} - 1 = 0$$
$$\tilde x^{2} = \frac{1}{6}$$
es la ecuación para el tipo dos planos paralelos
- está reducida a la forma canónica