x*(p-q)+4*(q-p)+(p-q) forma canónica

Se da la ecuación de superficie de 2 grado:
$$- 3 p + 3 q + x \left(p - q\right) = 0$$
Esta ecuación tiene la forma:
$$a_{11} x^{2} + 2 a_{12} q x + 2 a_{13} p x + 2 a_{14} x + a_{22} q^{2} + 2 a_{23} p q + 2 a_{24} q + a_{33} p^{2} + 2 a_{34} p + a_{44} = 0$$
donde
$$a_{11} = 0$$
$$a_{12} = - \frac{1}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 0$$
$$a_{34} = - \frac{3}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$

     | 0    -1/2|   |0  0|   | 0   1/2|
I2 = |          | + |    | + |        |
     |-1/2   0  |   |0  0|   |1/2   0 |

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

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

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

$$I_{1} = 0$$
$$I_{2} = - \frac{1}{2}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{2}$$
$$K_{2} = - \frac{9}{2}$$
$$K_{3} = 0$$
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
$$\lambda^{3} - \frac{\lambda}{2} = 0$$
$$\lambda_{1} = - \frac{\sqrt{2}}{2}$$
$$\lambda_{2} = \frac{\sqrt{2}}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde q^{2} \lambda_{2} + \tilde x^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\frac{\sqrt{2} \tilde q^{2}}{2} - \frac{\sqrt{2} \tilde x^{2}}{2} = 0$$
$$- \frac{\tilde q^{2}}{\left(\sqrt[4]{2}\right)^{2}} + \frac{\tilde x^{2}}{\left(\sqrt[4]{2}\right)^{2}} = 0$$
es la ecuación para el tipo dos planos intersectantes
- está reducida a la forma canónica