x1^2+3x2^2+3x3^2-2x1x2=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x_{1}^{2} - 2 x_{1} x_{2} + 3 x_{2}^{2} + 3 x_{3}^{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + 2 a_{14} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
donde
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = -1$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 7$$

     |3  0|   |3   -1|   |3  0|
I2 = |    | + |      | + |    |
     |0  3|   |-1  1 |   |0  1|

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

     |3  0|   |3  0|   |1  0|
K2 = |    | + |    | + |    |
     |0  0|   |0  0|   |0  0|

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

$$I_{1} = 7$$
$$I_{2} = 14$$
$$I_{3} = 6$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 14 \lambda + 6$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Como

I3 != 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
$$\lambda^{3} - 7 \lambda^{2} + 14 \lambda - 6 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2 - \sqrt{2}$$
$$\lambda_{3} = \sqrt{2} + 2$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x1^{2} \left(\sqrt{2} + 2\right) + \tilde x2^{2} \left(2 - \sqrt{2}\right) + 3 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{1}{\sqrt{\sqrt{2} + 2}}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{2 - \sqrt{2}}}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica