3x1^2+x2^2-3/2x3^2+2*√3*x1x2 forma canónica

Se da la ecuación de superficie de 2 grado:
$$3 x_{1}^{2} + 2 \sqrt{3} x_{1} x_{2} + x_{2}^{2} - \frac{3 x_{3}^{2}}{2} = 0$$
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
$$a_{11} = - \frac{3}{2}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = \sqrt{3}$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = \frac{5}{2}$$

                 |         ___|            
     |-3/2  0|   |  1    \/ 3 |   |-3/2  0|
I2 = |       | + |            | + |       |
     | 0    1|   |  ___       |   | 0    3|
                 |\/ 3     3  |            

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

     |-3/2  0|   |1  0|   |3  0|
K2 = |       | + |    | + |    |
     | 0    0|   |0  0|   |0  0|

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

$$I_{1} = \frac{5}{2}$$
$$I_{2} = -6$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{5 \lambda^{2}}{2} + 6 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - \frac{5 \lambda^{2}}{2} - 6 \lambda = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = - \frac{3}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \frac{3 \tilde x2^{2}}{2} + 4 \tilde x3^{2} = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{\sqrt{6}}{3}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
es la ecuación para el tipo dos planos intersectantes
- está reducida a la forma canónica