x^2-2*y^2+z^2+4*x*y+4*y*z-10*z*x+2*x+4*y-10*z-1=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} + 4 x y - 10 x z + 2 x - 2 y^{2} + 4 y z + 4 y + z^{2} - 10 z - 1 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = -5$$
$$a_{14} = 1$$
$$a_{22} = -2$$
$$a_{23} = 2$$
$$a_{24} = 2$$
$$a_{33} = 1$$
$$a_{34} = -5$$
$$a_{44} = -1$$
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
$$I_{1} = 0$$

     |1  2 |   |-2  2|   |1   -5|
I2 = |     | + |     | + |      |
     |2  -2|   |2   1|   |-5  1 |

$$I_{3} = \left|\begin{matrix}1 & 2 & -5\\2 & -2 & 2\\-5 & 2 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 2 & -5 & 1\\2 & -2 & 2 & 2\\-5 & 2 & 1 & -5\\1 & 2 & -5 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 2 & -5\\2 & - \lambda - 2 & 2\\-5 & 2 & 1 - \lambda\end{matrix}\right|$$

     |1  1 |   |-2  2 |   |1   -5|
K2 = |     | + |      | + |      |
     |1  -1|   |2   -1|   |-5  -1|

     |1  2   1 |   |-2  2   2 |   |1   -5  1 |
     |         |   |          |   |          |
K3 = |2  -2  2 | + |2   1   -5| + |-5  1   -5|
     |         |   |          |   |          |
     |1  2   -1|   |2   -5  -1|   |1   -5  -1|

$$I_{1} = 0$$
$$I_{2} = -36$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 36 \lambda$$
$$K_{2} = -30$$
$$K_{3} = 72$$
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
$$\lambda^{3} - 36 \lambda = 0$$
$$\lambda_{1} = -6$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- 6 \tilde x^{2} + 6 \tilde y^{2} - 2 = 0$$

        2           2     
\tilde x    \tilde y      
--------- - --------- = -1
   1/3         1/3        

es la ecuación para el tipo cilindro hiperbólico
- está reducida a la forma canónica