x^2-4*y^2+z^2-4*x*y+x*z-4*y*z+2*x+6*y-2*z+4=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} - 4 x y + x z + 2 x - 4 y^{2} - 4 y z + 6 y + z^{2} - 2 z + 4 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -2$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 1$$
$$a_{22} = -4$$
$$a_{23} = -2$$
$$a_{24} = 3$$
$$a_{33} = 1$$
$$a_{34} = -1$$
$$a_{44} = 4$$
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
$$I_{1} = -2$$

     |1   -2|   |-4  -2|   | 1   1/2|
I2 = |      | + |      | + |        |
     |-2  -4|   |-2  1 |   |1/2   1 |

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

     |1  1|   |-4  3|   |1   -1|
K2 = |    | + |     | + |      |
     |1  4|   |3   4|   |-1  4 |

     |1   -2  1|   |-4  -2  3 |   | 1   1/2  1 |
     |         |   |          |   |            |
K3 = |-2  -4  3| + |-2  1   -1| + |1/2   1   -1|
     |         |   |          |   |            |
     |1   3   4|   |3   -1  4 |   | 1   -1   4 |

$$I_{1} = -2$$
$$I_{2} = - \frac{61}{4}$$
$$I_{3} = -7$$
$$I_{4} = - \frac{27}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 2 \lambda^{2} + \frac{61 \lambda}{4} - 7$$
$$K_{2} = -19$$
$$K_{3} = -74$$
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
$$\lambda^{3} + 2 \lambda^{2} - \frac{61 \lambda}{4} + 7 = 0$$
$$\lambda_{1} = \frac{1}{2}$$
$$\lambda_{2} = - \frac{\sqrt{249}}{4} - \frac{5}{4}$$
$$\lambda_{3} = - \frac{5}{4} + \frac{\sqrt{249}}{4}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\frac{\tilde x^{2}}{2} + \tilde y^{2} \left(- \frac{\sqrt{249}}{4} - \frac{5}{4}\right) + \tilde z^{2} \left(- \frac{5}{4} + \frac{\sqrt{249}}{4}\right) + \frac{27}{28} = 0$$
$$- \frac{\tilde y^{2}}{\left(\frac{1}{\frac{2 \sqrt{21}}{9} \sqrt{\frac{5}{4} + \frac{\sqrt{249}}{4}}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{\frac{2}{9} \sqrt{21}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{\frac{2 \sqrt{21}}{9} \sqrt{- \frac{5}{4} + \frac{\sqrt{249}}{4}}}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica