x^2+y^2+z^2+2*x*y-2*x*z-y*z+4*x+3*y-5*z+4=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} + 2 x y - 2 x z + 4 x + y^{2} - y z + 3 y + z^{2} - 5 z + 4 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 1$$
$$a_{13} = -1$$
$$a_{14} = 2$$
$$a_{22} = 1$$
$$a_{23} = - \frac{1}{2}$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 1$$
$$a_{34} = - \frac{5}{2}$$
$$a_{44} = 4$$
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
$$I_{1} = 3$$

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

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

     |1  2|   | 1   3/2|   | 1    -5/2|
K2 = |    | + |        | + |          |
     |2  4|   |3/2   4 |   |-5/2   4  |

     |1   1    2 |   | 1    -1/2  3/2 |   |1    -1    2  |
     |           |   |                |   |              |
K3 = |1   1   3/2| + |-1/2   1    -5/2| + |-1   1    -5/2|
     |           |   |                |   |              |
     |2  3/2   4 |   |3/2   -5/2   4  |   |2   -5/2   4  |

$$I_{1} = 3$$
$$I_{2} = \frac{3}{4}$$
$$I_{3} = - \frac{1}{4}$$
$$I_{4} = \frac{1}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - \frac{3 \lambda}{4} - \frac{1}{4}$$
$$K_{2} = - \frac{1}{2}$$
$$K_{3} = - \frac{9}{4}$$
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
$$\lambda^{3} - 3 \lambda^{2} + \frac{3 \lambda}{4} + \frac{1}{4} = 0$$
$$\lambda_{1} = \frac{1}{2}$$
$$\lambda_{2} = \frac{5}{4} - \frac{\sqrt{33}}{4}$$
$$\lambda_{3} = \frac{5}{4} + \frac{\sqrt{33}}{4}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\frac{\tilde x^{2}}{2} + \tilde y^{2} \left(\frac{5}{4} - \frac{\sqrt{33}}{4}\right) + \tilde z^{2} \left(\frac{5}{4} + \frac{\sqrt{33}}{4}\right) - 1 = 0$$
$$- \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{4} + \frac{\sqrt{33}}{4}}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{\sqrt{\frac{5}{4} + \frac{\sqrt{33}}{4}}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica