2xx+5yy+2zz-2xy-4xz+2yz+2x-10y-2z-1=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$2 x^{2} - 2 x y - 4 x z + 2 x + 5 y^{2} + 2 y z - 10 y + 2 z^{2} - 2 z - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = -1$$
$$a_{13} = -2$$
$$a_{14} = 1$$
$$a_{22} = 5$$
$$a_{23} = 1$$
$$a_{24} = -5$$
$$a_{33} = 2$$
$$a_{34} = -1$$
$$a_{44} = -1$$
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
$$I_{1} = 9$$

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

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

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

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

$$I_{1} = 9$$
$$I_{2} = 18$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 9 \lambda^{2} - 18 \lambda$$
$$K_{2} = -36$$
$$K_{3} = -108$$
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
$$\lambda^{3} - 9 \lambda^{2} + 18 \lambda = 0$$
$$\lambda_{1} = 6$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$6 \tilde x^{2} + 3 \tilde y^{2} - 6 = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{6} \sqrt{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{6}}\right)^{2}} = 1$$
es la ecuación para el tipo cilindro elíptico
- está reducida a la forma canónica