9x^2+y^2+z^2-6xy+3x-y-2z+1=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$9 x^{2} - 6 x y + 3 x + y^{2} - y + z^{2} - 2 z + 1 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = -3$$
$$a_{13} = 0$$
$$a_{14} = \frac{3}{2}$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = - \frac{1}{2}$$
$$a_{33} = 1$$
$$a_{34} = -1$$
$$a_{44} = 1$$
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
$$I_{1} = 11$$

     |9   -3|   |1  0|   |9  0|
I2 = |      | + |    | + |    |
     |-3  1 |   |0  1|   |0  1|

$$I_{3} = \left|\begin{matrix}9 & -3 & 0\\-3 & 1 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}9 & -3 & 0 & \frac{3}{2}\\-3 & 1 & 0 & - \frac{1}{2}\\0 & 0 & 1 & -1\\\frac{3}{2} & - \frac{1}{2} & -1 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}9 - \lambda & -3 & 0\\-3 & 1 - \lambda & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$

     | 9   3/2|   | 1    -1/2|   |1   -1|
K2 = |        | + |          | + |      |
     |3/2   1 |   |-1/2   1  |   |-1  1 |

     | 9    -3   3/2 |   | 1    0   -1/2|   | 9   0   3/2|
     |               |   |              |   |            |
K3 = |-3    1    -1/2| + | 0    1    -1 | + | 0   1   -1 |
     |               |   |              |   |            |
     |3/2  -1/2   1  |   |-1/2  -1   1  |   |3/2  -1   1 |

$$I_{1} = 11$$
$$I_{2} = 10$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 11 \lambda^{2} - 10 \lambda$$
$$K_{2} = \frac{15}{2}$$
$$K_{3} = - \frac{5}{2}$$
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
$$\lambda^{3} - 11 \lambda^{2} + 10 \lambda = 0$$
$$\lambda_{1} = 10$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$10 \tilde x^{2} + \tilde y^{2} - \frac{1}{4} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{2}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2}\right)^{2}} = 1$$
es la ecuación para el tipo cilindro elíptico
- está reducida a la forma canónica