2x**2+2y**2-z**2-2xy+4yz+4xz-22x-4y+2z+16 forma canónica

Se da la ecuación de superficie de 2 grado:
$$2 x^{2} - 2 x y + 4 x z - 22 x + 2 y^{2} + 4 y z - 4 y - z^{2} + 2 z + 16 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = -1$$
$$a_{13} = 2$$
$$a_{14} = -11$$
$$a_{22} = 2$$
$$a_{23} = 2$$
$$a_{24} = -2$$
$$a_{33} = -1$$
$$a_{34} = 1$$
$$a_{44} = 16$$
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
$$I_{1} = 3$$

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

$$I_{3} = \left|\begin{matrix}2 & -1 & 2\\-1 & 2 & 2\\2 & 2 & -1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & -1 & 2 & -11\\-1 & 2 & 2 & -2\\2 & 2 & -1 & 1\\-11 & -2 & 1 & 16\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & -1 & 2\\-1 & 2 - \lambda & 2\\2 & 2 & - \lambda - 1\end{matrix}\right|$$

     | 2   -11|   |2   -2|   |-1  1 |
K2 = |        | + |      | + |      |
     |-11  16 |   |-2  16|   |1   16|

     | 2   -1  -11|   |2   2   -2|   | 2   2   -11|
     |            |   |          |   |            |
K3 = |-1   2   -2 | + |2   -1  1 | + | 2   -1   1 |
     |            |   |          |   |            |
     |-11  -2  16 |   |-2  1   16|   |-11  1   16 |

$$I_{1} = 3$$
$$I_{2} = -9$$
$$I_{3} = -27$$
$$I_{4} = 27$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} + 9 \lambda - 27$$
$$K_{2} = -78$$
$$K_{3} = -369$$
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
$$\lambda^{3} - 3 \lambda^{2} - 9 \lambda + 27 = 0$$
$$\lambda_{1} = -3$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 3$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- 3 \tilde x^{2} + 3 \tilde y^{2} + 3 \tilde z^{2} - 1 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{1}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{1}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica