x^2-2y^2+z^2+4*x*y-4*y*z-8*z*y-14*x-4*y+14*z+16 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} + 4 x y - 14 x - 2 y^{2} - 12 y z - 4 y + z^{2} + 14 z + 16 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{14} = -7$$
$$a_{22} = -2$$
$$a_{23} = -6$$
$$a_{24} = -2$$
$$a_{33} = 1$$
$$a_{34} = 7$$
$$a_{44} = 16$$
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
$$I_{1} = 0$$

     |1  2 |   |-2  -6|   |1  0|
I2 = |     | + |      | + |    |
     |2  -2|   |-6  1 |   |0  1|

$$I_{3} = \left|\begin{matrix}1 & 2 & 0\\2 & -2 & -6\\0 & -6 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 2 & 0 & -7\\2 & -2 & -6 & -2\\0 & -6 & 1 & 7\\-7 & -2 & 7 & 16\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 2 & 0\\2 & - \lambda - 2 & -6\\0 & -6 & 1 - \lambda\end{matrix}\right|$$

     |1   -7|   |-2  -2|   |1  7 |
K2 = |      | + |      | + |     |
     |-7  16|   |-2  16|   |7  16|

     |1   2   -7|   |-2  -6  -2|   |1   0  -7|
     |          |   |          |   |         |
K3 = |2   -2  -2| + |-6  1   7 | + |0   1  7 |
     |          |   |          |   |         |
     |-7  -2  16|   |-2  7   16|   |-7  7  16|

$$I_{1} = 0$$
$$I_{2} = -43$$
$$I_{3} = -42$$
$$I_{4} = 528$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 43 \lambda - 42$$
$$K_{2} = -102$$
$$K_{3} = -374$$
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
$$\lambda^{3} - 43 \lambda + 42 = 0$$
$$\lambda_{1} = 6$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -7$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$6 \tilde x^{2} + \tilde y^{2} - 7 \tilde z^{2} - \frac{88}{7} = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\frac{1}{7} \sqrt{7}}{\frac{1}{44} \sqrt{154}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{44} \sqrt{154}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{1}{44} \sqrt{154}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica