2x^2+2y^2+2z^2-xz-sqrt(3)yz=1 forma canónica

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

              |            ___ |               
              |         -\/ 3  |               
              |   2     -------|               
     |2  0|   |            2   |   | 2    -1/2|
I2 = |    | + |                | + |          |
     |0  2|   |   ___          |   |-1/2   2  |
              |-\/ 3           |               
              |-------     2   |               
              |   2            |               

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

     |2  0 |   |2  0 |   |2  0 |
K2 = |     | + |     | + |     |
     |0  -1|   |0  -1|   |0  -1|

                  |            ___     |                   
                  |         -\/ 3      |                   
                  |   2     -------  0 |                   
     |2  0  0 |   |            2       |   | 2    -1/2  0 |
     |        |   |                    |   |              |
K3 = |0  2  0 | + |   ___              | + |-1/2   2    0 |
     |        |   |-\/ 3               |   |              |
     |0  0  -1|   |-------     2     0 |   | 0     0    -1|
                  |   2                |                   
                  |                    |                   
                  |   0        0     -1|                   

$$I_{1} = 6$$
$$I_{2} = 11$$
$$I_{3} = 6$$
$$I_{4} = -6$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} - 11 \lambda + 6$$
$$K_{2} = -6$$
$$K_{3} = -11$$
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
$$\lambda^{3} - 6 \lambda^{2} + 11 \lambda - 6 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 1$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x^{2} + 2 \tilde y^{2} + \tilde z^{2} - 1 = 0$$
$$\frac{\tilde z^{2}}{\left(1^{-1}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{1}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{1}\right)^{2}}\right) = 1$$
es la ecuación para el tipo elipsoide
- está reducida a la forma canónica