3/2x^2+4xy-3/2y^2-4sqrt(5)xz-2sqrt(5)yz-5/2z^2 forma canónica

Se da la ecuación de superficie de 2 grado:
$$\frac{3 x^{2}}{2} + 4 x y - 4 \sqrt{5} x z - \frac{3 y^{2}}{2} - 2 \sqrt{5} y z - \frac{5 z^{2}}{2} = 0$$
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
$$a_{11} = \frac{3}{2}$$
$$a_{12} = 2$$
$$a_{13} = - 2 \sqrt{5}$$
$$a_{14} = 0$$
$$a_{22} = - \frac{3}{2}$$
$$a_{23} = - \sqrt{5}$$
$$a_{24} = 0$$
$$a_{33} = - \frac{5}{2}$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = - \frac{5}{2}$$

                   |           ___|   |               ___|
     |3/2   2  |   | -3/2   -\/ 5 |   |  3/2     -2*\/ 5 |
I2 = |         | + |              | + |                  |
     | 2   -3/2|   |   ___        |   |     ___          |
                   |-\/ 5    -5/2 |   |-2*\/ 5     -5/2  |

$$I_{3} = \left|\begin{matrix}\frac{3}{2} & 2 & - 2 \sqrt{5}\\2 & - \frac{3}{2} & - \sqrt{5}\\- 2 \sqrt{5} & - \sqrt{5} & - \frac{5}{2}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}\frac{3}{2} & 2 & - 2 \sqrt{5} & 0\\2 & - \frac{3}{2} & - \sqrt{5} & 0\\- 2 \sqrt{5} & - \sqrt{5} & - \frac{5}{2} & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}\frac{3}{2} - \lambda & 2 & - 2 \sqrt{5}\\2 & - \lambda - \frac{3}{2} & - \sqrt{5}\\- 2 \sqrt{5} & - \sqrt{5} & - \lambda - \frac{5}{2}\end{matrix}\right|$$

     |3/2  0|   |-3/2  0|   |-5/2  0|
K2 = |      | + |       | + |       |
     | 0   0|   | 0    0|   | 0    0|

                      |           ___   |   |               ___   |
     |3/2   2    0|   | -3/2   -\/ 5   0|   |  3/2     -2*\/ 5   0|
     |            |   |                 |   |                     |
K3 = | 2   -3/2  0| + |   ___           | + |     ___             |
     |            |   |-\/ 5    -5/2   0|   |-2*\/ 5     -5/2    0|
     | 0    0    0|   |                 |   |                     |
                      |  0       0     0|   |   0         0      0|

$$I_{1} = - \frac{5}{2}$$
$$I_{2} = - \frac{125}{4}$$
$$I_{3} = \frac{625}{8}$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \frac{5 \lambda^{2}}{2} + \frac{125 \lambda}{4} + \frac{625}{8}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} + \frac{5 \lambda^{2}}{2} - \frac{125 \lambda}{4} - \frac{625}{8} = 0$$
$$\lambda_{1} = - \frac{5}{2}$$
$$\lambda_{2} = - \frac{5 \sqrt{5}}{2}$$
$$\lambda_{3} = \frac{5 \sqrt{5}}{2}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \frac{5 \tilde x^{2}}{2} - \frac{5 \sqrt{5} \tilde y^{2}}{2} + \frac{5 \sqrt{5} \tilde z^{2}}{2} = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\sqrt{2} \sqrt[4]{5}}{5}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{10}}{5}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{2} \sqrt[4]{5}}{5}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono
- está reducida a la forma canónica