Sr Examen
x^2+y^2+z^2+xy-xz+yz+4y+2y+4z-6=0 forma canónica
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Ha introducido
[src]
2 2 2 -6 + x + y + z + 4*z + 6*y + x*y + y*z - x*z = 0
$$x^{2} + x y - x z + y^{2} + y z + 6 y + z^{2} + 4 z - 6 = 0$$
x^2 + x*y - x*z + y^2 + y*z + 6*y + z^2 + 4*z - 6 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$x^{2} + x y - x z + y^{2} + y z + 6 y + z^{2} + 4 z - 6 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = - \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = 3$$
$$a_{33} = 1$$
$$a_{34} = 2$$
$$a_{44} = -6$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
sustituimos coeficientes
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}1 & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & 1 & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & \frac{1}{2} & - \frac{1}{2} & 0\\\frac{1}{2} & 1 & \frac{1}{2} & 3\\- \frac{1}{2} & \frac{1}{2} & 1 & 2\\0 & 3 & 2 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & 1 - \lambda & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = \frac{9}{4}$$
$$I_{3} = 0$$
$$I_{4} = - \frac{3}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - \frac{9 \lambda}{4}$$
$$K_{2} = -31$$
$$K_{3} = - \frac{67}{2}$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - 3 \lambda^{2} + \frac{9 \lambda}{4} = 0$$
$$\lambda_{1} = \frac{3}{2}$$
$$\lambda_{2} = \frac{3}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} + \frac{2 \sqrt{3} \tilde z}{3} = 0$$
y
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} - \frac{2 \sqrt{3} \tilde z}{3} = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{2}{9} \sqrt{3}} + \frac{\tilde y^{2}}{\frac{2}{9} \sqrt{3}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{2}{9} \sqrt{3}} + \frac{\tilde y^{2}}{\frac{2}{9} \sqrt{3}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica
$$x^{2} + x y - x z + y^{2} + y z + 6 y + z^{2} + 4 z - 6 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = - \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = 3$$
$$a_{33} = 1$$
$$a_{34} = 2$$
$$a_{44} = -6$$
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
$$I_{1} = 3$$
| 1 1/2| | 1 1/2| | 1 -1/2|
I2 = | | + | | + | |
|1/2 1 | |1/2 1 | |-1/2 1 |$$I_{3} = \left|\begin{matrix}1 & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & 1 & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & \frac{1}{2} & - \frac{1}{2} & 0\\\frac{1}{2} & 1 & \frac{1}{2} & 3\\- \frac{1}{2} & \frac{1}{2} & 1 & 2\\0 & 3 & 2 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & 1 - \lambda & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & 1 - \lambda\end{matrix}\right|$$
|1 0 | |1 3 | |1 2 |
K2 = | | + | | + | |
|0 -6| |3 -6| |2 -6| | 1 1/2 0 | | 1 1/2 3 | | 1 -1/2 0 |
| | | | | |
K3 = |1/2 1 3 | + |1/2 1 2 | + |-1/2 1 2 |
| | | | | |
| 0 3 -6| | 3 2 -6| | 0 2 -6|$$I_{1} = 3$$
$$I_{2} = \frac{9}{4}$$
$$I_{3} = 0$$
$$I_{4} = - \frac{3}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - \frac{9 \lambda}{4}$$
$$K_{2} = -31$$
$$K_{3} = - \frac{67}{2}$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - 3 \lambda^{2} + \frac{9 \lambda}{4} = 0$$
$$\lambda_{1} = \frac{3}{2}$$
$$\lambda_{2} = \frac{3}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} + \frac{2 \sqrt{3} \tilde z}{3} = 0$$
y
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} - \frac{2 \sqrt{3} \tilde z}{3} = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{2}{9} \sqrt{3}} + \frac{\tilde y^{2}}{\frac{2}{9} \sqrt{3}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{2}{9} \sqrt{3}} + \frac{\tilde y^{2}}{\frac{2}{9} \sqrt{3}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica