Sr Examen
17(x1)^2+(14(x2)^2)+14(x3)^2-4(x1)(x2)-4(x1)(x3)-8(x2)(x3) forma canónica
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Ha introducido
[src]
2 2 2 14*x2 + 14*x3 + 17*x1 - 8*x2*x3 - 4*x1*x2 - 4*x1*x3 = 0
$$17 x_{1}^{2} - 4 x_{1} x_{2} - 4 x_{1} x_{3} + 14 x_{2}^{2} - 8 x_{2} x_{3} + 14 x_{3}^{2} = 0$$
17*x1^2 - 4*x1*x2 - 4*x1*x3 + 14*x2^2 - 8*x2*x3 + 14*x3^2 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$17 x_{1}^{2} - 4 x_{1} x_{2} - 4 x_{1} x_{3} + 14 x_{2}^{2} - 8 x_{2} x_{3} + 14 x_{3}^{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + 2 a_{14} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
donde
$$a_{11} = 14$$
$$a_{12} = -4$$
$$a_{13} = -2$$
$$a_{14} = 0$$
$$a_{22} = 14$$
$$a_{23} = -2$$
$$a_{24} = 0$$
$$a_{33} = 17$$
$$a_{34} = 0$$
$$a_{44} = 0$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
sustituimos coeficientes
$$I_{1} = 45$$
$$I_{3} = \left|\begin{matrix}14 & -4 & -2\\-4 & 14 & -2\\-2 & -2 & 17\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}14 & -4 & -2 & 0\\-4 & 14 & -2 & 0\\-2 & -2 & 17 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}14 - \lambda & -4 & -2\\-4 & 14 - \lambda & -2\\-2 & -2 & 17 - \lambda\end{matrix}\right|$$
$$I_{1} = 45$$
$$I_{2} = 648$$
$$I_{3} = 2916$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 45 \lambda^{2} - 648 \lambda + 2916$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Como
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
$$\lambda^{3} - 45 \lambda^{2} + 648 \lambda - 2916 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 18$$
$$\lambda_{3} = 18$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$18 \tilde x1^{2} + 18 \tilde x2^{2} + 9 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{2}}{6}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{6}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{3}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica
$$17 x_{1}^{2} - 4 x_{1} x_{2} - 4 x_{1} x_{3} + 14 x_{2}^{2} - 8 x_{2} x_{3} + 14 x_{3}^{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + 2 a_{14} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
donde
$$a_{11} = 14$$
$$a_{12} = -4$$
$$a_{13} = -2$$
$$a_{14} = 0$$
$$a_{22} = 14$$
$$a_{23} = -2$$
$$a_{24} = 0$$
$$a_{33} = 17$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 45$$
|14 -4| |14 -2| |14 -2|
I2 = | | + | | + | |
|-4 14| |-2 17| |-2 17|$$I_{3} = \left|\begin{matrix}14 & -4 & -2\\-4 & 14 & -2\\-2 & -2 & 17\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}14 & -4 & -2 & 0\\-4 & 14 & -2 & 0\\-2 & -2 & 17 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}14 - \lambda & -4 & -2\\-4 & 14 - \lambda & -2\\-2 & -2 & 17 - \lambda\end{matrix}\right|$$
|14 0| |14 0| |17 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |14 -4 0| |14 -2 0| |14 -2 0|
| | | | | |
K3 = |-4 14 0| + |-2 17 0| + |-2 17 0|
| | | | | |
|0 0 0| |0 0 0| |0 0 0|$$I_{1} = 45$$
$$I_{2} = 648$$
$$I_{3} = 2916$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 45 \lambda^{2} - 648 \lambda + 2916$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 45 \lambda^{2} + 648 \lambda - 2916 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 18$$
$$\lambda_{3} = 18$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$18 \tilde x1^{2} + 18 \tilde x2^{2} + 9 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{\sqrt{2}}{6}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{6}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{3}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica