Sr Examen
x1^2+3x2^2+3x3^2-2x1x2=0 forma canónica
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Ha introducido
[src]
2 2 2 x1 + 3*x2 + 3*x3 - 2*x1*x2 = 0
$$x_{1}^{2} - 2 x_{1} x_{2} + 3 x_{2}^{2} + 3 x_{3}^{2} = 0$$
x1^2 - 2*x1*x2 + 3*x2^2 + 3*x3^2 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$x_{1}^{2} - 2 x_{1} x_{2} + 3 x_{2}^{2} + 3 x_{3}^{2} = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = -1$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 7$$
$$I_{3} = \left|\begin{matrix}3 & 0 & 0\\0 & 3 & -1\\0 & -1 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 0 & 0 & 0\\0 & 3 & -1 & 0\\0 & -1 & 1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0 & 0\\0 & 3 - \lambda & -1\\0 & -1 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 7$$
$$I_{2} = 14$$
$$I_{3} = 6$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 14 \lambda + 6$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 7 \lambda^{2} + 14 \lambda - 6 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2 - \sqrt{2}$$
$$\lambda_{3} = \sqrt{2} + 2$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x1^{2} \left(\sqrt{2} + 2\right) + \tilde x2^{2} \left(2 - \sqrt{2}\right) + 3 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{1}{\sqrt{\sqrt{2} + 2}}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{2 - \sqrt{2}}}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica
$$x_{1}^{2} - 2 x_{1} x_{2} + 3 x_{2}^{2} + 3 x_{3}^{2} = 0$$
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
$$a_{11} = 3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = -1$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 7$$
|3 0| |3 -1| |3 0|
I2 = | | + | | + | |
|0 3| |-1 1 | |0 1|$$I_{3} = \left|\begin{matrix}3 & 0 & 0\\0 & 3 & -1\\0 & -1 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 0 & 0 & 0\\0 & 3 & -1 & 0\\0 & -1 & 1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & 0 & 0\\0 & 3 - \lambda & -1\\0 & -1 & 1 - \lambda\end{matrix}\right|$$
|3 0| |3 0| |1 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |3 0 0| |3 -1 0| |3 0 0|
| | | | | |
K3 = |0 3 0| + |-1 1 0| + |0 1 0|
| | | | | |
|0 0 0| |0 0 0| |0 0 0|$$I_{1} = 7$$
$$I_{2} = 14$$
$$I_{3} = 6$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 14 \lambda + 6$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 7 \lambda^{2} + 14 \lambda - 6 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2 - \sqrt{2}$$
$$\lambda_{3} = \sqrt{2} + 2$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x1^{2} \left(\sqrt{2} + 2\right) + \tilde x2^{2} \left(2 - \sqrt{2}\right) + 3 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{1}{\sqrt{\sqrt{2} + 2}}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{2 - \sqrt{2}}}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica