Sr Examen
3x1^2+3x2^2+3x3^2++2x1x2+2x1x3-2x2x3 forma canónica
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Solución
Ha introducido
[src]
2 2 2 3*x1 + 3*x2 + 3*x3 - 2*x2*x3 + 2*x1*x2 + 2*x1*x3 = 0
$$3 x_{1}^{2} + 2 x_{1} x_{2} + 2 x_{1} x_{3} + 3 x_{2}^{2} - 2 x_{2} x_{3} + 3 x_{3}^{2} = 0$$
3*x1^2 + 2*x1*x2 + 2*x1*x3 + 3*x2^2 - 2*x2*x3 + 3*x3^2 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$3 x_{1}^{2} + 2 x_{1} x_{2} + 2 x_{1} x_{3} + 3 x_{2}^{2} - 2 x_{2} x_{3} + 3 x_{3}^{2} = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -1$$
$$a_{13} = 1$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = 1$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 9$$
$$I_{3} = \left|\begin{matrix}3 & -1 & 1\\-1 & 3 & 1\\1 & 1 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & -1 & 1 & 0\\-1 & 3 & 1 & 0\\1 & 1 & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & -1 & 1\\-1 & 3 - \lambda & 1\\1 & 1 & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = 9$$
$$I_{2} = 24$$
$$I_{3} = 16$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 9 \lambda^{2} - 24 \lambda + 16$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 9 \lambda^{2} + 24 \lambda - 16 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 4$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$4 \tilde x1^{2} + 4 \tilde x2^{2} + \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{1}{2}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} + \frac{\tilde x3^{2}}{1^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica
$$3 x_{1}^{2} + 2 x_{1} x_{2} + 2 x_{1} x_{3} + 3 x_{2}^{2} - 2 x_{2} x_{3} + 3 x_{3}^{2} = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -1$$
$$a_{13} = 1$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = 1$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 9$$
|3 -1| |3 1| |3 1|
I2 = | | + | | + | |
|-1 3 | |1 3| |1 3|$$I_{3} = \left|\begin{matrix}3 & -1 & 1\\-1 & 3 & 1\\1 & 1 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & -1 & 1 & 0\\-1 & 3 & 1 & 0\\1 & 1 & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & -1 & 1\\-1 & 3 - \lambda & 1\\1 & 1 & 3 - \lambda\end{matrix}\right|$$
|3 0| |3 0| |3 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |3 -1 0| |3 1 0| |3 1 0|
| | | | | |
K3 = |-1 3 0| + |1 3 0| + |1 3 0|
| | | | | |
|0 0 0| |0 0 0| |0 0 0|$$I_{1} = 9$$
$$I_{2} = 24$$
$$I_{3} = 16$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 9 \lambda^{2} - 24 \lambda + 16$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 9 \lambda^{2} + 24 \lambda - 16 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 4$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$4 \tilde x1^{2} + 4 \tilde x2^{2} + \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{\left(\frac{1}{2}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} + \frac{\tilde x3^{2}}{1^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica