Sr Examen
x^2-2*x*y+2*x*z+y*z+z^2=0 forma canónica
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Solución
Ha introducido
[src]
2 2 x + z + y*z - 2*x*y + 2*x*z = 0
$$x^{2} - 2 x y + 2 x z + y z + z^{2} = 0$$
x^2 - 2*x*y + 2*x*z + y*z + z^2 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$x^{2} - 2 x y + 2 x z + y z + z^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -1$$
$$a_{13} = 1$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 2$$
$$I_{3} = \left|\begin{matrix}1 & -1 & 1\\-1 & 0 & \frac{1}{2}\\1 & \frac{1}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & -1 & 1 & 0\\-1 & 0 & \frac{1}{2} & 0\\1 & \frac{1}{2} & 1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -1 & 1\\-1 & - \lambda & \frac{1}{2}\\1 & \frac{1}{2} & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 2$$
$$I_{2} = - \frac{5}{4}$$
$$I_{3} = - \frac{9}{4}$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + \frac{5 \lambda}{4} - \frac{9}{4}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 2 \lambda^{2} - \frac{5 \lambda}{4} + \frac{9}{4} = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = \frac{1}{2} - \frac{\sqrt{10}}{2}$$
$$\lambda_{3} = \frac{1}{2} + \frac{\sqrt{10}}{2}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + \tilde y^{2} \left(\frac{1}{2} - \frac{\sqrt{10}}{2}\right) + \tilde z^{2} \left(\frac{1}{2} + \frac{\sqrt{10}}{2}\right) = 0$$
$$- \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{- \frac{1}{2} + \frac{\sqrt{10}}{2}}}\right)^{2}} + \left(\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{\sqrt{\frac{1}{2} + \frac{\sqrt{10}}{2}}}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono
- está reducida a la forma canónica
$$x^{2} - 2 x y + 2 x z + y z + z^{2} = 0$$
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
$$a_{11} = 1$$
$$a_{12} = -1$$
$$a_{13} = 1$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 2$$
|1 -1| | 0 1/2| |1 1|
I2 = | | + | | + | |
|-1 0 | |1/2 1 | |1 1|$$I_{3} = \left|\begin{matrix}1 & -1 & 1\\-1 & 0 & \frac{1}{2}\\1 & \frac{1}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & -1 & 1 & 0\\-1 & 0 & \frac{1}{2} & 0\\1 & \frac{1}{2} & 1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -1 & 1\\-1 & - \lambda & \frac{1}{2}\\1 & \frac{1}{2} & 1 - \lambda\end{matrix}\right|$$
|1 0| |0 0| |1 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |1 -1 0| | 0 1/2 0| |1 1 0|
| | | | | |
K3 = |-1 0 0| + |1/2 1 0| + |1 1 0|
| | | | | |
|0 0 0| | 0 0 0| |0 0 0|$$I_{1} = 2$$
$$I_{2} = - \frac{5}{4}$$
$$I_{3} = - \frac{9}{4}$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} + \frac{5 \lambda}{4} - \frac{9}{4}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 2 \lambda^{2} - \frac{5 \lambda}{4} + \frac{9}{4} = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = \frac{1}{2} - \frac{\sqrt{10}}{2}$$
$$\lambda_{3} = \frac{1}{2} + \frac{\sqrt{10}}{2}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + \tilde y^{2} \left(\frac{1}{2} - \frac{\sqrt{10}}{2}\right) + \tilde z^{2} \left(\frac{1}{2} + \frac{\sqrt{10}}{2}\right) = 0$$
$$- \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{- \frac{1}{2} + \frac{\sqrt{10}}{2}}}\right)^{2}} + \left(\frac{\tilde x^{2}}{1^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{\sqrt{\frac{1}{2} + \frac{\sqrt{10}}{2}}}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono
- está reducida a la forma canónica