Sr Examen
3x1^2+x2^2-3/2x3^2+2*√3*x1x2 forma canónica
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Solución
Ha introducido
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2
2 2 3*x3 ___
x2 + 3*x1 - ----- + 2*x1*x2*\/ 3 = 0
2
$$3 x_{1}^{2} + 2 \sqrt{3} x_{1} x_{2} + x_{2}^{2} - \frac{3 x_{3}^{2}}{2} = 0$$
3*x1^2 + 2*sqrt(3)*x1*x2 + x2^2 - 3*x3^2/2 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$3 x_{1}^{2} + 2 \sqrt{3} x_{1} x_{2} + x_{2}^{2} - \frac{3 x_{3}^{2}}{2} = 0$$
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
$$a_{11} = - \frac{3}{2}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = \sqrt{3}$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = \frac{5}{2}$$
$$I_{3} = \left|\begin{matrix}- \frac{3}{2} & 0 & 0\\0 & 1 & \sqrt{3}\\0 & \sqrt{3} & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}- \frac{3}{2} & 0 & 0 & 0\\0 & 1 & \sqrt{3} & 0\\0 & \sqrt{3} & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - \frac{3}{2} & 0 & 0\\0 & 1 - \lambda & \sqrt{3}\\0 & \sqrt{3} & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = \frac{5}{2}$$
$$I_{2} = -6$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{5 \lambda^{2}}{2} + 6 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - \frac{5 \lambda^{2}}{2} - 6 \lambda = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = - \frac{3}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \frac{3 \tilde x2^{2}}{2} + 4 \tilde x3^{2} = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{\sqrt{6}}{3}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
es la ecuación para el tipo dos planos intersectantes
- está reducida a la forma canónica
$$3 x_{1}^{2} + 2 \sqrt{3} x_{1} x_{2} + x_{2}^{2} - \frac{3 x_{3}^{2}}{2} = 0$$
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
$$a_{11} = - \frac{3}{2}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = \sqrt{3}$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = \frac{5}{2}$$
| ___|
|-3/2 0| | 1 \/ 3 | |-3/2 0|
I2 = | | + | | + | |
| 0 1| | ___ | | 0 3|
|\/ 3 3 | $$I_{3} = \left|\begin{matrix}- \frac{3}{2} & 0 & 0\\0 & 1 & \sqrt{3}\\0 & \sqrt{3} & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}- \frac{3}{2} & 0 & 0 & 0\\0 & 1 & \sqrt{3} & 0\\0 & \sqrt{3} & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - \frac{3}{2} & 0 & 0\\0 & 1 - \lambda & \sqrt{3}\\0 & \sqrt{3} & 3 - \lambda\end{matrix}\right|$$
|-3/2 0| |1 0| |3 0|
K2 = | | + | | + | |
| 0 0| |0 0| |0 0| | ___ |
|-3/2 0 0| | 1 \/ 3 0| |-3/2 0 0|
| | | | | |
K3 = | 0 1 0| + | ___ | + | 0 3 0|
| | |\/ 3 3 0| | |
| 0 0 0| | | | 0 0 0|
| 0 0 0| $$I_{1} = \frac{5}{2}$$
$$I_{2} = -6$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{5 \lambda^{2}}{2} + 6 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - \frac{5 \lambda^{2}}{2} - 6 \lambda = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = - \frac{3}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$- \frac{3 \tilde x2^{2}}{2} + 4 \tilde x3^{2} = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{\sqrt{6}}{3}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{2}\right)^{2}} = 0$$
es la ecuación para el tipo dos planos intersectantes
- está reducida a la forma canónica