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Forma canónica/ 2x^2+2y^2-z^2-2xy+4yz+4xz-22x-4y+2z+16=0

2x^2+2y^2-z^2-2xy+4yz+4xz-22x-4y+2z+16=0 forma canónica

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Solución

Ha introducido [src] 
      2                         2      2                            
16 - z  - 22*x - 4*y + 2*z + 2*x  + 2*y  - 2*x*y + 4*x*z + 4*y*z = 0
2x2−2xy+4xz−22x+2y2+4yz−4y−z2+2z+16=02 x^{2} - 2 x y + 4 x z - 22 x + 2 y^{2} + 4 y z - 4 y - z^{2} + 2 z + 16 = 0 
2*x^2 - 2*x*y + 4*x*z - 22*x + 2*y^2 + 4*y*z - 4*y - z^2 + 2*z + 16 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
2x2−2xy+4xz−22x+2y2+4yz−4y−z2+2z+16=02 x^{2} - 2 x y + 4 x z - 22 x + 2 y^{2} + 4 y z - 4 y - z^{2} + 2 z + 16 = 0
Esta ecuación tiene la forma:
a11x2+2a12xy+2a13xz+2a14x+a22y2+2a23yz+2a24y+a33z2+2a34z+a44=0a_{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
a11=2a_{11} = 2
a12=−1a_{12} = -1
a13=2a_{13} = 2
a14=−11a_{14} = -11
a22=2a_{22} = 2
a23=2a_{23} = 2
a24=−2a_{24} = -2
a33=−1a_{33} = -1
a34=1a_{34} = 1
a44=16a_{44} = 16
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
I1=a11+a22+a33I_{1} = a_{11} + a_{22} + a_{33}
     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

I3=∣a11a12a13a12a22a23a13a23a33∣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|
I4=∣a11a12a13a14a12a22a23a24a13a23a33a34a14a24a34a44∣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(λ)=∣a11−λa12a13a12a22−λa23a13a23a33−λ∣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
I1=3I_{1} = 3
     |2   -1|   |2  2 |   |2  2 |
I2 = |      | + |     | + |     |
     |-1  2 |   |2  -1|   |2  -1|

I3=∣2−12−12222−1∣I_{3} = \left|\begin{matrix}2 & -1 & 2\\-1 & 2 & 2\\2 & 2 & -1\end{matrix}\right|
I4=∣2−12−11−122−222−11−11−2116∣I_{4} = \left|\begin{matrix}2 & -1 & 2 & -11\\-1 & 2 & 2 & -2\\2 & 2 & -1 & 1\\-11 & -2 & 1 & 16\end{matrix}\right|
I(λ)=∣2−λ−12−12−λ222−λ−1∣I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & -1 & 2\\-1 & 2 - \lambda & 2\\2 & 2 & - \lambda - 1\end{matrix}\right|
     | 2   -11|   |2   -2|   |-1  1 |
K2 = |        | + |      | + |      |
     |-11  16 |   |-2  16|   |1   16|

     | 2   -1  -11|   |2   2   -2|   | 2   2   -11|
     |            |   |          |   |            |
K3 = |-1   2   -2 | + |2   -1  1 | + | 2   -1   1 |
     |            |   |          |   |            |
     |-11  -2  16 |   |-2  1   16|   |-11  1   16 |

I1=3I_{1} = 3
I2=−9I_{2} = -9
I3=−27I_{3} = -27
I4=27I_{4} = 27
I(λ)=−λ3+3λ2+9λ−27I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} + 9 \lambda - 27
K2=−78K_{2} = -78
K3=−369K_{3} = -369
Como
I3 != 0

entonces por razón de tipos de rectas:
hay que
Formulamos la ecuación característica para nuestra superficie:
−I1λ2+I2λ−I3+λ3=0- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0
o
λ3−3λ2−9λ+27=0\lambda^{3} - 3 \lambda^{2} - 9 \lambda + 27 = 0
λ1=−3\lambda_{1} = -3
λ2=3\lambda_{2} = 3
λ3=3\lambda_{3} = 3
entonces la forma canónica de la ecuación será
(z~2λ3+(x~2λ1+y~2λ2))+I4I3=0\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
−3x~2+3y~2+3z~2−1=0- 3 \tilde x^{2} + 3 \tilde y^{2} + 3 \tilde z^{2} - 1 = 0
−x~2(1331)2+(y~2(1331)2+z~2(1331)2)=1- \frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{1}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{1}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{1}\right)^{2}}\right) = 1
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica
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