Mister Exam

# x^2+y^2+4z^2−2xy+4xz−4yz−2x+2y+2z=0 canonical form

x: [, ]
y: [, ]
z: [, ]

#### Quality:

(Number of points on the axis)

### The solution

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 2    2                        2
x  + y  - 2*x + 2*y + 2*z + 4*z  - 4*y*z - 2*x*y + 4*x*z = 0
$$x^{2} - 2 x y + 4 x z - 2 x + y^{2} - 4 y z + 2 y + 4 z^{2} + 2 z = 0$$
x^2 - 2*x*y + 4*x*z - 2*x + y^2 - 4*y*z + 2*y + 4*z^2 + 2*z = 0
Invariants method
Given equation of the surface of 2-order:
$$x^{2} - 2 x y + 4 x z - 2 x + y^{2} - 4 y z + 2 y + 4 z^{2} + 2 z = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = -1$$
$$a_{13} = 2$$
$$a_{14} = -1$$
$$a_{22} = 1$$
$$a_{23} = -2$$
$$a_{24} = 1$$
$$a_{33} = 4$$
$$a_{34} = 1$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
     |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|

substitute coefficients
$$I_{1} = 6$$
     |1   -1|   |1   -2|   |1  2|
I2 = |      | + |      | + |    |
|-1  1 |   |-2  4 |   |2  4|

$$I_{3} = \left|\begin{matrix}1 & -1 & 2\\-1 & 1 & -2\\2 & -2 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & -1 & 2 & -1\\-1 & 1 & -2 & 1\\2 & -2 & 4 & 1\\-1 & 1 & 1 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -1 & 2\\-1 & 1 - \lambda & -2\\2 & -2 & 4 - \lambda\end{matrix}\right|$$
     |1   -1|   |1  1|   |4  1|
K2 = |      | + |    | + |    |
|-1  0 |   |1  0|   |1  0|

     |1   -1  -1|   |1   -2  1|   |1   2  -1|
|          |   |         |   |         |
K3 = |-1  1   1 | + |-2  4   1| + |2   4  1 |
|          |   |         |   |         |
|-1  1   0 |   |1   1   0|   |-1  1  0 |

$$I_{1} = 6$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2}$$
$$K_{2} = -3$$
$$K_{3} = -18$$
Because
$$I_{2} = 0 \wedge I_{3} = 0 \wedge I_{4} = 0 \wedge I_{1} \neq 0 \wedge K_{3} \neq 0$$
then by type of surface:
you need to
then the canonical form of the equation will be
$$I_{1} \tilde x^{2} + \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
and
$$I_{1} \tilde x^{2} - \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
$$6 \tilde x^{2} + 2 \sqrt{3} \tilde y = 0$$
and
$$6 \tilde x^{2} - 2 \sqrt{3} \tilde y = 0$$
$$\tilde x^{2} = \frac{\sqrt{3}}{3} \tilde y$$
and
$$\tilde x^{2} = - \frac{\sqrt{3}}{3} \tilde y$$
this equation is fora type parabolic cylinder
- reduced to canonical form
To see a detailed solution - share to all your student friends
To see a detailed solution,
share to all your student friends: