Mister Exam

# 6x^2+3y^2+3z^2-4xy-2yz+4xz canonical form

The teacher will be very surprised to see your correct solution 😉

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

#### Quality:

(Number of points on the axis)

### The solution

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   2      2      2
3*y  + 3*z  + 6*x  - 4*x*y - 2*y*z + 4*x*z = 0
$$6 x^{2} - 4 x y + 4 x z + 3 y^{2} - 2 y z + 3 z^{2} = 0$$
6*x^2 - 4*x*y + 4*x*z + 3*y^2 - 2*y*z + 3*z^2 = 0
Invariants method
Given equation of the surface of 2-order:
$$6 x^{2} - 4 x y + 4 x z + 3 y^{2} - 2 y z + 3 z^{2} = 0$$
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
$$a_{11} = 6$$
$$a_{12} = -2$$
$$a_{13} = 2$$
$$a_{14} = 0$$
$$a_{22} = 3$$
$$a_{23} = -1$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 12$$
     |6   -2|   |3   -1|   |6  2|
I2 = |      | + |      | + |    |
|-2  3 |   |-1  3 |   |2  3|

$$I_{3} = \left|\begin{matrix}6 & -2 & 2\\-2 & 3 & -1\\2 & -1 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}6 & -2 & 2 & 0\\-2 & 3 & -1 & 0\\2 & -1 & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}6 - \lambda & -2 & 2\\-2 & 3 - \lambda & -1\\2 & -1 & 3 - \lambda\end{matrix}\right|$$
     |6  0|   |3  0|   |3  0|
K2 = |    | + |    | + |    |
|0  0|   |0  0|   |0  0|

     |6   -2  0|   |3   -1  0|   |6  2  0|
|         |   |         |   |       |
K3 = |-2  3   0| + |-1  3   0| + |2  3  0|
|         |   |         |   |       |
|0   0   0|   |0   0   0|   |0  0  0|

$$I_{1} = 12$$
$$I_{2} = 36$$
$$I_{3} = 32$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 12 \lambda^{2} - 36 \lambda + 32$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
I3 != 0

then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - 12 \lambda^{2} + 36 \lambda - 32 = 0$$
Solve this equation
$$\lambda_{1} = 8$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 2$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$8 \tilde x^{2} + 2 \tilde y^{2} + 2 \tilde z^{2} = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{2}}{4}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}}\right) = 0$$
this equation is fora type imaginary cone
- reduced to canonical form