Mister Exam

# 2x1x2-2x1x3-2x2x3=0 canonical form

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

v

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

#### Quality:

(Number of points on the axis)

### The solution

You have entered [src]
-2*x1*x3 - 2*x2*x3 + 2*x1*x2 = 0
$$2 x_{1} x_{2} - 2 x_{1} x_{3} - 2 x_{2} x_{3} = 0$$
2*x1*x2 - 2*x1*x3 - 2*x2*x3 = 0
Invariants method
Given equation of the surface of 2-order:
$$2 x_{1} x_{2} - 2 x_{1} x_{3} - 2 x_{2} x_{3} = 0$$
This equation looks like:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + a_{33} x_{1}^{2} + 2 a_{14} x_{3} + 2 a_{24} x_{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = -1$$
$$a_{13} = -1$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 1$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
     |0   -1|   |0  1|   |0   -1|
I2 = |      | + |    | + |      |
|-1  0 |   |1  0|   |-1  0 |

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

     |0   -1  0|   |0  1  0|   |0   -1  0|
|         |   |       |   |         |
K3 = |-1  0   0| + |1  0  0| + |-1  0   0|
|         |   |       |   |         |
|0   0   0|   |0  0  0|   |0   0   0|

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

then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + \lambda^{3} + I_{2} \lambda - I_{3} = 0$$
or
$$\lambda^{3} - 3 \lambda - 2 = 0$$
Solve this equation
$$\lambda_{1} = 2$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -1$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \tilde x1^{2} - \tilde x2^{2} + 2 \tilde x3^{2} = 0$$
$$- \frac{\tilde x3^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \left(\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{1^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form