2x1^2+4x2^2+2x3^2-2x1x2-2x1x3+2x2×x3 canonical form

Given equation of the surface of 2-order:
$$2 x_{1}^{2} - 2 x_{1} x_{2} - 2 x_{1} x_{3} + 4 x_{2}^{2} + 2 x_{2} x_{3} + 2 x_{3}^{2} = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 1$$
$$a_{13} = -1$$
$$a_{14} = 0$$
$$a_{22} = 4$$
$$a_{23} = -1$$
$$a_{24} = 0$$
$$a_{33} = 2$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 8$$

     |2  1|   |4   -1|   |2   -1|
I2 = |    | + |      | + |      |
     |1  4|   |-1  2 |   |-1  2 |

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

     |2  0|   |4  0|   |2  0|
K2 = |    | + |    | + |    |
     |0  0|   |0  0|   |0  0|

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

$$I_{1} = 8$$
$$I_{2} = 17$$
$$I_{3} = 10$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 8 \lambda^{2} - 17 \lambda + 10$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 8 \lambda^{2} + 17 \lambda - 10 = 0$$
Solve this equation
$$\lambda_{1} = 5$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 1$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x1^{2} + 2 \tilde x2^{2} + 5 \tilde x3^{2} = 0$$
$$\frac{\tilde x1^{2}}{1^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{\sqrt{5}}{5}\right)^{2}}\right) = 0$$
this equation is fora type imaginary cone
- reduced to canonical form