4x1x2+x2^2+4x2x3+2x3^2-4x1-2x2-5=0 canonical form

Given equation of the surface of 2-order:
$$4 x_{1} x_{2} + x_{2}^{2} + 4 x_{2} x_{3} + 2 x_{3}^{2} - 4 x_{1} - 2 x_{2} - 5 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 2$$
$$a_{24} = -1$$
$$a_{33} = 0$$
$$a_{34} = -2$$
$$a_{44} = -5$$
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
$$I_{1} = 3$$

     |2  2|   |1  2|   |2  0|
I2 = |    | + |    | + |    |
     |2  1|   |2  0|   |0  0|

$$I_{3} = \left|\begin{matrix}2 & 2 & 0\\2 & 1 & 2\\0 & 2 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 2 & 0 & 0\\2 & 1 & 2 & -1\\0 & 2 & 0 & -2\\0 & -1 & -2 & -5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 2 & 2 & 0\\2 & - \lambda + 1 & 2\\0 & 2 & - \lambda\end{matrix}\right|$$

     |2  0 |   |1   -1|   |0   -2|
K2 = |     | + |      | + |      |
     |0  -5|   |-1  -5|   |-2  -5|

     |2  2   0 |   |1   2   -1|   |2  0   0 |
     |         |   |          |   |         |
K3 = |2  1   -1| + |2   0   -2| + |0  0   -2|
     |         |   |          |   |         |
     |0  -1  -5|   |-1  -2  -5|   |0  -2  -5|

$$I_{1} = 3$$
$$I_{2} = -6$$
$$I_{3} = -8$$
$$I_{4} = 64$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} + 6 \lambda - 8$$
$$K_{2} = -20$$
$$K_{3} = 24$$
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
$$\lambda^{3} - 3 \lambda^{2} - 6 \lambda + 8 = 0$$
Solve this equation
$$\lambda_{1} = 4$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -2$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- 2 \tilde x1^{2} + \tilde x2^{2} + 4 \tilde x3^{2} - 8 = 0$$
$$- \frac{\tilde x1^{2}}{2^{2}} + \left(\frac{\tilde x2^{2}}{\left(2 \sqrt{2}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\sqrt{2}\right)^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form