5x^2+4xy-10xz-2y^2+4yz+5z^2-8x+2y-8z+1=0 canonical form

Given equation of the surface of 2-order:
$$5 x^{2} + 4 x y - 10 x z - 8 x - 2 y^{2} + 4 y z + 2 y + 5 z^{2} - 8 z + 1 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = -5$$
$$a_{14} = -4$$
$$a_{22} = -2$$
$$a_{23} = 2$$
$$a_{24} = 1$$
$$a_{33} = 5$$
$$a_{34} = -4$$
$$a_{44} = 1$$
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
$$I_{1} = 8$$

     |5  2 |   |-2  2|   |5   -5|
I2 = |     | + |     | + |      |
     |2  -2|   |2   5|   |-5  5 |

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

     |5   -4|   |-2  1|   |5   -4|
K2 = |      | + |     | + |      |
     |-4  1 |   |1   1|   |-4  1 |

     |5   2   -4|   |-2  2   1 |   |5   -5  -4|
     |          |   |          |   |          |
K3 = |2   -2  1 | + |2   5   -4| + |-5  5   -4|
     |          |   |          |   |          |
     |-4  1   1 |   |1   -4  1 |   |-4  -4  1 |

$$I_{1} = 8$$
$$I_{2} = -28$$
$$I_{3} = -80$$
$$I_{4} = 240$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 8 \lambda^{2} + 28 \lambda - 80$$
$$K_{2} = -25$$
$$K_{3} = -326$$
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
$$\lambda^{3} - 8 \lambda^{2} - 28 \lambda + 80 = 0$$
Solve this equation
$$\lambda_{1} = 10$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = -4$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$10 \tilde x^{2} + 2 \tilde y^{2} - 4 \tilde z^{2} - 3 = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\sqrt{3}}{2}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{30}}{10}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{6}}{2}\right)^{2}}\right) = 1$$
this equation is fora type one-sided hyperboloid
- reduced to canonical form