Given equation of the surface of 2-order:
$$x^{2} - 2 x - y^{2} - z^{2} + z = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -1$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = -1$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = -1$$
|1 0 | |-1 0 | |1 0 |
I2 = | | + | | + | |
|0 -1| |0 -1| |0 -1|
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & -1\\0 & -1 & 0 & 0\\0 & 0 & -1 & \frac{1}{2}\\-1 & 0 & \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & - \lambda - 1 & 0\\0 & 0 & - \lambda - 1\end{matrix}\right|$$
|1 -1| |-1 0| |-1 1/2|
K2 = | | + | | + | |
|-1 0 | |0 0| |1/2 0 |
|1 0 -1| |-1 0 0 | |1 0 -1 |
| | | | | |
K3 = |0 -1 0 | + |0 -1 1/2| + |0 -1 1/2|
| | | | | |
|-1 0 0 | |0 1/2 0 | |-1 1/2 0 |
$$I_{1} = -1$$
$$I_{2} = -1$$
$$I_{3} = 1$$
$$I_{4} = - \frac{3}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \lambda^{2} + \lambda + 1$$
$$K_{2} = - \frac{5}{4}$$
$$K_{3} = 2$$
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
$$\lambda^{3} + \lambda^{2} - \lambda - 1 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -1$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} - \tilde y^{2} - \tilde z^{2} - \frac{3}{4} = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{1}{\frac{2}{3} \sqrt{3}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{3} \sqrt{3}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{1}{\frac{2}{3} \sqrt{3}}\right)^{2}}\right) = -1$$
this equation is fora type two-sided hyperboloid
- reduced to canonical form