Given equation of the surface of 2-order:
$$3 x^{2} + 2 x y + 2 x z + 5 y^{2} + 2 y z + 3 z^{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + a_{22} y^{2} + 2 a_{23} y z + a_{33} z^{2} + 2 a_{14} x + 2 a_{24} y + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 3$$
$$a_{12} = 1$$
$$a_{13} = 1$$
$$a_{14} = 0$$
$$a_{22} = 5$$
$$a_{23} = 1$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 11$$
|3 1| |5 1| |3 1|
I2 = | | + | | + | |
|1 5| |1 3| |1 3|
$$I_{3} = \left|\begin{matrix}3 & 1 & 1\\1 & 5 & 1\\1 & 1 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & 1 & 1 & 0\\1 & 5 & 1 & 0\\1 & 1 & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 3 & 1 & 1\\1 & - \lambda + 5 & 1\\1 & 1 & - \lambda + 3\end{matrix}\right|$$
|3 0| |5 0| |3 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0|
|3 1 0| |5 1 0| |3 1 0|
| | | | | |
K3 = |1 5 0| + |1 3 0| + |1 3 0|
| | | | | |
|0 0 0| |0 0 0| |0 0 0|
$$I_{1} = 11$$
$$I_{2} = 36$$
$$I_{3} = 36$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 11 \lambda^{2} - 36 \lambda + 36$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - 11 \lambda^{2} + 36 \lambda - 36 = 0$$
Solve this equation$$\lambda_{1} = 6$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 2$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$6 \tilde x^{2} + 3 \tilde y^{2} + 2 \tilde z^{2} = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
this equation is fora type imaginary cone
- reduced to canonical form