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8x^2+7y^2+3z^2-12xy+4xz-8yz canonical form

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   2      2      2                             
3*z  + 7*y  + 8*x  - 12*x*y - 8*y*z + 4*x*z = 0
$$8 x^{2} - 12 x y + 4 x z + 7 y^{2} - 8 y z + 3 z^{2} = 0$$
8*x^2 - 12*x*y + 4*x*z + 7*y^2 - 8*y*z + 3*z^2 = 0
Invariants method
Given equation of the surface of 2-order:
$$8 x^{2} - 12 x y + 4 x z + 7 y^{2} - 8 y z + 3 z^{2} = 0$$
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
$$a_{11} = 8$$
$$a_{12} = -6$$
$$a_{13} = 2$$
$$a_{14} = 0$$
$$a_{22} = 7$$
$$a_{23} = -4$$
$$a_{24} = 0$$
$$a_{33} = 3$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 18$$
     |8   -6|   |7   -4|   |8  2|
I2 = |      | + |      | + |    |
     |-6  7 |   |-4  3 |   |2  3|

$$I_{3} = \left|\begin{matrix}8 & -6 & 2\\-6 & 7 & -4\\2 & -4 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}8 & -6 & 2 & 0\\-6 & 7 & -4 & 0\\2 & -4 & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 8 & -6 & 2\\-6 & - \lambda + 7 & -4\\2 & -4 & - \lambda + 3\end{matrix}\right|$$
     |8  0|   |7  0|   |3  0|
K2 = |    | + |    | + |    |
     |0  0|   |0  0|   |0  0|

     |8   -6  0|   |7   -4  0|   |8  2  0|
     |         |   |         |   |       |
K3 = |-6  7   0| + |-4  3   0| + |2  3  0|
     |         |   |         |   |       |
     |0   0   0|   |0   0   0|   |0  0  0|

$$I_{1} = 18$$
$$I_{2} = 45$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 18 \lambda^{2} - 45 \lambda$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 18 \lambda^{2} + 45 \lambda = 0$$
Solve this equation
$$\lambda_{1} = 15$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$15 \tilde x^{2} + 3 \tilde y^{2} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\sqrt{15}}{15}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0$$
this equation is fora type two imaginary intersecting planes
- reduced to canonical form