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25x^2+9y^2-9z^2=225 canonical form

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          2      2       2    
-225 - 9*z  + 9*y  + 25*x  = 0
$$25 x^{2} + 9 y^{2} - 9 z^{2} - 225 = 0$$
25*x^2 + 9*y^2 - 9*z^2 - 225 = 0
Invariants method
Given equation of the surface of 2-order:
$$25 x^{2} + 9 y^{2} - 9 z^{2} - 225 = 0$$
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
$$a_{11} = 25$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = -9$$
$$a_{34} = 0$$
$$a_{44} = -225$$
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
$$I_{1} = 25$$
     |25  0|   |9  0 |   |25  0 |
I2 = |     | + |     | + |      |
     |0   9|   |0  -9|   |0   -9|

$$I_{3} = \left|\begin{matrix}25 & 0 & 0\\0 & 9 & 0\\0 & 0 & -9\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}25 & 0 & 0 & 0\\0 & 9 & 0 & 0\\0 & 0 & -9 & 0\\0 & 0 & 0 & -225\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}25 - \lambda & 0 & 0\\0 & 9 - \lambda & 0\\0 & 0 & - \lambda - 9\end{matrix}\right|$$
     |25   0  |   |9   0  |   |-9   0  |
K2 = |        | + |       | + |        |
     |0   -225|   |0  -225|   |0   -225|

     |25  0   0  |   |9  0    0  |   |25  0    0  |
     |           |   |           |   |            |
K3 = |0   9   0  | + |0  -9   0  | + |0   -9   0  |
     |           |   |           |   |            |
     |0   0  -225|   |0  0   -225|   |0   0   -225|

$$I_{1} = 25$$
$$I_{2} = -81$$
$$I_{3} = -2025$$
$$I_{4} = 455625$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 25 \lambda^{2} + 81 \lambda - 2025$$
$$K_{2} = -5625$$
$$K_{3} = 18225$$
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
$$\lambda^{3} - 25 \lambda^{2} - 81 \lambda + 2025 = 0$$
$$\lambda_{1} = 25$$
$$\lambda_{2} = 9$$
$$\lambda_{3} = -9$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$25 \tilde x^{2} + 9 \tilde y^{2} - 9 \tilde z^{2} - 225 = 0$$
        2           2           2    
\tilde x    \tilde y    \tilde z     
--------- + --------- - --------- = 1
        2           2           2    
/  1   \    /  1   \    /  1   \     
|------|    |------|    |------|     
\5*1/15/    \3*1/15/    \3*1/15/     

this equation is fora type one-sided hyperboloid
- reduced to canonical form