(x^2)/25-(y^2)/16+(z^2)/4=-1 canonical form

Given equation of the surface of 2-order:
$$\frac{x^{2}}{25} - \frac{y^{2}}{16} + \frac{z^{2}}{4} + 1 = 0$$
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
$$a_{11} = \frac{1}{25}$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = - \frac{1}{16}$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = \frac{1}{4}$$
$$a_{34} = 0$$
$$a_{44} = 1$$
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
$$I_{1} = \frac{91}{400}$$

     |1/25    0  |   |-1/16   0 |   |1/25   0 |
I2 = |           | + |          | + |         |
     | 0    -1/16|   |  0    1/4|   | 0    1/4|

$$I_{3} = \left|\begin{matrix}\frac{1}{25} & 0 & 0\\0 & - \frac{1}{16} & 0\\0 & 0 & \frac{1}{4}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}\frac{1}{25} & 0 & 0 & 0\\0 & - \frac{1}{16} & 0 & 0\\0 & 0 & \frac{1}{4} & 0\\0 & 0 & 0 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + \frac{1}{25} & 0 & 0\\0 & - \lambda - \frac{1}{16} & 0\\0 & 0 & - \lambda + \frac{1}{4}\end{matrix}\right|$$

     |1/25  0|   |-1/16  0|   |1/4  0|
K2 = |       | + |        | + |      |
     | 0    1|   |  0    1|   | 0   1|

     |1/25    0    0|   |-1/16   0   0|   |1/25   0   0|
     |              |   |             |   |            |
K3 = | 0    -1/16  0| + |  0    1/4  0| + | 0    1/4  0|
     |              |   |             |   |            |
     | 0      0    1|   |  0     0   1|   | 0     0   1|

$$I_{1} = \frac{91}{400}$$
$$I_{2} = - \frac{13}{1600}$$
$$I_{3} = - \frac{1}{1600}$$
$$I_{4} = - \frac{1}{1600}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{91 \lambda^{2}}{400} + \frac{13 \lambda}{1600} - \frac{1}{1600}$$
$$K_{2} = \frac{91}{400}$$
$$K_{3} = - \frac{13}{1600}$$
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
$$\lambda^{3} - \frac{91 \lambda^{2}}{400} - \frac{13 \lambda}{1600} + \frac{1}{1600} = 0$$
Solve this equation
$$\lambda_{1} = \frac{1}{4}$$
$$\lambda_{2} = - \frac{1}{16}$$
$$\lambda_{3} = \frac{1}{25}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{16} + \frac{\tilde z^{2}}{25} + 1 = 0$$
$$- \frac{\tilde y^{2}}{4^{2}} + \left(\frac{\tilde x^{2}}{2^{2}} + \frac{\tilde z^{2}}{5^{2}}\right) = -1$$
this equation is fora type two-sided hyperboloid
- reduced to canonical form