Mister Exam
x^2+7*y^2+2*z^2+8*x−20*z+200=0 canonical form
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2 2 2 200 + x - 20*z + 2*z + 7*y + 8*x = 0
$$x^{2} + 8 x + 7 y^{2} + 2 z^{2} - 20 z + 200 = 0$$
x^2 + 8*x + 7*y^2 + 2*z^2 - 20*z + 200 = 0
Invariants method
Given equation of the surface of 2-order:
$$x^{2} + 8 x + 7 y^{2} + 2 z^{2} - 20 z + 200 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 4$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 2$$
$$a_{34} = -10$$
$$a_{44} = 200$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 10$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 7 & 0\\0 & 0 & 2\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & 4\\0 & 7 & 0 & 0\\0 & 0 & 2 & -10\\4 & 0 & -10 & 200\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 7 - \lambda & 0\\0 & 0 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 10$$
$$I_{2} = 23$$
$$I_{3} = 14$$
$$I_{4} = 1876$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 10 \lambda^{2} - 23 \lambda + 14$$
$$K_{2} = 1884$$
$$K_{3} = 3656$$
Because
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
$$\lambda^{3} - 10 \lambda^{2} + 23 \lambda - 14 = 0$$
$$\lambda_{1} = 7$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 1$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$7 \tilde x^{2} + 2 \tilde y^{2} + \tilde z^{2} + 134 = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{1}{\frac{1}{134} \sqrt{134}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{7} \sqrt{7}}{\frac{1}{134} \sqrt{134}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{134} \sqrt{134}}\right)^{2}}\right) = -1$$
this equation is fora type imaginary ellipsoid
- reduced to canonical form
$$x^{2} + 8 x + 7 y^{2} + 2 z^{2} - 20 z + 200 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 4$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 2$$
$$a_{34} = -10$$
$$a_{44} = 200$$
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
$$I_{1} = 10$$
|1 0| |7 0| |1 0|
I2 = | | + | | + | |
|0 7| |0 2| |0 2|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 7 & 0\\0 & 0 & 2\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & 4\\0 & 7 & 0 & 0\\0 & 0 & 2 & -10\\4 & 0 & -10 & 200\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 7 - \lambda & 0\\0 & 0 & 2 - \lambda\end{matrix}\right|$$
|1 4 | |7 0 | | 2 -10|
K2 = | | + | | + | |
|4 200| |0 200| |-10 200| |1 0 4 | |7 0 0 | |1 0 4 |
| | | | | |
K3 = |0 7 0 | + |0 2 -10| + |0 2 -10|
| | | | | |
|4 0 200| |0 -10 200| |4 -10 200|$$I_{1} = 10$$
$$I_{2} = 23$$
$$I_{3} = 14$$
$$I_{4} = 1876$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 10 \lambda^{2} - 23 \lambda + 14$$
$$K_{2} = 1884$$
$$K_{3} = 3656$$
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
$$\lambda^{3} - 10 \lambda^{2} + 23 \lambda - 14 = 0$$
$$\lambda_{1} = 7$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 1$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$7 \tilde x^{2} + 2 \tilde y^{2} + \tilde z^{2} + 134 = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{1}{\frac{1}{134} \sqrt{134}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{7} \sqrt{7}}{\frac{1}{134} \sqrt{134}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{134} \sqrt{134}}\right)^{2}}\right) = -1$$
this equation is fora type imaginary ellipsoid
- reduced to canonical form