Mister Exam
-2x^2-y^2+8z^2+6x+4y+2z-9=0 canonical form
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2 2 2 -9 - y - 2*x + 2*z + 4*y + 6*x + 8*z = 0
$$- 2 x^{2} + 6 x - y^{2} + 4 y + 8 z^{2} + 2 z - 9 = 0$$
-2*x^2 + 6*x - y^2 + 4*y + 8*z^2 + 2*z - 9 = 0
Invariants method
Given equation of the surface of 2-order:
$$- 2 x^{2} + 6 x - y^{2} + 4 y + 8 z^{2} + 2 z - 9 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 3$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{24} = 2$$
$$a_{33} = 8$$
$$a_{34} = 1$$
$$a_{44} = -9$$
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
$$I_{1} = 5$$
$$I_{3} = \left|\begin{matrix}-2 & 0 & 0\\0 & -1 & 0\\0 & 0 & 8\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-2 & 0 & 0 & 3\\0 & -1 & 0 & 2\\0 & 0 & 8 & 1\\3 & 2 & 1 & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & 0 & 0\\0 & - \lambda - 1 & 0\\0 & 0 & 8 - \lambda\end{matrix}\right|$$
$$I_{1} = 5$$
$$I_{2} = -22$$
$$I_{3} = 16$$
$$I_{4} = -10$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 5 \lambda^{2} + 22 \lambda + 16$$
$$K_{2} = -59$$
$$K_{3} = 114$$
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
$$\lambda^{3} - 5 \lambda^{2} - 22 \lambda - 16 = 0$$
$$\lambda_{1} = 8$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -2$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$8 \tilde x^{2} - \tilde y^{2} - 2 \tilde z^{2} - \frac{5}{8} = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{\frac{1}{4} \sqrt{2}}{\frac{2}{5} \sqrt{10}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{5} \sqrt{10}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{2}{5} \sqrt{10}}\right)^{2}}\right) = -1$$
this equation is fora type two-sided hyperboloid
- reduced to canonical form
$$- 2 x^{2} + 6 x - y^{2} + 4 y + 8 z^{2} + 2 z - 9 = 0$$
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
$$a_{11} = -2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 3$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{24} = 2$$
$$a_{33} = 8$$
$$a_{34} = 1$$
$$a_{44} = -9$$
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
$$I_{1} = 5$$
|-2 0 | |-1 0| |-2 0|
I2 = | | + | | + | |
|0 -1| |0 8| |0 8|$$I_{3} = \left|\begin{matrix}-2 & 0 & 0\\0 & -1 & 0\\0 & 0 & 8\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-2 & 0 & 0 & 3\\0 & -1 & 0 & 2\\0 & 0 & 8 & 1\\3 & 2 & 1 & -9\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 2 & 0 & 0\\0 & - \lambda - 1 & 0\\0 & 0 & 8 - \lambda\end{matrix}\right|$$
|-2 3 | |-1 2 | |8 1 |
K2 = | | + | | + | |
|3 -9| |2 -9| |1 -9| |-2 0 3 | |-1 0 2 | |-2 0 3 |
| | | | | |
K3 = |0 -1 2 | + |0 8 1 | + |0 8 1 |
| | | | | |
|3 2 -9| |2 1 -9| |3 1 -9|$$I_{1} = 5$$
$$I_{2} = -22$$
$$I_{3} = 16$$
$$I_{4} = -10$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 5 \lambda^{2} + 22 \lambda + 16$$
$$K_{2} = -59$$
$$K_{3} = 114$$
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
$$\lambda^{3} - 5 \lambda^{2} - 22 \lambda - 16 = 0$$
$$\lambda_{1} = 8$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -2$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$8 \tilde x^{2} - \tilde y^{2} - 2 \tilde z^{2} - \frac{5}{8} = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{\frac{1}{4} \sqrt{2}}{\frac{2}{5} \sqrt{10}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{1}{\frac{2}{5} \sqrt{10}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{2}{5} \sqrt{10}}\right)^{2}}\right) = -1$$
this equation is fora type two-sided hyperboloid
- reduced to canonical form