Mister Exam
x^2−4x+9y^2+18y−4z^2+24z+49=0 canonical form
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2 2 2 49 + x - 4*x - 4*z + 9*y + 18*y + 24*z = 0
$$x^{2} + 9 y^{2} - 4 z^{2} - 4 x + 18 y + 24 z + 49 = 0$$
x^2 - 4*x + 9*y^2 + 18*y - 4*z^2 + 24*z + 49 = 0
Invariants method
Given equation of the surface of 2-order:
$$x^{2} + 9 y^{2} - 4 z^{2} - 4 x + 18 y + 24 z + 49 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -2$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{24} = 9$$
$$a_{33} = -4$$
$$a_{34} = 12$$
$$a_{44} = 49$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 9 & 0\\0 & 0 & -4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & -2\\0 & 9 & 0 & 9\\0 & 0 & -4 & 12\\-2 & 9 & 12 & 49\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 1 & 0 & 0\\0 & - \lambda + 9 & 0\\0 & 0 & - \lambda - 4\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = -31$$
$$I_{3} = -36$$
$$I_{4} = -2592$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} + 31 \lambda - 36$$
$$K_{2} = 65$$
$$K_{3} = -2736$$
Because
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
$$\lambda^{3} - 6 \lambda^{2} - 31 \lambda + 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -4$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + \tilde y^{2} - 4 \tilde z^{2} + 72 = 0$$
$$- \frac{\tilde z^{2}}{\left(3 \sqrt{2}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(2 \sqrt{2}\right)^{2}} + \frac{\tilde y^{2}}{\left(6 \sqrt{2}\right)^{2}}\right) = -1$$
this equation is fora type two-sided hyperboloid
- reduced to canonical form
$$x^{2} + 9 y^{2} - 4 z^{2} - 4 x + 18 y + 24 z + 49 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -2$$
$$a_{22} = 9$$
$$a_{23} = 0$$
$$a_{24} = 9$$
$$a_{33} = -4$$
$$a_{34} = 12$$
$$a_{44} = 49$$
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
$$I_{1} = 6$$
|1 0| |9 0 | |1 0 |
I2 = | | + | | + | |
|0 9| |0 -4| |0 -4|$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 9 & 0\\0 & 0 & -4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & -2\\0 & 9 & 0 & 9\\0 & 0 & -4 & 12\\-2 & 9 & 12 & 49\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 1 & 0 & 0\\0 & - \lambda + 9 & 0\\0 & 0 & - \lambda - 4\end{matrix}\right|$$
|1 -2| |9 9 | |-4 12|
K2 = | | + | | + | |
|-2 49| |9 49| |12 49| |1 0 -2| |9 0 9 | |1 0 -2|
| | | | | |
K3 = |0 9 9 | + |0 -4 12| + |0 -4 12|
| | | | | |
|-2 9 49| |9 12 49| |-2 12 49|$$I_{1} = 6$$
$$I_{2} = -31$$
$$I_{3} = -36$$
$$I_{4} = -2592$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} + 31 \lambda - 36$$
$$K_{2} = 65$$
$$K_{3} = -2736$$
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
$$\lambda^{3} - 6 \lambda^{2} - 31 \lambda + 36 = 0$$
Solve this equation
$$\lambda_{1} = 9$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -4$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + \tilde y^{2} - 4 \tilde z^{2} + 72 = 0$$
$$- \frac{\tilde z^{2}}{\left(3 \sqrt{2}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(2 \sqrt{2}\right)^{2}} + \frac{\tilde y^{2}}{\left(6 \sqrt{2}\right)^{2}}\right) = -1$$
this equation is fora type two-sided hyperboloid
- reduced to canonical form