Mister Exam
5x^2+6y^2+7z^2-4xy+4yz-10x+8y+14z-6=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 2 -6 - 10*x + 5*x + 6*y + 7*z + 8*y + 14*z - 4*x*y + 4*y*z = 0
$$5 x^{2} - 4 x y - 10 x + 6 y^{2} + 4 y z + 8 y + 7 z^{2} + 14 z - 6 = 0$$
5*x^2 - 4*x*y - 10*x + 6*y^2 + 4*y*z + 8*y + 7*z^2 + 14*z - 6 = 0
Invariants method
Given equation of the surface of 2-order:
$$5 x^{2} - 4 x y - 10 x + 6 y^{2} + 4 y z + 8 y + 7 z^{2} + 14 z - 6 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = -2$$
$$a_{13} = 0$$
$$a_{14} = -5$$
$$a_{22} = 6$$
$$a_{23} = 2$$
$$a_{24} = 4$$
$$a_{33} = 7$$
$$a_{34} = 7$$
$$a_{44} = -6$$
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
$$I_{1} = 18$$
$$I_{3} = \left|\begin{matrix}5 & -2 & 0\\-2 & 6 & 2\\0 & 2 & 7\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}5 & -2 & 0 & -5\\-2 & 6 & 2 & 4\\0 & 2 & 7 & 7\\-5 & 4 & 7 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & -2 & 0\\-2 & 6 - \lambda & 2\\0 & 2 & 7 - \lambda\end{matrix}\right|$$
$$I_{1} = 18$$
$$I_{2} = 99$$
$$I_{3} = 162$$
$$I_{4} = -2916$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 18 \lambda^{2} - 99 \lambda + 162$$
$$K_{2} = -198$$
$$K_{3} = -1458$$
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
$$\lambda^{3} - 18 \lambda^{2} + 99 \lambda - 162 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = 3$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + 6 \tilde y^{2} + 3 \tilde z^{2} - 18 = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{2}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{6} \sqrt{2}}\right)^{2}}\right) = 1$$
this equation is fora type ellipsoid
- reduced to canonical form
$$5 x^{2} - 4 x y - 10 x + 6 y^{2} + 4 y z + 8 y + 7 z^{2} + 14 z - 6 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = -2$$
$$a_{13} = 0$$
$$a_{14} = -5$$
$$a_{22} = 6$$
$$a_{23} = 2$$
$$a_{24} = 4$$
$$a_{33} = 7$$
$$a_{34} = 7$$
$$a_{44} = -6$$
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
$$I_{1} = 18$$
|5 -2| |6 2| |5 0|
I2 = | | + | | + | |
|-2 6 | |2 7| |0 7|$$I_{3} = \left|\begin{matrix}5 & -2 & 0\\-2 & 6 & 2\\0 & 2 & 7\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}5 & -2 & 0 & -5\\-2 & 6 & 2 & 4\\0 & 2 & 7 & 7\\-5 & 4 & 7 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & -2 & 0\\-2 & 6 - \lambda & 2\\0 & 2 & 7 - \lambda\end{matrix}\right|$$
|5 -5| |6 4 | |7 7 |
K2 = | | + | | + | |
|-5 -6| |4 -6| |7 -6| |5 -2 -5| |6 2 4 | |5 0 -5|
| | | | | |
K3 = |-2 6 4 | + |2 7 7 | + |0 7 7 |
| | | | | |
|-5 4 -6| |4 7 -6| |-5 7 -6|$$I_{1} = 18$$
$$I_{2} = 99$$
$$I_{3} = 162$$
$$I_{4} = -2916$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 18 \lambda^{2} - 99 \lambda + 162$$
$$K_{2} = -198$$
$$K_{3} = -1458$$
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
$$\lambda^{3} - 18 \lambda^{2} + 99 \lambda - 162 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 6$$
$$\lambda_{3} = 3$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + 6 \tilde y^{2} + 3 \tilde z^{2} - 18 = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{2}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{\sqrt{2}}{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{1}{6} \sqrt{2}}\right)^{2}}\right) = 1$$
this equation is fora type ellipsoid
- reduced to canonical form