Mister Exam
xy-xz+yz=0 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
Invariants method
Given equation of the surface of 2-order:
$$x y - x z + y z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = - \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & 0 & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & \frac{1}{2} & - \frac{1}{2} & 0\\\frac{1}{2} & 0 & \frac{1}{2} & 0\\- \frac{1}{2} & \frac{1}{2} & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & - \lambda & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = - \frac{3}{4}$$
$$I_{3} = - \frac{1}{4}$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{3 \lambda}{4} - \frac{1}{4}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - \frac{3 \lambda}{4} + \frac{1}{4} = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = \frac{1}{2}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \tilde x^{2} + \frac{\tilde y^{2}}{2} + \frac{\tilde z^{2}}{2} = 0$$
$$- \frac{\tilde x^{2}}{1^{2}} + \left(\frac{\tilde y^{2}}{\left(\sqrt{2}\right)^{2}} + \frac{\tilde z^{2}}{\left(\sqrt{2}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form
$$x y - x z + y z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = - \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
| 0 1/2| | 0 1/2| | 0 -1/2|
I2 = | | + | | + | |
|1/2 0 | |1/2 0 | |-1/2 0 |$$I_{3} = \left|\begin{matrix}0 & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & 0 & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & \frac{1}{2} & - \frac{1}{2} & 0\\\frac{1}{2} & 0 & \frac{1}{2} & 0\\- \frac{1}{2} & \frac{1}{2} & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & \frac{1}{2} & - \frac{1}{2}\\\frac{1}{2} & - \lambda & \frac{1}{2}\\- \frac{1}{2} & \frac{1}{2} & - \lambda\end{matrix}\right|$$
|0 0| |0 0| |0 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| | 0 1/2 0| | 0 1/2 0| | 0 -1/2 0|
| | | | | |
K3 = |1/2 0 0| + |1/2 0 0| + |-1/2 0 0|
| | | | | |
| 0 0 0| | 0 0 0| | 0 0 0|$$I_{1} = 0$$
$$I_{2} = - \frac{3}{4}$$
$$I_{3} = - \frac{1}{4}$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{3 \lambda}{4} - \frac{1}{4}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - \frac{3 \lambda}{4} + \frac{1}{4} = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = \frac{1}{2}$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \tilde x^{2} + \frac{\tilde y^{2}}{2} + \frac{\tilde z^{2}}{2} = 0$$
$$- \frac{\tilde x^{2}}{1^{2}} + \left(\frac{\tilde y^{2}}{\left(\sqrt{2}\right)^{2}} + \frac{\tilde z^{2}}{\left(\sqrt{2}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form