Mister Exam
-x^2-10x^2-6x1x2 canonical form
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2 - 11*x - 6*x1*x2 = 0
$$- 11 x^{2} - 6 x_{1} x_{2} = 0$$
-11*x^2 - 6*x1*x2 = 0
Invariants method
Given equation of the surface of 2-order:
$$- 11 x^{2} - 6 x_{1} x_{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x x_{2} + 2 a_{13} x x_{1} + 2 a_{14} x + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = -11$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = -3$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = -11$$
$$I_{3} = \left|\begin{matrix}-11 & 0 & 0\\0 & 0 & -3\\0 & -3 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-11 & 0 & 0 & 0\\0 & 0 & -3 & 0\\0 & -3 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 11 & 0 & 0\\0 & - \lambda & -3\\0 & -3 & - \lambda\end{matrix}\right|$$
$$I_{1} = -11$$
$$I_{2} = -9$$
$$I_{3} = 99$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 11 \lambda^{2} + 9 \lambda + 99$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} + 11 \lambda^{2} - 9 \lambda - 99 = 0$$
Solve this equation
$$\lambda_{1} = 3$$
$$\lambda_{2} = -3$$
$$\lambda_{3} = -11$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde x2^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x^{2} - 11 \tilde x1^{2} - 3 \tilde x2^{2} = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \left(\frac{\tilde x1^{2}}{\left(\frac{\sqrt{11}}{11}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form
$$- 11 x^{2} - 6 x_{1} x_{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x x_{2} + 2 a_{13} x x_{1} + 2 a_{14} x + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = -11$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = -3$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = -11$$
|-11 0| |0 -3| |-11 0|
I2 = | | + | | + | |
| 0 0| |-3 0 | | 0 0|$$I_{3} = \left|\begin{matrix}-11 & 0 & 0\\0 & 0 & -3\\0 & -3 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-11 & 0 & 0 & 0\\0 & 0 & -3 & 0\\0 & -3 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 11 & 0 & 0\\0 & - \lambda & -3\\0 & -3 & - \lambda\end{matrix}\right|$$
|-11 0| |0 0| |0 0|
K2 = | | + | | + | |
| 0 0| |0 0| |0 0| |-11 0 0| |0 -3 0| |-11 0 0|
| | | | | |
K3 = | 0 0 0| + |-3 0 0| + | 0 0 0|
| | | | | |
| 0 0 0| |0 0 0| | 0 0 0|$$I_{1} = -11$$
$$I_{2} = -9$$
$$I_{3} = 99$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 11 \lambda^{2} + 9 \lambda + 99$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} + 11 \lambda^{2} - 9 \lambda - 99 = 0$$
Solve this equation
$$\lambda_{1} = 3$$
$$\lambda_{2} = -3$$
$$\lambda_{3} = -11$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde x2^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x^{2} - 11 \tilde x1^{2} - 3 \tilde x2^{2} = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \left(\frac{\tilde x1^{2}}{\left(\frac{\sqrt{11}}{11}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form