Mister Exam

-x1^2-2x2^2-2x1x2-2x1x3 canonical form

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The solution

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    2       2
- x1  - 2*x2  - 2*x1*x2 - 2*x1*x3 = 0
$$- x_{1}^{2} - 2 x_{1} x_{2} - 2 x_{1} x_{3} - 2 x_{2}^{2} = 0$$
-x1^2 - 2*x1*x2 - 2*x1*x3 - 2*x2^2 = 0
Invariants method
Given equation of the surface of 2-order:
$$- x_{1}^{2} - 2 x_{1} x_{2} - 2 x_{1} x_{3} - 2 x_{2}^{2} = 0$$
This equation looks like:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + 2 a_{14} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + 2 a_{24} x_{2} + a_{33} x_{1}^{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = -1$$
$$a_{14} = 0$$
$$a_{22} = -2$$
$$a_{23} = -1$$
$$a_{24} = 0$$
$$a_{33} = -1$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = -3$$
     |0  0 |   |-2  -1|   |0   -1|
I2 = |     | + |      | + |      |
|0  -2|   |-1  -1|   |-1  -1|

$$I_{3} = \left|\begin{matrix}0 & 0 & -1\\0 & -2 & -1\\-1 & -1 & -1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & -1 & 0\\0 & -2 & -1 & 0\\-1 & -1 & -1 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & -1\\0 & - \lambda - 2 & -1\\-1 & -1 & - \lambda - 1\end{matrix}\right|$$
     |0  0|   |-2  0|   |-1  0|
K2 = |    | + |     | + |     |
|0  0|   |0   0|   |0   0|

     |0  0   0|   |-2  -1  0|   |0   -1  0|
|        |   |         |   |         |
K3 = |0  -2  0| + |-1  -1  0| + |-1  -1  0|
|        |   |         |   |         |
|0  0   0|   |0   0   0|   |0   0   0|

$$I_{1} = -3$$
$$I_{2} = 0$$
$$I_{3} = 2$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 3 \lambda^{2} + 2$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} + 3 \lambda^{2} - 2 = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = - \sqrt{3} - 1$$
$$\lambda_{3} = -1 + \sqrt{3}$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x1^{2} \left(-1 + \sqrt{3}\right) + \tilde x2^{2} \left(- \sqrt{3} - 1\right) - \tilde x3^{2} = 0$$
$$- \frac{\tilde x1^{2}}{\left(\frac{1}{\sqrt{-1 + \sqrt{3}}}\right)^{2}} + \left(\frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{1 + \sqrt{3}}}\right)^{2}} + \frac{\tilde x3^{2}}{1^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form