Mister Exam

# -x^2-10x^2-6x1x2 canonical form

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### The solution

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- 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:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
     |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