Mister Exam

# x1x2+x1x3−x2x3 canonical form

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### The solution

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x1*x2 + x1*x3 - x2*x3 = 0
$$x_{1} x_{2} + x_{1} x_{3} - x_{2} x_{3} = 0$$
x1*x2 + x1*x3 - x2*x3 = 0
Invariants method
Given equation of the surface of 2-order:
$$x_{1} x_{2} + x_{1} x_{3} - x_{2} x_{3} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = - \frac{1}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\frac{\tilde x1^{2}}{2} + \frac{\tilde x2^{2}}{2} - \tilde x3^{2} = 0$$
$$- \frac{\tilde x3^{2}}{1^{2}} + \left(\frac{\tilde x1^{2}}{\left(\sqrt{2}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\sqrt{2}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form