Mister Exam
Other calculators
- Canonical form of a elliptical paraboloid
- Canonical form of a double hyperboloid
- Canonical form of a imaginary ellipsoid
- Canonical form of a degenerate ellipse
- Canonical form of a parabola
-
How to use it?
- Canonical form:
-
8x^2+5y^2+15x+6y-7=0
- 0x^2+xy+0y^2+2x+y+2=0
- 4x^2+y+20x-11=0
- z^2=x^2+y^2
- Plot:
- (2*y-2*x)*(y-1)=0
- tan(x*factorial(y))=1
-
x^2+y^2
- tan(x*y!)=1
-
Identical expressions
- 𝑥1^ two +𝑥 two ^2+5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3
- 𝑥1 squared plus 𝑥2 squared plus 5𝑥32−6𝑥1𝑥2 plus 2𝑥1𝑥3−2𝑥2𝑥3
- 𝑥1 to the power of two plus 𝑥 two squared plus 5𝑥32−6𝑥1𝑥2 plus 2𝑥1𝑥3−2𝑥2𝑥3
- 𝑥12+𝑥22+5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3
- 𝑥1²+𝑥2²+5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3
- 𝑥1 to the power of 2+𝑥2 to the power of 2+5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3
-
Similar expressions
- 𝑥1^2+𝑥2^2-5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3
- 𝑥1^2+𝑥2^2+5𝑥32−6𝑥1𝑥2-2𝑥1𝑥3−2𝑥2𝑥3
- 𝑥1^2-𝑥2^2+5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3
𝑥1^2+𝑥2^2+5𝑥32−6𝑥1𝑥2+2𝑥1𝑥3−2𝑥2𝑥3 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 2 x1 + x2 + 5*x32 - 6*x1*x2 - 2*x2*x3 + 2*x1*x3 = 0
$$x_{1}^{2} - 6 x_{1} x_{2} + 2 x_{1} x_{3} + x_{2}^{2} - 2 x_{2} x_{3} + 5 x_{32} = 0$$
x1^2 - 6*x1*x2 + 2*x1*x3 + x2^2 - 2*x2*x3 + 5*x32 = 0
Invariants method
Given equation of the surface of 2-order:
$$x_{1}^{2} - 6 x_{1} x_{2} + 2 x_{1} x_{3} + x_{2}^{2} - 2 x_{2} x_{3} + 5 x_{32} = 0$$
This equation looks like:
$$a_{11} x_{32}^{2} + 2 a_{12} x_{3} x_{32} + 2 a_{13} x_{2} x_{32} + 2 a_{14} x_{32} + a_{22} x_{3}^{2} + 2 a_{23} x_{2} x_{3} + 2 a_{24} x_{3} + a_{33} x_{2}^{2} + 2 a_{34} x_{2} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = -1$$
$$a_{24} = x_{1}$$
$$a_{33} = 1$$
$$a_{34} = - 3 x_{1}$$
$$a_{44} = x_{1}^{2}$$
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
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 0 & -1\\0 & -1 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & \frac{5}{2}\\0 & 0 & -1 & x_{1}\\0 & -1 & 1 & - 3 x_{1}\\\frac{5}{2} & x_{1} & - 3 x_{1} & x_{1}^{2}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & - \lambda & -1\\0 & -1 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = -1$$
$$I_{3} = 0$$
$$I_{4} = \frac{25}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + \lambda$$
$$K_{2} = - 9 x_{1}^{2} - \frac{25}{4}$$
$$K_{3} = 4 x_{1}^{2} - \frac{25}{4}$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \lambda^{2} - \lambda = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{\sqrt{5}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{\sqrt{5}}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x3^{2} \lambda_{2} + \tilde x32^{2} \lambda_{1}\right) = 0$$
and
$$- \tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x3^{2} \lambda_{2} + \tilde x32^{2} \lambda_{1}\right) = 0$$
$$5 \tilde x2 + \tilde x3^{2} \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) + \tilde x32^{2} \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) = 0$$
and
$$- 5 \tilde x2 + \tilde x3^{2} \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) + \tilde x32^{2} \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) = 0$$
$$- 2 \tilde x2 + \left(- \frac{\tilde x3^{2}}{\frac{5}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{5}}{2}}} + \frac{\tilde x32^{2}}{\frac{5}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{5}}{2}}}\right) = 0$$
and
$$2 \tilde x2 + \left(- \frac{\tilde x3^{2}}{\frac{5}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{5}}{2}}} + \frac{\tilde x32^{2}}{\frac{5}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{5}}{2}}}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form
$$x_{1}^{2} - 6 x_{1} x_{2} + 2 x_{1} x_{3} + x_{2}^{2} - 2 x_{2} x_{3} + 5 x_{32} = 0$$
This equation looks like:
$$a_{11} x_{32}^{2} + 2 a_{12} x_{3} x_{32} + 2 a_{13} x_{2} x_{32} + 2 a_{14} x_{32} + a_{22} x_{3}^{2} + 2 a_{23} x_{2} x_{3} + 2 a_{24} x_{3} + a_{33} x_{2}^{2} + 2 a_{34} x_{2} + a_{44} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = -1$$
$$a_{24} = x_{1}$$
$$a_{33} = 1$$
$$a_{34} = - 3 x_{1}$$
$$a_{44} = x_{1}^{2}$$
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
$$I_{1} = 1$$
|0 0| |0 -1| |0 0|
I2 = | | + | | + | |
|0 0| |-1 1 | |0 1|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 0 & -1\\0 & -1 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & \frac{5}{2}\\0 & 0 & -1 & x_{1}\\0 & -1 & 1 & - 3 x_{1}\\\frac{5}{2} & x_{1} & - 3 x_{1} & x_{1}^{2}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & - \lambda & -1\\0 & -1 & 1 - \lambda\end{matrix}\right|$$
| 0 5/2| |0 x1 | | 1 -3*x1|
| | | | | |
K2 = | 2| + | 2| + | 2 |
|5/2 x1 | |x1 x1 | |-3*x1 x1 |
| 0 0 5/2| |0 -1 x1 | | 0 0 5/2 |
| | | | | |
| 0 0 x1 | |-1 1 -3*x1| | 0 1 -3*x1|
K3 = | | + | | + | |
| 2| | 2 | | 2 |
|5/2 x1 x1 | |x1 -3*x1 x1 | |5/2 -3*x1 x1 |
$$I_{1} = 1$$
$$I_{2} = -1$$
$$I_{3} = 0$$
$$I_{4} = \frac{25}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + \lambda$$
$$K_{2} = - 9 x_{1}^{2} - \frac{25}{4}$$
$$K_{3} = 4 x_{1}^{2} - \frac{25}{4}$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \lambda^{2} - \lambda = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{\sqrt{5}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{\sqrt{5}}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x3^{2} \lambda_{2} + \tilde x32^{2} \lambda_{1}\right) = 0$$
and
$$- \tilde x2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x3^{2} \lambda_{2} + \tilde x32^{2} \lambda_{1}\right) = 0$$
$$5 \tilde x2 + \tilde x3^{2} \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) + \tilde x32^{2} \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) = 0$$
and
$$- 5 \tilde x2 + \tilde x3^{2} \left(\frac{1}{2} + \frac{\sqrt{5}}{2}\right) + \tilde x32^{2} \left(\frac{1}{2} - \frac{\sqrt{5}}{2}\right) = 0$$
$$- 2 \tilde x2 + \left(- \frac{\tilde x3^{2}}{\frac{5}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{5}}{2}}} + \frac{\tilde x32^{2}}{\frac{5}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{5}}{2}}}\right) = 0$$
and
$$2 \tilde x2 + \left(- \frac{\tilde x3^{2}}{\frac{5}{2} \frac{1}{\frac{1}{2} + \frac{\sqrt{5}}{2}}} + \frac{\tilde x32^{2}}{\frac{5}{2} \frac{1}{- \frac{1}{2} + \frac{\sqrt{5}}{2}}}\right) = 0$$
this equation is fora type hyperbolic paraboloid
- reduced to canonical form