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
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How to use it?
- Canonical form:
- x^2+y^2-6x+4y-12=0
- 6x1^2-3x2^2-3x3^2
- 4𝑥1𝑥3−6𝑥2𝑥4
- 2x^2+5y^2+3z^2+4xy=0
- Plot:
- z=3/sqrt(x^2+y-9)
- tan(sin(e−1x))−x=0
- (9,27-(x+y+z))*x*y*z=9,27
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4x^2+4y^2+8x*y+4x+4y+1==0
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Identical expressions
- 6x1^ two -3x two ^ two -3x3^2
- 6x1 squared minus 3x2 squared minus 3x3 squared
- 6x1 to the power of two minus 3x two to the power of two minus 3x3 squared
- 6x12-3x22-3x32
- 6x1²-3x2²-3x3²
- 6x1 to the power of 2-3x2 to the power of 2-3x3 to the power of 2
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Similar expressions
- 6x1^2+3x2^2-3x3^2
- 6x1^2-3x2^2+3x3^2
6x1^2-3x2^2-3x3^2 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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2 2 2 - 3*x2 - 3*x3 + 6*x1 = 0
$$6 x_{1}^{2} - 3 x_{2}^{2} - 3 x_{3}^{2} = 0$$
6*x1^2 - 3*x2^2 - 3*x3^2 = 0
Invariants method
Given equation of the surface of 2-order:
$$6 x_{1}^{2} - 3 x_{2}^{2} - 3 x_{3}^{2} = 0$$
This equation looks like:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + a_{33} x_{1}^{2} + 2 a_{14} x_{3} + 2 a_{24} x_{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = -3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 6$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}-3 & 0 & 0\\0 & -3 & 0\\0 & 0 & 6\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-3 & 0 & 0 & 0\\0 & -3 & 0 & 0\\0 & 0 & 6 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 3 & 0 & 0\\0 & - \lambda - 3 & 0\\0 & 0 & - \lambda + 6\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -27$$
$$I_{3} = 54$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 27 \lambda + 54$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + \lambda^{3} + I_{2} \lambda - I_{3} = 0$$
or
$$\lambda^{3} - 27 \lambda - 54 = 0$$
Solve this equation
$$\lambda_{1} = 6$$
$$\lambda_{2} = -3$$
$$\lambda_{3} = -3$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- 3 \tilde x1^{2} - 3 \tilde x2^{2} + 6 \tilde x3^{2} = 0$$
$$- \frac{\tilde x3^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} + \left(\frac{\tilde x1^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form
$$6 x_{1}^{2} - 3 x_{2}^{2} - 3 x_{3}^{2} = 0$$
This equation looks like:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{1} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{1} x_{2} + a_{33} x_{1}^{2} + 2 a_{14} x_{3} + 2 a_{24} x_{2} + 2 a_{34} x_{1} + a_{44} = 0$$
where
$$a_{11} = -3$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 6$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
|-3 0 | |-3 0| |-3 0|
I2 = | | + | | + | |
|0 -3| |0 6| |0 6|$$I_{3} = \left|\begin{matrix}-3 & 0 & 0\\0 & -3 & 0\\0 & 0 & 6\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-3 & 0 & 0 & 0\\0 & -3 & 0 & 0\\0 & 0 & 6 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 3 & 0 & 0\\0 & - \lambda - 3 & 0\\0 & 0 & - \lambda + 6\end{matrix}\right|$$
|-3 0| |-3 0| |6 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |-3 0 0| |-3 0 0| |-3 0 0|
| | | | | |
K3 = |0 -3 0| + |0 6 0| + |0 6 0|
| | | | | |
|0 0 0| |0 0 0| |0 0 0|$$I_{1} = 0$$
$$I_{2} = -27$$
$$I_{3} = 54$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 27 \lambda + 54$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
I3 != 0
then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + \lambda^{3} + I_{2} \lambda - I_{3} = 0$$
or
$$\lambda^{3} - 27 \lambda - 54 = 0$$
Solve this equation
$$\lambda_{1} = 6$$
$$\lambda_{2} = -3$$
$$\lambda_{3} = -3$$
then the canonical form of the equation will be
$$\left(\tilde x1^{2} \lambda_{3} + \left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- 3 \tilde x1^{2} - 3 \tilde x2^{2} + 6 \tilde x3^{2} = 0$$
$$- \frac{\tilde x3^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} + \left(\frac{\tilde x1^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form