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+z^2-xy+xz+yz+3x+3y-3z=0
-
5x^2+6xy+5y^2-16x+16y-16=0
-
x^4-y^4=xy
-
3x^2+10xy+3y^2-2x-14y-13
- Plot:
- (x^2+y^2-1)-x^2*y^3=0
- 3(5a-b+6c)=3(5a−b+6c)
- x^2+(y-sqrt(|x|))^2=1
- xy^2
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Identical expressions
- x^ two +y^ two +z^ two -xy+xz+yz+3x+3y-3z= zero
- x squared plus y squared plus z squared minus xy plus xz plus yz plus 3x plus 3y minus 3z equally 0
- x to the power of two plus y to the power of two plus z to the power of two minus xy plus xz plus yz plus 3x plus 3y minus 3z equally zero
- x2+y2+z2-xy+xz+yz+3x+3y-3z=0
- x²+y²+z²-xy+xz+yz+3x+3y-3z=0
- x to the power of 2+y to the power of 2+z to the power of 2-xy+xz+yz+3x+3y-3z=0
- x^2+y^2+z^2-xy+xz+yz+3x+3y-3z=O
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Similar expressions
- x^2+y^2+z^2-xy+xz+yz-3x+3y-3z=0
- x^2+y^2+z^2-xy-xz+yz+3x+3y-3z=0
- x^2+y^2+z^2+xy+xz+yz+3x+3y-3z=0
- x^2-y^2+z^2-xy+xz+yz+3x+3y-3z=0
- x^2+y^2+z^2-xy+xz+yz+3x-3y-3z=0
- x^2+y^2+z^2-xy+xz+yz+3x+3y+3z=0
- x^2+y^2+z^2-xy+xz-yz+3x+3y-3z=0
- x^2+y^2-z^2-xy+xz+yz+3x+3y-3z=0
x^2+y^2+z^2-xy+xz+yz+3x+3y-3z=0 canonical form
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2 2 2 x + y + z - 3*z + 3*x + 3*y + x*z + y*z - x*y = 0
$$x^{2} - x y + x z + 3 x + y^{2} + y z + 3 y + z^{2} - 3 z = 0$$
x^2 - x*y + x*z + 3*x + y^2 + y*z + 3*y + z^2 - 3*z = 0
Invariants method
Given equation of the surface of 2-order:
$$x^{2} - x y + x z + 3 x + y^{2} + y z + 3 y + z^{2} - 3 z = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = - \frac{1}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = \frac{3}{2}$$
$$a_{22} = 1$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 1$$
$$a_{34} = - \frac{3}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}1 & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & 1 & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & - \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\- \frac{1}{2} & 1 & \frac{1}{2} & \frac{3}{2}\\\frac{1}{2} & \frac{1}{2} & 1 & - \frac{3}{2}\\\frac{3}{2} & \frac{3}{2} & - \frac{3}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & 1 - \lambda & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = \frac{9}{4}$$
$$I_{3} = 0$$
$$I_{4} = - \frac{243}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - \frac{9 \lambda}{4}$$
$$K_{2} = - \frac{27}{4}$$
$$K_{3} = - \frac{81}{4}$$
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
$$\lambda^{3} - 3 \lambda^{2} + \frac{9 \lambda}{4} = 0$$
$$\lambda_{1} = \frac{3}{2}$$
$$\lambda_{2} = \frac{3}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} + 3 \sqrt{3} \tilde z = 0$$
and
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} - 3 \sqrt{3} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\sqrt{3}} + \frac{\tilde y^{2}}{\sqrt{3}}\right) = 0$$
and
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\sqrt{3}} + \frac{\tilde y^{2}}{\sqrt{3}}\right) = 0$$
this equation is fora type elliptical paraboloid
- reduced to canonical form
$$x^{2} - x y + x z + 3 x + y^{2} + y z + 3 y + z^{2} - 3 z = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = - \frac{1}{2}$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = \frac{3}{2}$$
$$a_{22} = 1$$
$$a_{23} = \frac{1}{2}$$
$$a_{24} = \frac{3}{2}$$
$$a_{33} = 1$$
$$a_{34} = - \frac{3}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 3$$
| 1 -1/2| | 1 1/2| | 1 1/2|
I2 = | | + | | + | |
|-1/2 1 | |1/2 1 | |1/2 1 |$$I_{3} = \left|\begin{matrix}1 & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & 1 & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & - \frac{1}{2} & \frac{1}{2} & \frac{3}{2}\\- \frac{1}{2} & 1 & \frac{1}{2} & \frac{3}{2}\\\frac{1}{2} & \frac{1}{2} & 1 & - \frac{3}{2}\\\frac{3}{2} & \frac{3}{2} & - \frac{3}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{1}{2} & \frac{1}{2}\\- \frac{1}{2} & 1 - \lambda & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & 1 - \lambda\end{matrix}\right|$$
| 1 3/2| | 1 3/2| | 1 -3/2|
K2 = | | + | | + | |
|3/2 0 | |3/2 0 | |-3/2 0 | | 1 -1/2 3/2| | 1 1/2 3/2 | | 1 1/2 3/2 |
| | | | | |
K3 = |-1/2 1 3/2| + |1/2 1 -3/2| + |1/2 1 -3/2|
| | | | | |
|3/2 3/2 0 | |3/2 -3/2 0 | |3/2 -3/2 0 |$$I_{1} = 3$$
$$I_{2} = \frac{9}{4}$$
$$I_{3} = 0$$
$$I_{4} = - \frac{243}{16}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - \frac{9 \lambda}{4}$$
$$K_{2} = - \frac{27}{4}$$
$$K_{3} = - \frac{81}{4}$$
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
$$\lambda^{3} - 3 \lambda^{2} + \frac{9 \lambda}{4} = 0$$
$$\lambda_{1} = \frac{3}{2}$$
$$\lambda_{2} = \frac{3}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} + 3 \sqrt{3} \tilde z = 0$$
and
$$\frac{3 \tilde x^{2}}{2} + \frac{3 \tilde y^{2}}{2} - 3 \sqrt{3} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\sqrt{3}} + \frac{\tilde y^{2}}{\sqrt{3}}\right) = 0$$
and
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\sqrt{3}} + \frac{\tilde y^{2}}{\sqrt{3}}\right) = 0$$
this equation is fora type elliptical paraboloid
- reduced to canonical form