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:
- 5*x^2-6*y^2+10*x-12*y-31=0
- -6x1x2+3x1x3
- 5x1^2+13x2^2+5x3^2+4*x1*x2+8*x2*x3
- 4x^2-5y^2+4z^2=-20
- Plot:
-
((x+5/2)^2-(y-11^(1/2)/2)^2+2.5)*((x+5/2)^2-(y+11^(1/2)/2)^2+2.5)=444-36*x^2
- z=x^2+y^2/4
- z=(x-1)^2+2*y^2
- y^2=4*x
-
Identical expressions
- -6x1x2+3x1x3
- minus 6x1x2 plus 3x1x3
-
Similar expressions
- 6x1x2+3x1x3
- -6x1x2-3x1x3
-6x1x2+3x1x3 canonical form
The teacher will be very surprised to see your correct solution 😉
The solution
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-6*x1*x2 + 3*x1*x3 = 0
$$- 6 x_{1} x_{2} + 3 x_{1} x_{3} = 0$$
-6*x1*x2 + 3*x1*x3 = 0
Invariants method
Given equation of the surface of 2-order:
$$- 6 x_{1} x_{2} + 3 x_{1} x_{3} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = -3$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
substitute coefficients
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{3}{2}\\0 & 0 & -3\\\frac{3}{2} & -3 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & \frac{3}{2} & 0\\0 & 0 & -3 & 0\\\frac{3}{2} & -3 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & \frac{3}{2}\\0 & - \lambda & -3\\\frac{3}{2} & -3 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = - \frac{45}{4}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{45 \lambda}{4}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - \frac{45 \lambda}{4} = 0$$
$$\lambda_{1} = - \frac{3 \sqrt{5}}{2}$$
$$\lambda_{2} = \frac{3 \sqrt{5}}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\frac{3 \sqrt{5} \tilde x2^{2}}{2} - \frac{3 \sqrt{5} \tilde x3^{2}}{2} = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{5^{\frac{3}{4}} \sqrt{6}}{15}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{5^{\frac{3}{4}} \sqrt{6}}{15}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form
$$- 6 x_{1} x_{2} + 3 x_{1} x_{3} = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{3}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = -3$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
|0 0| |0 -3| | 0 3/2|
I2 = | | + | | + | |
|0 0| |-3 0 | |3/2 0 |$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{3}{2}\\0 & 0 & -3\\\frac{3}{2} & -3 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & \frac{3}{2} & 0\\0 & 0 & -3 & 0\\\frac{3}{2} & -3 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & \frac{3}{2}\\0 & - \lambda & -3\\\frac{3}{2} & -3 & - \lambda\end{matrix}\right|$$
|0 0| |0 0| |0 0|
K2 = | | + | | + | |
|0 0| |0 0| |0 0| |0 0 0| |0 -3 0| | 0 3/2 0|
| | | | | |
K3 = |0 0 0| + |-3 0 0| + |3/2 0 0|
| | | | | |
|0 0 0| |0 0 0| | 0 0 0|$$I_{1} = 0$$
$$I_{2} = - \frac{45}{4}$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{45 \lambda}{4}$$
$$K_{2} = 0$$
$$K_{3} = 0$$
Because
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - \frac{45 \lambda}{4} = 0$$
$$\lambda_{1} = - \frac{3 \sqrt{5}}{2}$$
$$\lambda_{2} = \frac{3 \sqrt{5}}{2}$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\left(\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$\frac{3 \sqrt{5} \tilde x2^{2}}{2} - \frac{3 \sqrt{5} \tilde x3^{2}}{2} = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{5^{\frac{3}{4}} \sqrt{6}}{15}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{5^{\frac{3}{4}} \sqrt{6}}{15}\right)^{2}} = 0$$
this equation is fora type two intersecting planes
- reduced to canonical form