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Canonical form/ -3x1^2-3x2^2-2x1x2

-3x1^2-3x2^2-2x1x2 canonical form

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The solution

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      2       2              
- 3*x1  - 3*x2  - 2*x1*x2 = 0
−3x12−2x1x2−3x22=0- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0 
-3*x1^2 - 2*x1*x2 - 3*x2^2 = 0
Detail solution
Given line equation of 2-order:
−3x12−2x1x2−3x22=0- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0
This equation looks like:
a11x22+2a12x1x2+2a13x2+a22x12+2a23x1+a33=0a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0
where
a11=−3a_{11} = -3
a12=−1a_{12} = -1
a13=0a_{13} = 0
a22=−3a_{22} = -3
a23=0a_{23} = 0
a33=0a_{33} = 0
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣−3−1−1−3∣\Delta = \left|\begin{matrix}-3 & -1\\-1 & -3\end{matrix}\right|
Δ=8\Delta = 8
Because
Δ\Delta
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
a11x20+a12x10+a13=0a_{11} x_{20} + a_{12} x_{10} + a_{13} = 0
a12x20+a22x10+a23=0a_{12} x_{20} + a_{22} x_{10} + a_{23} = 0
substitute coefficients
−x10−3x20=0- x_{10} - 3 x_{20} = 0
−3x10−x20=0- 3 x_{10} - x_{20} = 0
then
x20=0x_{20} = 0
x10=0x_{10} = 0
Thus, we have the equation in the coordinate system O'x'y'
a33′+a11x2′2+2a12x1′x2′+a22x1′2=0a'_{33} + a_{11} x2'^{2} + 2 a_{12} x1' x2' + a_{22} x1'^{2} = 0
where
a33′=a13x20+a23x10+a33a'_{33} = a_{13} x_{20} + a_{23} x_{10} + a_{33}
or
a33′=0a'_{33} = 0
a33′=0a'_{33} = 0
then equation turns into
−3x1′2−2x1′x2′−3x2′2=0- 3 x1'^{2} - 2 x1' x2' - 3 x2'^{2} = 0
Rotate the resulting coordinate system by an angle φ
x2′=−x~1sin⁡(ϕ)+x~2cos⁡(ϕ)x2' = - \tilde x1 \sin{\left(\phi \right)} + \tilde x2 \cos{\left(\phi \right)}
x1′=x~1cos⁡(ϕ)+x~2sin⁡(ϕ)x1' = \tilde x1 \cos{\left(\phi \right)} + \tilde x2 \sin{\left(\phi \right)}
φ - determined from the formula
cot⁡(2ϕ)=a11−a222a12\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}
substitute coefficients
cot⁡(2ϕ)=0\cot{\left(2 \phi \right)} = 0
then
ϕ=π4\phi = \frac{\pi}{4}
sin⁡(2ϕ)=1\sin{\left(2 \phi \right)} = 1
cos⁡(2ϕ)=0\cos{\left(2 \phi \right)} = 0
cos⁡(ϕ)=cos⁡(2ϕ)2+12\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}
sin⁡(ϕ)=1−cos⁡2(ϕ)\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}
cos⁡(ϕ)=22\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}
sin⁡(ϕ)=22\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}
substitute coefficients
x2′=−2x~12+2x~22x2' = - \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}
x1′=2x~12+2x~22x1' = \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}
then the equation turns from
−3x1′2−2x1′x2′−3x2′2=0- 3 x1'^{2} - 2 x1' x2' - 3 x2'^{2} = 0
to
−3(−2x~12+2x~22)2−2(−2x~12+2x~22)(2x~12+2x~22)−3(2x~12+2x~22)2=0- 3 \left(- \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right)^{2} - 2 \left(- \frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right) \left(\frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right) - 3 \left(\frac{\sqrt{2} \tilde x1}{2} + \frac{\sqrt{2} \tilde x2}{2}\right)^{2} = 0
simplify
−2x~12−4x~22=0- 2 \tilde x1^{2} - 4 \tilde x2^{2} = 0
2x~12+4x~22=02 \tilde x1^{2} + 4 \tilde x2^{2} = 0
Given equation is degenerate ellipse
x~12(22)2+x~22(12)2=0\frac{\tilde x1^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{1}{2}\right)^{2}} = 0
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e⃗1=(22, 22)\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)
e⃗2=(−22, 22)\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)
Invariants method
Given line equation of 2-order:
−3x12−2x1x2−3x22=0- 3 x_{1}^{2} - 2 x_{1} x_{2} - 3 x_{2}^{2} = 0
This equation looks like:
a11x22+2a12x1x2+2a13x2+a22x12+2a23x1+a33=0a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0
where
a11=−3a_{11} = -3
a12=−1a_{12} = -1
a13=0a_{13} = 0
a22=−3a_{22} = -3
a23=0a_{23} = 0
a33=0a_{33} = 0
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=∣a11a12a13a12a22a23a13a23a33∣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(λ)=∣a11−λa12a12a22−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|
     |a11  a13|   |a22  a23|
K2 = |        | + |        |
     |a13  a33|   |a23  a33|

substitute coefficients
I1=−6I_{1} = -6
     |-3  -1|
I2 = |      |
     |-1  -3|

I3=∣−3−10−1−30000∣I_{3} = \left|\begin{matrix}-3 & -1 & 0\\-1 & -3 & 0\\0 & 0 & 0\end{matrix}\right|
I(λ)=∣−λ−3−1−1−λ−3∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 3 & -1\\-1 & - \lambda - 3\end{matrix}\right|
     |-3  0|   |-3  0|
K2 = |     | + |     |
     |0   0|   |0   0|

I1=−6I_{1} = -6
I2=8I_{2} = 8
I3=0I_{3} = 0
I(λ)=λ2+6λ+8I{\left(\lambda \right)} = \lambda^{2} + 6 \lambda + 8
K2=0K_{2} = 0
Because
I3=0∧I2>0I_{3} = 0 \wedge I_{2} > 0
then by line type:
this equation is of type : degenerate ellipse
Make the characteristic equation for the line:
−I1λ+I2+λ2=0- I_{1} \lambda + I_{2} + \lambda^{2} = 0
or
λ2+6λ+8=0\lambda^{2} + 6 \lambda + 8 = 0
λ1=−2\lambda_{1} = -2
λ2=−4\lambda_{2} = -4
then the canonical form of the equation will be
x~12λ2+x~22λ1+I3I2=0\tilde x1^{2} \lambda_{2} + \tilde x2^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0
or
−4x~12−2x~22=0- 4 \tilde x1^{2} - 2 \tilde x2^{2} = 0
x~12(12)2+x~22(22)2=0\frac{\tilde x1^{2}}{\left(\frac{1}{2}\right)^{2}} + \frac{\tilde x2^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}} = 0
- reduced to canonical form
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