Mister Exam

Other calculators

Canonical form/ x1^2+x2^2

x1^2+x2^2 canonical form

The teacher will be very surprised to see your correct solution 😉

v

The graph:

x: [, ]
y: [, ]
z: [, ]

Quality:

 (Number of points on the axis)

Plot type:

The solution

You have entered [src] 
  2     2    
x1  + x2  = 0
x12+x22=0x_{1}^{2} + x_{2}^{2} = 0 
x1^2 + x2^2 = 0
Detail solution
Given line equation of 2-order:
x12+x22=0x_{1}^{2} + 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=1a_{11} = 1
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=1a_{22} = 1
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=∣1001∣\Delta = \left|\begin{matrix}1 & 0\\0 & 1\end{matrix}\right|
Δ=1\Delta = 1
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
x20=0x_{20} = 0
x10=0x_{10} = 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
x1′2+x2′2=0x1'^{2} + x2'^{2} = 0
Given equation is degenerate ellipse
x~1212+x~2212=0\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{1^{2}} = 0
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e⃗1=(1, 0)\vec e_1 = \left( 1, \ 0\right)
e⃗2=(0, 1)\vec e_2 = \left( 0, \ 1\right)
Invariants method
Given line equation of 2-order:
x12+x22=0x_{1}^{2} + 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=1a_{11} = 1
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=1a_{22} = 1
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=2I_{1} = 2
     |1  0|
I2 = |    |
     |0  1|

I3=∣100010000∣I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|
I(λ)=∣1−λ001−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & 1 - \lambda\end{matrix}\right|
     |1  0|   |1  0|
K2 = |    | + |    |
     |0  0|   |0  0|

I1=2I_{1} = 2
I2=1I_{2} = 1
I3=0I_{3} = 0
I(λ)=λ2−2λ+1I{\left(\lambda \right)} = \lambda^{2} - 2 \lambda + 1
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
λ2−2λ+1=0\lambda^{2} - 2 \lambda + 1 = 0
λ1=1\lambda_{1} = 1
λ2=1\lambda_{2} = 1
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
x~12+x~22=0\tilde x1^{2} + \tilde x2^{2} = 0
x~1212+x~2212=0\frac{\tilde x1^{2}}{1^{2}} + \frac{\tilde x2^{2}}{1^{2}} = 0
- reduced to canonical form
    © Mister Exam – Calculator