196 canonical form

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False

False

False

Invariants method

Given line equation of 2-order:

False

This equation looks like:

a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0

where

a_{11} = 0

a_{12} = 0

a_{13} = 0

a_{22} = 0

a_{23} = 0

a_{33} = 196

The invariants of the equation when converting coordinates are determinants:

I_{1} = a_{11} + a_{22}

     |a11  a12|
I2 = |        |
     |a12  a22|

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{\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

I_{1} = 0

     |0  0|
I2 = |    |
     |0  0|

I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 196\end{matrix}\right|

I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|

     |0   0 |   |0   0 |
K2 = |      | + |      |
     |0  196|   |0  196|

I_{1} = 0

I_{2} = 0

I_{3} = 0

I{\left(\lambda \right)} = \lambda^{2}

K_{2} = 0

Because

I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)

then by line type:
this equation is of type : two coincident straight lines

I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0

False

None

- reduced to canonical form