Respuesta :
Answer:
[tex]A = \left[\begin{array}{ccc}1&0\\-2&1\end{array}\right][/tex]
Step-by-step explanation:
A is a 2 x 2 coefficient matrix with repeated eigenvalues
[tex]A = \left[\begin{array}{ccc}a_{11} &a_{12} \\a_{21} &a_{22} \end{array}\right][/tex]
The linear equation is given by the relationship y' = Ay
[tex]\left[\begin{array}{ccc}y_{1}\\y_{2} \end{array}\right]' = \left[\begin{array}{ccc}a_{11} &a_{12} \\a_{21} &a_{22} \end{array}\right] + \left[\begin{array}{ccc}y_{1}\\y_{2} \end{array}\right][/tex]
[tex]y_{1}' = a_{11}y_{1} + a_{12}y_{2}\\[/tex]...............(1)
[tex]y_{2}' = a_{21}y_{1} + a_{22}y_{2}[/tex]................(2)
Vertical tangents on the line, [tex]y_{1} = 0[/tex]
[tex]y_{1}' = a_{11}y_{1} + a_{12}y_{2} = 0[/tex]
[tex]a_{12} y_{2} = 0[/tex], [tex]a_{12} = 0[/tex]
[tex]a_{21} = -6[/tex], [tex]y_{2} = 2y_{1}[/tex]
[tex]y_{2}' = a_{21}y_{1} + a_{22}y_{2} = 0[/tex]...........(2)
Substitute these values into equation (2)
[tex]-6y_{1} + a_{22}(2y_{1}) = 0\\(-3 + a_{22})y_{1} = 0\\ y_{1} =0\\ -3 + a_{22} = 0\\a_{22} = 3[/tex]
[tex]A = \left[\begin{array}{ccc}3&0\\-6&3\end{array}\right][/tex]
[tex]A = \left[\begin{array}{ccc}1&0\\-2&1\end{array}\right][/tex]
