Found transition matrix and state matrix

Question

I tried to found a solution for this problem but I can't! Any suggestion?Thank you.

Given that A is a 2x2 matrix and that dx/dt=Ax(t)

suppose that x(0)=[1 ; -3] implies x(t)=[e^-3t ; -3e^-3t]

and that x(0)=[1 ; 1] implies x(t)=[e^t ; e^t]

find the transition matrix for the system and find A.

Answer 1

Let $X(t) = \begin{bmatrix} x_1(t) & x_2(t)\end{bmatrix}$, where $x_1(t) = e^{-3t} (1,-3)^T$, $x_2(t) = e^{t} (1,1)^T$.

Since $X$ is a solution, we have $\dot{X}(t) = A X(t)$, and, in particular, $ \dot{X}(0) = A X(0)$. Hence we get $A = \dot{X}(0) X(0)^{-1}$. We have $\dot{X}(0) = \begin{bmatrix} -3 & 1 \\ 9 & 1 \end{bmatrix}$, ${X}(0) = \begin{bmatrix} 1 & 1 \\ -3 & 1 \end{bmatrix}$, and so $A = \begin{bmatrix} 0 & 1 \\ 3 & -2 \end{bmatrix}$.

Since $Y(t)=X(t)X(t_0)^{-1}$ also solves the system, and $Y(t_0) = I$, we see that $\Phi(t,t_0) = X(t)X(t_0)^{-1}$ is the state transition matrix.

Since $X(t) = X(0) \begin{bmatrix} e^{-3t} & 0 \\ 0 & e^{t} \end{bmatrix}$, we have \begin{eqnarray} \Phi(t,t_0) &=& X(0) \begin{bmatrix} e^{-3t} & 0 \\ 0 & e^{t} \end{bmatrix} \begin{bmatrix} e^{3t_0} & 0 \\ 0 & e^{-t_0} \end{bmatrix} X(0)^{-1} \\ &=& X(0) \begin{bmatrix} e^{-3(t-t_0)} & 0 \\ 0 & e^{t-t_0} \end{bmatrix} X(0)^{-1} \\ &=& \frac{1}{4} \left( e^{-3(t-t_0)} \begin{bmatrix} 1 & -1 \\ -3 & 3 \end{bmatrix} + e^{t-t_0} \begin{bmatrix} 3 & 1 \\ 3 & 1 \end{bmatrix} \right) \end{eqnarray}