Let

$$\boldsymbol{\vec{y}^{\,\prime}} = \left\lbrack \begin{array}{rr} 1 & 1 \\ 0 & -2 \end{array} \right\rbrack \boldsymbol{\vec{y}},$$ $$\Psi(t) = \left\lbrack \begin{array}{rr} e^{t} & e^{-2t} \\ 0 & -3e^{-2t} \end{array} \right\rbrack, \ \ \ \hat{\Psi}(t) = \left\lbrack \begin{array}{rr} 2e^{-2t} & 0 \\ -6e^{-2t} & 0 \end{array} \right\rbrack.$$

1. Verify that the matrix $\Psi(t)$ is a fundamental matrix of the given linear system.
   
   
2. Determine a constant matrix $C$ such that the given matrix $\hat{\Psi}(t)$ can be represented as $\hat{\Psi}(t) = \Psi(t) C$.

   $C =$

   $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right[$ $\left.\vphantom{\begin{array}{c}\!\strut\\\!\strut\\\!\strut\\\end{array}}\right]$

   
   
   
3. The determinant of the matrix $C$ is , which is Choose zero nonzero one . Therefore, the matrix $\hat{\Psi}(t)$ is Choose also a fundamental matrix simply a solution matrix .