91Ó°ÊÓ

To find the eigenvalues, we need to find the roots of the characteristic equation given by the determinant: \(\det(A - \lambda I) = 0\), where \(A = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix}\) and \(I\) is the identity matrix. The characteristic equation is : \[ \begin{vmatrix} 3-\lambda & -2 \\ 2 & -2-\lambda \end{vmatrix} = (3-\lambda)(-2-\lambda) - (-2)(2) = \lambda^2-\lambda-2 \] Now, we find the roots (eigenvalues) of the characteristic equation: \[ \lambda^2-\lambda-2 = 0 \]

The eigenvalues of the characteristic equation are \(\lambda_1=2\) and \(\lambda_2=-1\). Now, we find the eigenvectors corresponding to each eigenvalue by solving the system \((A-\lambda I)v=0\). For \(\lambda_1=2\) : \[ (A - \lambda_1 I)v_1 = \begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] We can see that the second row is a scalar multiple of the first row, so we only need to focus on one row: \[ v_{11} - 2 v_{12} = 0 \] We can choose \(v_{12} = 1\), which gives us \(v_{11} = 2\). So, \(v_1=\begin{pmatrix} 2 \\ 1 \end{pmatrix}\). For \(\lambda_2=-1\) : \[ (A - \lambda_2 I)v_2 = \begin{pmatrix} 4 & -2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] Again, the second row is a scalar multiple of the first row, so we only need to focus on one row: \[ 4 v_{21} - 2 v_{22} = 0 \] We can simplify this to \(2v_{21} - v_{22} = 0\). Let \(v_{21}=1\). Then \(v_{22}=2\). So, \(v_2=\begin{pmatrix} 1 \\ 2 \end{pmatrix}\).

The stability of the critical point \((0,0)\) can be determined using the eigenvalues found earlier. We have \(\lambda_1=2\) and \(\lambda_2=-1\). Since one eigenvalue is positive and the other is negative, the critical point \((0,0)\) is a saddle point and is unstable.

To sketch the trajectories, we need to find the general solution of the system and simplify the expression. The general solution is given by: \[ \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \] Substituting the eigenvalues and eigenvectors, we get: \[ \mathbf{x}(t) = c_1 e^{2t} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + c_2 e^{-t} \begin{pmatrix} 1 \\ 2 \end{pmatrix} \] Since the system is unstable, the trajectories will move away from the origin. For the given system, the trajectories will be stretched along the eigenvector corresponding to the positive eigenvalue (\(\begin{pmatrix} 2 \\ 1 \end{pmatrix}\)), and compressed along the eigenvector corresponding to the negative eigenvalue (\(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\)).