91Ó°ÊÓ

We have the matrix A: $$ A = \begin{pmatrix} 5 & -1 \\ 3 & 1 \end{pmatrix} $$ To find the eigenvalues, we will solve the characteristic equation: $$ det(A - \lambda I) = \begin{vmatrix} 5-\lambda & -1 \\ 3 & 1-\lambda \end{vmatrix} = (5-\lambda)(1-\lambda) - (-1)(3) = \lambda^2 - 6\lambda + 8. $$ The eigenvalues are the roots of this quadratic equation, which are \(\lambda_1 = 2\) and \(\lambda_2 = 4\). Next, we find the eigenvectors corresponding to these eigenvalues: For \(\lambda_1 = 2\): $$ (A - 2I)\mathbf{v_1} = \begin{pmatrix} 3 & -1 \\ 3 & -1 \end{pmatrix}\mathbf{v_1} = \mathbf{0} \Rightarrow \mathbf{v_1} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} $$ For \(\lambda_2 = 4\): $$ (A - 4I)\mathbf{v_2} = \begin{pmatrix} 1 & -1 \\ 3 & -3 \end{pmatrix}\mathbf{v_2} = \mathbf{0} \Rightarrow \mathbf{v_2} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$

Using the eigenvalues and eigenvectors, we can obtain the general solution of the system: $$ \mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ 3 \end{pmatrix} e^{2t} + c_2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{4t} $$

Now, we will create the fundamental matrix by putting the solutions as columns: $$ \Psi(t) = \begin{pmatrix} e^{2t} & e^{4t} \\ 3e^{2t} & e^{4t} \end{pmatrix} $$

To find the particular fundamental matrix \(\mathbf{\Phi}(t)\) satisfying the initial condition \(\Phi(0) = \mathbf{1}\), we will multiply the fundamental matrix \(\Psi(t)\) by a constant matrix \(\mathbf{C}\): $$ \mathbf{\Phi}(t) = \Psi(t) \cdot \mathbf{C} $$ Evaluate \(\Psi(0)\) first: $$ \Psi(0) = \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} $$ Now, we want \(\Phi(0) = \Psi(0) \cdot \mathbf{C} = \mathbf{1}\), so we solve for \(\mathbf{C}\): $$ \begin{pmatrix} 1 & 1 \\ 3 & 1 \end{pmatrix} \cdot \mathbf{C} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \Rightarrow \mathbf{C} = \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} $$ Now, we have the particular fundamental matrix \(\mathbf{\Phi}(t)\): $$ \mathbf{\Phi}(t) = \begin{pmatrix} e^{2t} & e^{4t} \\ 3e^{2t} & e^{4t} \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} e^{2t}+2e^{4t} & -e^{2t}+e^{4t} \\ 3e^{2t}+2e^{4t} & -3e^{2t}+e^{4t} \end{pmatrix} $$ So the fundamental matrix is: $$ \Psi(t) = \begin{pmatrix} e^{2t} & e^{4t} \\ 3e^{2t} & e^{4t} \end{pmatrix} $$ And the particular fundamental matrix satisfying the initial condition is: $$ \mathbf{\Phi}(t) = \begin{pmatrix} e^{2t}+2e^{4t} & -e^{2t}+e^{4t} \\ 3e^{2t}+2e^{4t} & -3e^{2t}+e^{4t} \end{pmatrix} $$