91Ó°ÊÓ

The given matrix equation is \( X^{\prime} = A X \), where \[ A = \begin{pmatrix} 4 & 0 & 1 \ 0 & 6 & 0 \ -4 & 0 & 4 \end{pmatrix}. \] This represents a system of linear differential equations.

To find the eigenvalues of the matrix \( A \), solve the characteristic equation \( \det(A - \lambda I) = 0 \). Thus, \[ \det \begin{pmatrix} 4 - \lambda & 0 & 1 \ 0 & 6 - \lambda & 0 \ -4 & 0 & 4 - \lambda \end{pmatrix} = 0. \]Calculating the determinant:\( (4-\lambda)[(6-\lambda)(4-\lambda)] = 0 \) which simplifies to \( (4 - \lambda)^2(6 - \lambda) = 0 \).Therefore, the eigenvalues are \( \lambda_1 = 4, \lambda_2 = 4, \lambda_3 = 6 \).

Let's find the eigenvectors for the eigenvalues:- For \( \lambda_1 = 6 \): Solve \( (A - 6I)\mathbf{v} = 0 \) which leads to \( \begin{pmatrix} -2 & 0 & 1 \ 0 & 0 & 0 \ -4 & 0 & -2 \end{pmatrix} \mathbf{v} = \mathbf{0} \). Solving this system, we get \( \mathbf{v_1} = \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} \).- For \( \lambda_2 = 4 \): Solve \( (A - 4I)\mathbf{v} = 0 \) leading to \( \begin{pmatrix} 0 & 0 & 1 \ 0 & 2 & 0 \ -4 & 0 & 0 \end{pmatrix} \mathbf{v} = \mathbf{0} \). A solution to this system is \( \mathbf{v_2} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \).- The second eigenvector for \( \lambda_3 = 4 \) results from the same calculation as \( \lambda_2 \), so it is \( \mathbf{v_3} = \begin{pmatrix} -1 \ 0 \ 0 \end{pmatrix} \).

Using the eigenvalues and eigenvectors, the general solution to the system is a linear combination of solutions of the form \( e^{\lambda t} \mathbf{v} \).Thus, the general solution is:\[ X(t) = c_1 e^{6t} \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix} + c_2 e^{4t} \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} + c_3 e^{4t} \begin{pmatrix} -1 \ 0 \ 0 \end{pmatrix} \]where \( c_1, c_2, \) and \( c_3 \) are arbitrary constants.