91Ó°ÊÓ

First, find the eigenvalues of the matrix \(A\) by solving the characteristic equation \(\det(A - \lambda I) = 0 \). The matrix \(A\) is \(\begin{pmatrix}0 & 6 & 0 \1 & 0 & 1 \1 & 1 & 0 \\end{pmatrix}\). We set up \(\det(A - \lambda I)\) as \\[\det\left(\begin{pmatrix}0-\lambda & 6 & 0 \1 & 0-\lambda & 1 \1 & 1 & 0-\lambda \\end{pmatrix}\right)=0.\\]Finding the determinant gives the characteristic polynomial \(-\lambda^3 + 9\lambda - 12 = 0\). Solving this polynomial we find the eigenvalues \(\lambda_1 = -1, \lambda_2 = -2\), and \(\lambda_3 = 3\).

Next, find the eigenvectors for each eigenvalue by solving \((A - \lambda_i I)\mathbf{v} = \mathbf{0}\) for each \(\lambda_i\).1. For \(\lambda_1 = -1\): Solve \((A + I)\mathbf{v} = \mathbf{0}\) to find\[\begin{pmatrix}1 & 6 & 0 \1 & 1 & 1 \1 & 1 & 1 \\end{pmatrix}\begin{pmatrix}v_1 \v_2 \v_3 \\end{pmatrix} = \begin{pmatrix}0 \0 \0 \\end{pmatrix}.\]This gives \(v_1 = 6v_2\), \(v_2 = -v_3\). A solution vector is \( \begin{pmatrix} 6 \ -1 \ -5 \end{pmatrix} \).2. For \(\lambda_2 = -2\): Solve \((A + 2I)\mathbf{v} = \mathbf{0}\) to find\[\begin{pmatrix}2 & 6 & 0 \1 & 2 & 1 \1 & 1 & 2 \\end{pmatrix}\begin{pmatrix}v_1 \v_2 \v_3 \\end{pmatrix} =\begin{pmatrix}0 \0 \0 \\end{pmatrix}.\]This gives \(v_1 = -3v_2\), \(v_2 = v_3\). A solution vector is \( \begin{pmatrix} -3 \ 1 \ 1 \end{pmatrix} \).3. For \(\lambda_3 = 3\): Solve \((A - 3I)\mathbf{v} = \mathbf{0}\) to find\[\begin{pmatrix}-3 & 6 & 0 \1 & -3 & 1 \1 & 1 & -3 \\end{pmatrix}\begin{pmatrix}v_1 \v_2 \v_3 \\end{pmatrix} =\begin{pmatrix}0 \0 \0 \\end{pmatrix}.\]This gives \(v_1 = 2v_3\), \(v_2 = v_3\). A solution vector is \( \begin{pmatrix} 2 \ 1 \ 1 \end{pmatrix} \).

Combine the solutions from Step 2 to form the general solution of the homogeneous system. Each solution corresponds to an eigenvalue and includes a constant of integration \(c_i\) and an exponential term matching the eigenvalue:- For \(\lambda_1 = -1\), the solution is \(c_1 \begin{pmatrix} 6 \ -1 \ -5 \end{pmatrix} e^{-t}\).- For \(\lambda_2 = -2\), the solution is \(c_2 \begin{pmatrix} -3 \ 1 \ 1 \end{pmatrix} e^{-2t}\).- For \(\lambda_3 = 3\), the solution is \(c_3 \begin{pmatrix} 2 \ 1 \ 1 \end{pmatrix} e^{3t}\).Thus, the general solution is \[\mathbf{X}=c_{1}\begin{pmatrix} 6 \ -1 \ -5 \end{pmatrix} e^{-t}+c_{2}\begin{pmatrix} -3 \ 1 \ 1 \end{pmatrix} e^{-2 t}+c_{3} \begin{pmatrix} 2 \ 1 \ 1 \end{pmatrix} e^{3t}.\]

Check that substituting \(\mathbf{X}\) into the original differential equation \(\mathbf{X}' = A\mathbf{X}\) gives a true statement. This ensures the general solution satisfies the system.