91Ó°ÊÓ

The given system of differential equations can be expressed as a matrix equation. Write the system as \( \mathbf{x}' = \mathbf{A} \mathbf{x} \), where \( \mathbf{A} \) is the coefficient matrix:\[ \mathbf{A} = \begin{pmatrix} 11 & -15 \ 6 & -8 \end{pmatrix} \]

To find the eigenvalues of \( \mathbf{A} \), solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Calculate the determinant:\[ \det(\begin{pmatrix} 11 - \lambda & -15 \ 6 & -8 - \lambda \end{pmatrix}) = (11 - \lambda)(-8 - \lambda) - (-15)(6) = 0 \]Simplify and solve the resulting quadratic equation for \( \lambda \).

The characteristic equation becomes:\[ (11 - \lambda)(-8 - \lambda) + 90 = \lambda^2 - 3\lambda + 2 = 0 \]Solve for \( \lambda \) using the quadratic formula:\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{1}}{2} \]The eigenvalues are \( \lambda_1 = 2 \) and \( \lambda_2 = 1 \).

Since \( \mathbf{A} \) has distinct eigenvalues, we can use the formula for diagonalizable matrices. If \( \mathbf{A} = \mathbf{PDP}^{-1} \), where \( \mathbf{D} \) is the diagonal matrix of eigenvalues, then \( e^{\mathbf{A}t} = \mathbf{P} e^{\mathbf{D}t} \mathbf{P}^{-1} \).First, confirm that \( \mathbf{A} \) is diagonalizable:1. Find eigenvectors corresponding to \( \lambda = 2 \) and \( \lambda = 1 \).2. Form matrix \( \mathbf{P} \) with these eigenvectors as columns.3. Compute \( e^{\mathbf{D}t} \) where \( \mathbf{D} = \begin{pmatrix} 2 & 0 \ 0 & 1 \end{pmatrix} \) and \( e^{\mathbf{D}t} = \begin{pmatrix} e^{2t} & 0 \ 0 & e^{t} \end{pmatrix} \).4. Finally calculate \( e^{\mathbf{A}t} = \mathbf{P} e^{\mathbf{D}t} \mathbf{P}^{-1} \).