7.5 p 405,# 12
Solve the following equation.
\begin{equation*}
x' = \begin{pmatrix} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{pmatrix}x
\end{equation*}
First, find the eigenvalues.
\begin{equation*}
\begin{pmatrix} 3-\lambda & 2 & 4 \\2 & -\lambda & 2 \\ 4 & 2 & 3-\lambda\end{pmatrix} = 0 \notag
\end{equation*}
The characteristic equation is:
\begin{gather}
\lambda^3 - 6\lambda^2 - 15\lambda - 8 = 0\\
(1 + \lambda)( (1 + \lambda)(8 - \lambda) = 0\\
\end{gather}
Therefore, the eigenvalues are $ \lambda = 8,\ \lambda = -1,\ \lambda = -1$.
For $ \lambda = 8$:
\begin{equation*}
\begin{pmatrix}
-5 & 2 & 4
\\2 & -8 & 2
\\ 4 & 2 & -5\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
The corresponding eigenvector is $v_1 = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$
For $ \lambda = -1$:
\begin{equation*}
\begin{pmatrix}
4 & 2 & 4
\\2 & 1 & 2
\\ 4 & 2 & 4\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{equation*}
The corresponding eigenvectors are $v_2 = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$ and $v_3 = \begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}$
Therefore, the general solution is:
\begin{equation*}x = c_1\begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}e^{8t} + c_2\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}e^{-t} + c_3\begin{pmatrix} 0 \\ -2 \\ 1 \end{pmatrix}e^{-t} \end{equation*}
