$$\textbf{x}'=\begin{pmatrix}3&-2\\4&-1\end{pmatrix}\textbf{x}$$
(a)$\\$
Find eigenvalues with $\det(\boldsymbol{A}-\lambda\boldsymbol{I_2})=0:$
$$\begin{aligned}\begin{array}{|c c|}3-\lambda&-2\\4&-1-\lambda\end{array}=0\\(3-\lambda)(-1-\lambda)-(-2)(4)=0\\\lambda^2-
2\lambda+5=0\end{aligned}$$
Since $\lambda={2\pm\sqrt{(-2)^2-4\cdot1\cdot 5}\over2},$ we have $\cases{\lambda_1=1+2i\\\lambda_2=1-2i}$ $\\$
Find egienvectors with $(\boldsymbol{A}-\lambda\boldsymbol{I_2})\textbf{x}=\boldsymbol{0}$, where $\textbf{x}$ represents the
eigenvectors.$\\$
When $\lambda=1+2i,$ $$\begin{pmatrix}2-2i&-2\\4&-2-
2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-2i&-2\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1-i\end{pmatrix}\text{where the eigenvector is}\space
\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1-i\end{pmatrix}.$$ Thus one solution of the given system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1-i\end{pmatrix}e^{(1+2i)t}.$ $\\$
When $\lambda=1+2i,$ $$\begin{pmatrix}2-2i&-2\\4&-2-
2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-2i&-2\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1-i\end{pmatrix}\text{where the eigenvector is}\space
\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1-i\end{pmatrix}.$$ Thus one solution of the given system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1-i\end{pmatrix}e^{(1+2i)t}.$ $\\$
When $\lambda=1-2i,$ $$\begin{pmatrix}2-2i&-2\\4&-2-
2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-2i&-2\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1-i\end{pmatrix}\text{where the eigenvector is}\space
\boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\1-i\end{pmatrix}.$$ Thus one solution of the given system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1-i\end{pmatrix}e^{(1+2i)t}.$ $\\$
To obtain a set of real-valued solution:
$$\begin{align}\textbf{x}^{(1)}(t)&=\begin{pmatrix}1\\1-i\end{pmatrix}e^t(\cos t+i\sin t)\\&=\begin{pmatrix}\cos 2t\\\cos 2t+\sin
2t\end{pmatrix}e^t+i\begin{pmatrix}\sin 2t\\\sin 2t-\cos 2t\end{pmatrix}e^t \\&=\boldsymbol{u}(t)+i\boldsymbol{v}(t)\end{align}$$
By the Theorem$\space 7.4.5$, we know that $\boldsymbol{u}(t)\space\text{and}\space\boldsymbol{v}(t)$ are also solutions of given
equation. $\\$
Therefore, the general solution of the given system equation expressed in terms of real-valued functions is
$$\textbf{x}(t)=c_1e^t\begin{pmatrix}\cos 2t\\\cos 2t+\sin 2t\end{pmatrix}+c_2e^t\begin{pmatrix}\sin 2t\\\sin 2t-\cos
2t\end{pmatrix} $$
(b)$\\$
All solutions approach to $\infty$ (unbounded) as $t\rightarrow \infty$.
