Given Information
We are given with the matrix differential equation $\mathbf { x } ^ { \prime } = A \mathbf { x }$ where, \[ A = \left[ \begin{array} { c c } { 3 } & { 1 } \\ { - 2 } & { 1 } \end{array} \right] \] We have to find the general solution of the differential equation.

Step-1:
The characteristic polynomial \[ \begin{aligned} \operatorname { det } ( A - \lambda I ) & = \left| \begin{array} { c c } { 3 - \lambda } & { 1 } \\ { - 2 } & { 1 - \lambda } \end{array} \right| \\ & = ( 3 - \lambda ) ( 1 - \lambda ) + 2 \\ & = 3 - \lambda - 3 \lambda + \lambda ^ { 2 } + 2 \\ & = \lambda ^ { 2 } - 4 \lambda + 5 \end{aligned} \] The eigenvalues can be found as: \[\begin{array}{l} \lambda = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)(5)} }}{{2(1)}}\\ = \dfrac{{4 \pm \sqrt {16 - 20} }}{2}\\ = \dfrac{{4 \pm \sqrt { - 4} }}{2}\\ = \dfrac{{4 \pm 2i}}{2}\\ = 2 \pm i \end{array}\] Thus, the eigenvalues are 2+i and 2-i

Step-2: The eigenvector for eigenvalue 2-i
\[\begin{array}{l} (A - (2 - i)I){{\bf{v}}_1} = {\bf{0}}\\ \left[ {\begin{array}{*{20}{c}} {3 - (2 - i)}&1\\ { - 2}&{1 - (2 - i)} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} {1 + i}&1\\ { - 2}&{ - 1 + i} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right] \end{array}\] The system of equation are: \[\begin{array}{l} (1 + i){x_1} + {x_2} = 0\\ - 2{x_1} + ( - 1 + i){x_2} = 0 \end{array}\] The general solution of above equations: \[ \begin{aligned} \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = \left[ \begin{array} { c } { \dfrac { ( - 1 + i ) } { 2 } x _ { 2 } } \\ { x _ { 2 } } \end{array} \right] \\ & = \left[ \dfrac { ( - 1 + i ) } { 2 } \right] x _ { 2 } \end{aligned} \] Therefore, the eigenvector is: \[{v_1} = \left[ {\begin{array}{*{20}{r}} {-1+ i }\\ 2 \end{array}} \right]\]

Step-3: The eigenvector for eigenvalue 2-i
By the conjugate theorem, the eigenvector is: \[{v_2} = \left[ {\begin{array}{*{20}{r}} {-1 - i }\\ 2 \end{array}} \right]\]

Step-4: The general solution
Using the eigenvalues and eigenvectors: \[\begin{array}{l} {{\bf{x}}_1} = {{\bf{v}}_1}{e^{(2 + i)t}}\\ {{\bf{x}}_2} = {{\bf{v}}_2}{e^{(2 - i)t}} \end{array}\]

Step-5: Upon substitution for $x_1$
\[\begin{array}{l} {{\bf{v}}_1}{e^{(2 + i)t}} = \left[ {\begin{array}{*{20}{c}} { - 1 - i}\\ 2 \end{array}} \right]{e^{2t}}{e^{it}}\\ = \left[ {\begin{array}{*{20}{c}} { - 1 - i}\\ 2 \end{array}} \right]{e^{2t}}(\cos t + i\sin t)\\ = \left[ {\begin{array}{*{20}{c}} { - \cos t - i\cos t - i\sin t - {i^2}\sin t}\\ {2\cos t + 2i\sin t} \end{array}} \right]{e^{2t}}\\ = \left[ {\begin{array}{*{20}{c}} { - \cos t - i\cos t - i\sin t + \sin t}\\ {2\cos t + 2i\sin t} \end{array}} \right]{e^{2t}}\\ = \left[ {\begin{array}{*{20}{c}} { - \cos t + \sin t}\\ {2\cos t} \end{array}} \right]{e^{2t}} + i\left[ {\begin{array}{*{20}{c}} { - \cos t - \sin t}\\ {2\sin t} \end{array}} \right]{e^{2t}} \end{array}\] Thus, the general solution is: \[ \mathbf { x } ( t ) = c _ { 1 } \left[ \begin{array} { c } { - \cos t + \sin t } \\ { 2 \cos t } \end{array} \right] e ^ { 2 t } + c _ { 2 } \left[ \begin{array} { c } { - \cos t - \sin t } \\ { 2 \sin t } \end{array} \right] e ^ { 2 t } \] The eigenvalues are complex and the real parts of the eigenvalues are positive, thus, $e ^ { 2 t } \rightarrow \infty$ as $t \rightarrow \infty$. Therefore,

The trajectories are spiral and the spirals tend away from the origin.