Question

For the linear system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left[\begin{array}{cc}-1.5& -1\\ 2& 0.5\end{array}\right]\overrightarrow{\mathbf{x}}\left(t\right)$find the matching phase portrait.

Short answer

The phase portrait corresponds to IV.

Step-by-step solution

To find the eigenvalues

Consider the linear system as follows.

$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}-1.5& -1\\ 2& 0.5\end{array}\right]\overrightarrow{x}\left(t\right)$

To find the eigenvalues, evaluate $\left|A-\lambda I\right|=0$as follows.

$$\begin{gathered} \left|A-\lambda I\right|=0 \\ \left[\begin{array}{cc}-1.5-\lambda & -1\\ 2& 0.5-\lambda \end{array}\right]=0 \\ \left(-1.5-\lambda \right)\left(0.5-\lambda \right)-2\left(-1\right)=0 \\ -0.75+1.5\lambda -0.5\lambda +{\lambda }^2+2=0 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\lambda }^2+\lambda +1.25=0 \\ \lambda =\frac{-1\pm \sqrt{{\left(1\right)}^2-4\left(1\right)\left(1.25\right)}}{2\left(1\right)} \\ \lambda =\frac{-1\pm \sqrt{1-5}}{2} \\ \lambda =\frac{-1\pm i\sqrt{2}}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\lambda }_1=\frac{-1+2i}{2} \\ {\lambda }_1=\frac{-1}{2}+2i \\ {\lambda }_2=\frac{-1-2i}{2} \\ {\lambda }_2=\frac{-1}{2}-2i \end{gathered}$$

Therefore, the eigenvalues are ${\lambda }_1=\frac{-1}{2}+2iand{\lambda }_2=\frac{-1}{2}-2i$

Observation of the phase portrait

The eigenvalues are ${\lambda }_1=\frac{-1}{2}+2i$and ${\lambda }_2=\frac{-1}{2}-2i$also in the form as follows.

$$\begin{gathered} p+iq \\ p=-\frac{1}{2} \\ q=1 \end{gathered}$$

Since, the real part of the eigenvalues are negative.

So, the phase portrait is spiral inward in clockwise directions towards the origin.

Therefore, the phase portrait is in figure 4 as follows.

Hence, the phase portrait to the linear system$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}-1.5& -1\\ 2& 0.5\end{array}\right]\overrightarrow{x}\left(t\right)$is IV.