Question

Sketch rough phase portraits for the dynamical systems given in Exercise 32 through 39.

$\overrightarrow{\mathbf{x}}\left(t+1\right)\mathbf{=}\left[\begin{array}{cc}0.8& -0.4\\ 0.3& 1.6\end{array}\right]\overrightarrow{\mathbf{x}}\left(t\right)$

Short answer

Thus, the rough sketch of the system in the explanation.

Step-by-step solution

Given in the question.

The given systems are:

$\overrightarrow{x}\left(t+1\right)=\left[\begin{array}{cc}0.8& -0.4\\ 0.3& 1.6\end{array}\right]\overrightarrow{x}\left(t\right)$

Solve the given system.

Solve the given system as follows:

$$\begin{gathered} \left[\begin{array}{c}{t}_1+1\\ t_2+1\end{array}\right]=\left[\begin{array}{cc}0.8& -0.4\\ 0.3& 1.6\end{array}\right]\left[\begin{array}{c}{t}_1\\ t_2\end{array}\right] \\ \left[\begin{array}{c}{t}_1+1\\ t_2+1\end{array}\right]=\left[\begin{array}{c}0.8t_1-0.4t_2\\ 0.3t_1+1.6t_2\end{array}\right] \end{gathered}$$

Now, equating both sides gives:

$$\begin{gathered} \left\{{\mathbf{t}}_{\mathbf{1}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}{\mathbf{t}}_{\mathbf{1}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{4}{\mathbf{t}}_{\mathbf{2}} \\ {\mathbf{t}}_{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}{\mathbf{t}}_{\mathbf{1}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{6}{\mathbf{t}}_{\mathbf{2}}\right. \\ \left\{\mathbf{1}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{2}{\mathbf{t}}_{\mathbf{1}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{4}{\mathbf{t}}_{\mathbf{2}} \\ \mathbf{1}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}{\mathbf{t}}_{\mathbf{1}}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{6}{\mathbf{t}}_{\mathbf{2}}\right. \end{gathered}$$

Thus the solutions is the span of$\left[\begin{array}{c}2\\ -1\end{array}\right]\left[\begin{array}{c}2\\ -3\end{array}\right]$

Determine the graph of the given system.

The graph is plotted below: