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Found in: Page 440

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# For the linear system $\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}\left[\begin{array}{cc}-2& 0\\ 3& 1\end{array}\right]\stackrel{\mathbf{\to }}{\mathbf{x}}\left(t\right)$find the matching phase portrait.

The phase portrait corresponds to $V$

See the step by step solution

## Step 1: To find the eigenvalues

Consider the linear system as follows.

$\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}-2& 0\\ 3& 1\end{array}\right]\stackrel{\to }{x}\left(t\right)$

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

$\left|A-\lambda I\right|=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}-2-\lambda & 0\\ 3& 1-\lambda \end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left(-2-\lambda \right)\left(1-\lambda \right)-3\left(0\right)=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-2+2\lambda -\lambda -0=0$

Simplify further as follows.

${\lambda }^{2}+\lambda -2=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-\lambda +2\lambda -2=0\phantom{\rule{0ex}{0ex}}\lambda \left(\lambda -1\right)+2\left(\lambda -1\right)=0\phantom{\rule{0ex}{0ex}}\left(\lambda +2\right)\left(\lambda -1\right)=0$

Simplify further as follows.

$\lambda +2=0\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=-2\phantom{\rule{0ex}{0ex}}\lambda -1=0\phantom{\rule{0ex}{0ex}}{\lambda }_{2}=1$

Therefore, the eigenvalues are ${\lambda }_{1}=-2$ and ${\lambda }_{2}=1$

## Step 2: Observation of the phase portrait

The eigenvalues are ${\lambda }_{1}=-2$ and ${\lambda }_{2}=1$

Since, the eigenvalues are real and distinct are of opposite sign.

Therefore, the phase portrait is in figure 5 as follows.

Hence, the phase portrait to the linear system $\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}-2& 0\\ 3& 1\end{array}\right]\stackrel{\to }{x}\left(t\right)$ is V.