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Expert-verified Found in: Page 426 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Question: Consider the system $\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]{\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}$ where${\mathbit{A}}{\mathbf{=}}\mathbf{\left[}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ 0& \mathbf{0}\end{array}\mathbf{\right]}$ . Sketch a direction field for Base on your sketch, describe the trajectories geometrically. Can you find the solution analytically?

The solution of the system is $\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}{c}_{1}+{c}_{2}+{c}_{2}t\\ {c}_{2}\end{array}\right]$ .

See the step by step solution

## Step 1: Determine the Eigen values of the matrix.

Consider the equation $\frac{d\stackrel{\to }{x}}{dt}=A\stackrel{\to }{x}$with $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$.

Assume is an Eigen value of the matrix $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ implies .

Substitute the values $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$for A androle="math" localid="1660643543411" $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for in the equation as follows$\left|A-\lambda I\right|=0$.

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

Simplify the equation$\left|\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$ as follows.

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

Therefore, the Eigen values of A are $\lambda =0$.

## Step 2: Determine the Eigen vector corresponding to the Eigen value λ=0

Assume ${v}_{1}=\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]and{v}_{2}=\left[\begin{array}{c}{x}_{2}\\ {y}_{2}\end{array}\right]$ are Eigen vector corresponding to$\lambda =0,0$ implies $\left|A-{\lambda }_{1}I\right|{v}_{1}=0and\left|A-{\lambda }_{2}I\right|{v}_{2}=0$ .

Substitute the values $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$for A,0 for role="math" localid="1660644995182" ${\lambda }_{1}\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]$, for v1 and for in the equation$\left|A-{\lambda }_{1}I\right|{v}_{1}=0$ as follows.

$\left|A-{\lambda }_{1}I\right|{v}_{1}=0\phantom{\rule{0ex}{0ex}}\left|\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]-0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left|\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right|\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}{y}_{1}\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}$

As is chosen to be arbitrary, assume x1 = 1 implies $\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}$

Therefore, the Eigen vector corresponding to role="math" localid="1660645072341" ${\lambda }_{1}=0is\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ .

Substitute the values$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ for A, 0 for , for and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for in the equation as follows.

As is chosen to be arbitrary, assume implies $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

Therefore, the Eigen vector corresponding to ${\lambda }_{2}=0is\left[\begin{array}{c}{x}_{2}\\ {y}_{2}\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right]$ .

The Eigen vectors are and corresponding to the Eigen valu${\lambda }_{2}\left[\begin{array}{c}{x}_{2}\\ {y}_{2}\end{array}\right]$es ${v}_{1}=\left[\begin{array}{c}1\\ 0\end{array}\right]and{v}_{2}\left[\begin{array}{c}1\\ 1\end{array}\right]$respectively.

## Step 3: Find the general solution for x→(t)

The general solution of$\stackrel{\to }{x}\left(t\right)$ is$\stackrel{\to }{x}\left(t\right)={c}_{1}{v}_{1}+{c}_{2}\left\{t{v}_{2}+{v}_{1}\right\}$ .

Substitute the value $\left[\begin{array}{c}1\\ 0\end{array}\right]$for v1 and $\left[\begin{array}{c}1\\ 1\end{array}\right]$for v2 in the equation $\stackrel{\to }{x}\left(t\right)={c}_{1}{v}_{1}+{c}_{2}\left\{t{v}_{2}+{v}_{1}\right\}$as follows.

$\stackrel{\to }{x}\left(t\right)={c}_{1}{v}_{1}+{c}_{2}\left\{t{v}_{2}+{v}_{1}\right\}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}\left(t\right)={c}_{1}\left[\begin{array}{c}1\\ 0\end{array}\right]+{c}_{2}\left\{t\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}1\\ 0\end{array}\right]\right\}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}{c}_{1}\\ 0\end{array}\right]+{c}_{2}\left\{t\left[\begin{array}{c}{c}_{2}t\\ {c}_{2}\end{array}\right]+\left[\begin{array}{c}{c}_{2}\\ 0\end{array}\right]\right\}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}{c}_{1}+{c}_{2}+{c}_{2}t\\ {c}_{2}\end{array}\right]$

## Step 3: Sketch the direction field graph ofAx→ .

As${\lambda }_{1,2}=0$ , draw the direction field graph of the function as follows. Hence, the formula for the solution of the system$\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\stackrel{\to }{x}$ is and the direction field graph is$\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}{c}_{1}+{c}_{2}+{c}_{2}t\\ {c}_{2}\end{array}\right]$ sketched. ### Want to see more solutions like these? 