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Q32E

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Found in: Page 441

Linear Algebra With Applications

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

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Solve the system$\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}\mathbf{\left[}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{4}& \mathbf{0}\end{array}\mathbf{\right]}\stackrel{\mathbf{\to }}{\mathbf{x}}$ with$\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{0}\end{array}\mathbf{\right]}$ . Give the solution in real form. Sketch the solution.

The solution of the system is and the graph is

See the step by step solution

Step 1: Find the Eigen values of the matrix.

Consider the equation $\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]\stackrel{\to }{x}$with the initial value$\stackrel{\to }{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$ .

Compare the equations $\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]\stackrel{\to }{x}$and$\frac{d\stackrel{\to }{x}}{dt}=A\stackrel{\to }{x}$ as follows.

$A=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]$

Assume$\lambda$ is an Eigen value of the matrix $\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]$implies$|A-\lambda I|=0$ .

Substitute the values $\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]$for$A$ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for$I$ in the equation$|A-\lambda I|=0$ as follows.

$\begin{array}{c}|A-\lambda I|=0\\ |\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\end{array}$

Simplify the equation $|\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0$as follows.

$\begin{array}{c}|\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\\ |\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]|=0\\ |\left[\begin{array}{cc}-\lambda & 1\\ -4& -\lambda \end{array}\right]|=0\\ {\lambda }^{2}+4=0\end{array}$

Therefore, the Eigen values of $A$are $\lambda =±i2$.

Step 2: Determine the Eigen vector corresponding to the Eigen valueλ=i2 .

Substitute the values$i2$ for$\lambda$ in the equation$|\left[\begin{array}{cc}-\lambda & 1\\ -4& -\lambda \end{array}\right]|=0$ as follows.

$\begin{array}{c}|\left[\begin{array}{cc}-\lambda & 1\\ -4& -\lambda \end{array}\right]|=0\\ |\left[\begin{array}{cc}-2i& 1\\ -4& -2i\end{array}\right]|=0\end{array}$

As ${E}_{-1+2i}=\mathrm{ker}\left[\begin{array}{cc}-2i& 1\\ -4& -2i\end{array}\right]=span\left[\begin{array}{c}-2i\\ 1\end{array}\right]$, the values $\stackrel{\to }{v}+i\stackrel{\to }{w}$is defined as follows.

$\stackrel{\to }{v}+i\stackrel{\to }{w}=\left[\begin{array}{c}0\\ 1\end{array}\right]+i\left[\begin{array}{c}-2\\ 0\end{array}\right]$

Therefore, the value of $S$is$S=\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]$ .

Step 3: Determine the solution for dx→dt=[01-40]x→

The inverse of the matrix$S=\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]$ is defined as follows.

$\begin{array}{c}{S}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]\\ {S}^{-1}=\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]\end{array}$

As $\stackrel{\to }{x}\left(t\right)={e}^{pt}S\left[\begin{array}{cc}\mathrm{cos}\left(qt\right)& -\mathrm{sin}\left(qt\right)\\ \mathrm{sin}\left(qt\right)& \mathrm{cos}\left(qt\right)\end{array}\right]{S}^{-1}{\stackrel{\to }{x}}_{0}$, Substitute the value$\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]$ for$S$ , $\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]$for ${S}^{-1}$, $\left[\begin{array}{c}1\\ 0\end{array}\right]$for ${\stackrel{\to }{x}}_{0}$,$0$ for$p$ and$2$ for$q$ in the equation

$\stackrel{\to }{x}\left(t\right)={e}^{pt}S\left[\begin{array}{cc}\mathrm{cos}\left(qt\right)& -\mathrm{sin}\left(qt\right)\\ \mathrm{sin}\left(qt\right)& \mathrm{cos}\left(qt\right)\end{array}\right]{S}^{-1}{\stackrel{\to }{x}}_{0}$as follows.

$\begin{array}{c}\stackrel{\to }{x}\left(t\right)={e}^{pt}S\left[\begin{array}{cc}\mathrm{cos}\left(qt\right)& -\mathrm{sin}\left(qt\right)\\ \mathrm{sin}\left(qt\right)& \mathrm{cos}\left(qt\right)\end{array}\right]{S}^{-1}{\stackrel{\to }{x}}_{0}\\ \stackrel{\to }{x}\left(t\right)={e}^{\left(0\right)t}\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\mathrm{cos}\left(2t\right)& -\mathrm{sin}\left(2t\right)\\ \mathrm{sin}\left(2t\right)& \mathrm{cos}\left(2t\right)\end{array}\right]\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{\to }{x}\left(t\right)=\left[\begin{array}{cc}-2\mathrm{cos}\left(2t\right)& 2\mathrm{sin}\left(2t\right)\\ \mathrm{sin}\left(2t\right)& \mathrm{cos}\left(2t\right)\end{array}\right]\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{\to }{x}\left(t\right)=\left[\begin{array}{cc}\mathrm{cos}\left(2t\right)& 2\mathrm{sin}\left(2t\right)\\ -\frac{\mathrm{sin}\left(2t\right)}{2}& \mathrm{cos}\left(2t\right)\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Further, simplify the equation as follows.

$\begin{array}{l}\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{cc}\mathrm{cos}\left(2t\right)& 2\mathrm{sin}\left(2t\right)\\ -\frac{\mathrm{sin}\left(2t\right)}{2}& \mathrm{cos}\left(2t\right)\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(2t\right)\\ -\frac{\mathrm{sin}\left(2t\right)}{2}\end{array}\right]\end{array}$

Therefore, the solution of the system is$\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(2t\right)\\ -\frac{\mathrm{sin}\left(2t\right)}{2}\end{array}\right]$ .

Step 4: Sketch the solution.

As${\lambda }_{1,2}=±2i$ , draw the graph of the solution$\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(2t\right)\\ -\frac{\mathrm{sin}\left(2t\right)}{2}\end{array}\right]$ as follows.

Hence, the solution of the system$\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]\stackrel{\to }{x}$ with the initial value$\stackrel{\to }{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$ is$\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}\mathrm{cos}\left(2t\right)\\ -\frac{\mathrm{sin}\left(2t\right)}{2}\end{array}\right]$ and the graph of the solution is an ellipse in counterclockwise direction.

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