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Expert-verified Found in: Page 441 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Solve the system $\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}\mathbf{\left[}\begin{array}{cc}\mathbf{-}\mathbf{1}& \mathbf{1}\\ \mathbf{-}\mathbf{2}& \mathbf{1}\end{array}\mathbf{\right]}\stackrel{\mathbf{\to }}{\mathbf{x}}$ with $\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{1}\end{array}\mathbf{\right]}$. Give the solution in real form. Sketch the solution.

The solution of the system is$\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{c}\mathrm{sin}\left(t\right)\\ \mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right)\end{array}\right]$ and the graph is See the step by step solution

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

Consider the equation$\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]\stackrel{\to }{x}$ with the initial value$\stackrel{\to }{x}\left(0\right)=\left[\begin{array}{c}0\\ 1\end{array}\right]$ .

Compare the equations$\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]\stackrel{\to }{x}$ and $\frac{d\stackrel{\to }{x}}{dt}=A\stackrel{\to }{x}$as follows.

$A=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]$

Assume $\lambda$is an Eigen value of the matrix$\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]$ implies $|A-\lambda I|=0$.

Substitute the values$\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]$ 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.

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

Simplify the equation$|\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0$ as follows.

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

Further, simplify the equation as follows.

$\begin{array}{c}\left(-1-\lambda \right)\left(1-\lambda \right)+2=0\\ -1+{\lambda }^{2}+2=0\\ {\lambda }^{2}+1=0\\ {\lambda }^{2}=-1\end{array}$

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

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

Substitute the values$i$ for$\lambda$ in the equation $|\left[\begin{array}{cc}-1-\lambda & 1\\ -2& 1-\lambda \end{array}\right]|=0$as follows.

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

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

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

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

## Step 3: Determine the solution for dx→dt=[-11-21]x→.

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

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

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$\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$ for $S$, $\left[\begin{array}{cc}-1& 1\\ 1& 0\end{array}\right]$for ,${S}^{-1}$ $\left[\begin{array}{c}0\\ 1\end{array}\right]$for ,${\stackrel{\to }{x}}_{0}$

$0$ for$p$ and$1$ 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.

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

Further, simplify the equation as follows.

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

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

## Step 4: Sketch the solution.

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

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