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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 $\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{-}\mathbf{1}& \mathbf{-}\mathbf{2}\\ \mathbf{2}& \mathbf{-}\mathbf{1}\end{array}\mathbf{\right]}\stackrel{\mathbf{\to }}{\mathbf{x}}$with$\stackrel{\mathbf{\to }}{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{-}\mathbf{1}\end{array}\mathbf{\right]}$ . Give the solution in real form. Sketch the solution.

The solution of the system is$\stackrel{\to }{x}\left(t\right)={e}^{t}\left[\begin{array}{c}\mathrm{cos}\left(2t\right)+\mathrm{sin}\left(2t\right)\\ \mathrm{sin}\left(2t\right)-\mathrm{cos}\left(2t\right)\end{array}\right]$ and the graph is: See the step by step solution

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

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

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

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

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

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

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

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

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

Further, simplify the equation as follows.

$\begin{array}{c}\left(-1-\lambda \right)\left(-1-\lambda \right)+4=0\\ 1+2\lambda +{\lambda }^{2}+4=0\\ {\lambda }^{2}+2\lambda +5=0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}\lambda =\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\\ \lambda =\frac{-\left(2\right)±\sqrt{{\left(2\right)}^{2}-4\left(1\right)\left(5\right)}}{2\left(1\right)}\\ \lambda =\frac{-2±\sqrt{4-20}}{2}\\ \lambda =\frac{-2±4i}{2}\end{array}$

Therefore, the Eigen values of $A$are $\lambda =-1±i2$.

## Step 2: Determine the Eigen vector corresponding to the Eigen value λ=-1+i2

Substitute the values$-1+i2$ for$\lambda$ in the equation $|\left[\begin{array}{cc}-1-\lambda & -2\\ 2& -1-\lambda \end{array}\right]|=0$as follows.

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

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

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

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

## Step 3: Determine the solution for  dx→dt=[-1-22-1]x→

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

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

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

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

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

## Step 4: Sketch the solution

As ${\lambda }_{1,2}=-1±2i$and$-1<0$ , draw the graph of the solution$\stackrel{\to }{x}\left(t\right)={e}^{-t}\left[\begin{array}{c}\mathrm{cos}\left(2t\right)+\mathrm{sin}\left(2t\right)\\ \mathrm{sin}\left(2t\right)-\mathrm{cos}\left(2t\right)\end{array}\right]$ as follows. Hence the solution of the system $\frac{d\stackrel{\to }{x}}{dt}=\left[\begin{array}{cc}-1& -2\\ 2& -1\end{array}\right]\stackrel{\to }{x}$with the initial value$\stackrel{\to }{x}\left(0\right)=\left[\begin{array}{c}1\\ -1\end{array}\right]$ is $\stackrel{\to }{x}\left(t\right)={e}^{-t}\left[\begin{array}{c}\mathrm{cos}\left(2t\right)+\mathrm{sin}\left(2t\right)\\ \mathrm{sin}\left(2t\right)-\mathrm{cos}\left(2t\right)\end{array}\right]$and the graph of the solution is a spirals in counterclockwise direction toward the origin. ### Want to see more solutions like these? 