Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Solve the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] \vec{x}$ with $\vec{x} (0) = [\begin{matrix} 1 \\ 0 \end{matrix}]$ . 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 by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the Eigen values of the matrix.

Consider the equation $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] \vec{x}$ with the initial value $\vec{x} (0) = [\begin{matrix} 1 \\ 0 \end{matrix}]$ .

Compare the equations $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] \vec{x}$ and $\frac{d \vec{x}}{d t} = A \vec{x}$ as follows.

$A = [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}]$

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

Substitute the values $[\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}]$ for $A$ and $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ for $I$ in the equation $| A - λ I | = 0$ as follows.

$\begin{matrix} | A - λ I | = 0 \\ | [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] - λ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] | = 0 \end{matrix}$

Simplify the equation $| [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] - λ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] | = 0$ as follows.

$\begin{matrix} | [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] - λ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] | = 0 \\ | [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] - [\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}] | = 0 \\ | [\begin{matrix} - λ & 1 \\ - 4 & - λ \end{matrix}] | = 0 \\ λ^{2} + 4 = 0 \end{matrix}$

Therefore, the Eigen values of $A$ are $λ = \pm i 2$ .

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

Substitute the values $i 2$ for $λ$ in the equation $| [\begin{matrix} - λ & 1 \\ - 4 & - λ \end{matrix}] | = 0$ as follows.

$\begin{matrix} | [\begin{matrix} - λ & 1 \\ - 4 & - λ \end{matrix}] | = 0 \\ | [\begin{matrix} - 2 i & 1 \\ - 4 & - 2 i \end{matrix}] | = 0 \end{matrix}$

As $E_{- 1 + 2 i} = \ker [\begin{matrix} - 2 i & 1 \\ - 4 & - 2 i \end{matrix}] = s p a n [\begin{matrix} - 2 i \\ 1 \end{matrix}]$ , the values $\vec{v} + i \vec{w}$ is defined as follows.

$\vec{v} + i \vec{w} = [\begin{matrix} 0 \\ 1 \end{matrix}] + i [\begin{matrix} - 2 \\ 0 \end{matrix}]$

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

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

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

$\begin{matrix} S^{- 1} = \frac{1}{- 2} [\begin{matrix} 1 & 0 \\ 0 & - 2 \end{matrix}] \\ S^{- 1} = [\begin{matrix} \frac{1}{- 2} & 0 \\ 0 & 1 \end{matrix}] \end{matrix}$

As $\vec{x} (t) = e^{p t} S [\begin{matrix} \cos (q t) & - \sin (q t) \\ \sin (q t) & \cos (q t) \end{matrix}] S^{- 1} {\vec{x}}_{0}$ , Substitute the value $[\begin{matrix} - 2 & 0 \\ 0 & 1 \end{matrix}]$ for $S$ , $[\begin{matrix} \frac{1}{- 2} & 0 \\ 0 & 1 \end{matrix}]$ for $S^{- 1}$ , $[\begin{matrix} 1 \\ 0 \end{matrix}]$ for ${\vec{x}}_{0}$ , $0$ for $p$ and $2$ for $q$ in the equation

$\vec{x} (t) = e^{p t} S [\begin{matrix} \cos (q t) & - \sin (q t) \\ \sin (q t) & \cos (q t) \end{matrix}] S^{- 1} {\vec{x}}_{0}$ as follows.

$\begin{matrix} \vec{x} (t) = e^{p t} S [\begin{matrix} \cos (q t) & - \sin (q t) \\ \sin (q t) & \cos (q t) \end{matrix}] S^{- 1} {\vec{x}}_{0} \\ \vec{x} (t) = e^{(0) t} [\begin{matrix} - 2 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} \cos (2 t) & - \sin (2 t) \\ \sin (2 t) & \cos (2 t) \end{matrix}] [\begin{matrix} \frac{1}{- 2} & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 \\ 0 \end{matrix}] \\ \vec{x} (t) = [\begin{matrix} - 2 \cos (2 t) & 2 \sin (2 t) \\ \sin (2 t) & \cos (2 t) \end{matrix}] [\begin{matrix} \frac{1}{- 2} & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 \\ 0 \end{matrix}] \\ \vec{x} (t) = [\begin{matrix} \cos (2 t) & 2 \sin (2 t) \\ - \frac{\sin (2 t)}{2} & \cos (2 t) \end{matrix}] [\begin{matrix} 1 \\ 0 \end{matrix}] \end{matrix}$

Further, simplify the equation as follows.

$\begin{array}{l} \vec{x} (t) = [\begin{matrix} \cos (2 t) & 2 \sin (2 t) \\ - \frac{\sin (2 t)}{2} & \cos (2 t) \end{matrix}] [\begin{matrix} 1 \\ 0 \end{matrix}] \\ \vec{x} (t) = [\begin{matrix} \cos (2 t) \\ - \frac{\sin (2 t)}{2} \end{matrix}] \end{array}$

Therefore, the solution of the system is $\vec{x} (t) = [\begin{matrix} \cos (2 t) \\ - \frac{\sin (2 t)}{2} \end{matrix}]$ .

Step 4: Sketch the solution.

As $λ_{1, 2} = \pm 2 i$ , draw the graph of the solution $\vec{x} (t) = [\begin{matrix} \cos (2 t) \\ - \frac{\sin (2 t)}{2} \end{matrix}]$ as follows.

Hence, the solution of the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] \vec{x}$ with the initial value $\vec{x} (0) = [\begin{matrix} 1 \\ 0 \end{matrix}]$ is $\vec{x} (t) = [\begin{matrix} \cos (2 t) \\ - \frac{\sin (2 t)}{2} \end{matrix}]$ and the graph of the solution is an ellipse in counterclockwise direction.

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Linear Algebra With Applications

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Short Answer

Solve the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] \vec{x}$ with $\vec{x} (0) = [\begin{matrix} 1 \\ 0 \end{matrix}]$ . Give the solution in real form. Sketch the solution.

Step by Step Solution

Step 1: Find the Eigen values of the matrix.

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

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

Step 4: Sketch the solution.

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Linear Algebra With Applications

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Short Answer

Solve the systemdx→dt=[01-40]x→ withx→(0)=[10] . Give the solution in real form. Sketch the solution.

Step by Step Solution

Step 1: Find the Eigen values of the matrix.

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

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

Step 4: Sketch the solution.

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Solve the system $\frac{d \vec{x}}{d t} = [\begin{matrix} 0 & 1 \\ - 4 & 0 \end{matrix}] \vec{x}$ with $\vec{x} (0) = [\begin{matrix} 1 \\ 0 \end{matrix}]$ . Give the solution in real form. Sketch the solution.