Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Solve the system $\frac{d \vec{x}}{d t} = [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] \vec{x}$ with $\vec{x} (0) = [\begin{matrix} 0 \\ 1 \end{matrix}]$ . Give the solution in real form. Sketch the solution.

The solution of the system is $\vec{x} (t) = [\begin{matrix} \sin (t) \\ \sin (t) + \cos (t) \end{matrix}]$ and the graph is

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the Eigen values of the matrix.

Consider the equation $\frac{d \vec{x}}{d t} = [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] \vec{x}$ with the initial value $\vec{x} (0) = [\begin{matrix} 0 \\ 1 \end{matrix}]$ .

Compare the equations $\frac{d \vec{x}}{d t} = [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] \vec{x}$ and $\frac{d \vec{x}}{d t} = A \vec{x}$ as follows.

$A = [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}]$

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

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

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

Simplify the equation $| [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] - λ [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] | = 0$ as follows.

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

Further, simplify the equation as follows.

$\begin{matrix} (- 1 - λ) (1 - λ) + 2 = 0 \\ - 1 + λ^{2} + 2 = 0 \\ λ^{2} + 1 = 0 \\ λ^{2} = - 1 \end{matrix}$

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

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

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

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

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

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

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

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

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

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

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 $[\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}]$ for $S$ , $[\begin{matrix} - 1 & 1 \\ 1 & 0 \end{matrix}]$ for , $S^{- 1}$ $[\begin{matrix} 0 \\ 1 \end{matrix}]$ for , ${\vec{x}}_{0}$

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

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

Further, simplify the equation as follows.

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

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

Step 4: Sketch the solution.

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

Hence, the solution of the system $\frac{d \vec{x}}{d t} = [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] \vec{x}$ with the initial value $\vec{x} (0) = [\begin{matrix} 0 \\ 1 \end{matrix}]$ is $\vec{x} (t) = [\begin{matrix} \sin (t) \\ \sin (t) + \cos (t) \end{matrix}]$ 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 $\frac{d \vec{x}}{d t} = [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] \vec{x}$ with $\vec{x} (0) = [\begin{matrix} 0 \\ 1 \end{matrix}]$ . 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 value λ=i.

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

Step 4: Sketch the solution.

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Linear Algebra With Applications

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Short Answer

Solve the system dx→dt=[-11-21]x→ with x→(0)=[01]. 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 value λ=i.

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

Step 4: Sketch the solution.

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Solve the system $\frac{d \vec{x}}{d t} = [\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}] \vec{x}$ with $\vec{x} (0) = [\begin{matrix} 0 \\ 1 \end{matrix}]$ . Give the solution in real form. Sketch the solution.