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Q1E

Expert-verified

Linear Algebra and its Applications

Found in: Page 267

Book edition 5th

Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald

Pages 483 pages

ISBN 978-03219822384

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Short Answer

A particle moving in a planar force field has a position vector .\(x\). that satisfies \(x' = Ax\). The \(2 \times 2\) matrix \(A\) has eigenvalues 4 and 2, with corresponding eigenvectors \({v_1} = \left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right)\) and \({v_2} = \left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right)\). Find the position of the particle at a time \(t\), assuming that \(x\left( 0 \right) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\).

The required position is:

\({\rm{x}}\left( t \right) = \left( {\begin{aligned}{{20}{c}}{ - 7.5{e^{4t}} + 1.5{e^{2t}}}\\{2.5{e^{4t}} - 1.5{e^{2t}}}\end{aligned}} \right)\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

System of Differential Equations

The general solution for any system of differential equations with the eigenvalues \({\lambda _1}\) and \({\lambda _2}\) with the respective eigenvectors \({v_1}\) and \({v_2}\) is given by:

\(x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\)

Here, \({c_1}\) and \({c_2}\) are the constants from the initial condition.

Find the position of the particle

According to the question;

Consider the eigenvalues of \(A\) be \({\lambda _1} = 4\) and \({\lambda _2} = 2\) with the respective eigenvectors:

\({v_1} = \left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right)\) and \({v_2} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\1\end{aligned}} \right)\)

Then the general solution of the equation \(x' = A{\rm{x}}\) is:

\(\begin{aligned}{c}x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\\ = {c_1}\left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right){e^{4t}} + {c_2}\left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right){e^{2t}}\\ = \left( {\begin{aligned}{*{20}{c}}{ - 3{c_1}{e^{4t}} - {c_2}{e^{2t}}}\\{{c_1}{e^{4t}} + {c_2}{e^{2t}}}\end{aligned}} \right)\end{aligned}\)

Now, substituting the initial condition \(x(0) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\), we get:

\(\left( {\begin{aligned}{{20}{c}}{ - 3{c_1} - {c_2}}\\{{c_1} + {c_2}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 6}\\1\end{aligned}} \right)\)

Solving this system, we have:

\({c_1} = 2.5,\,\,\,\,{c_2} = - 1.5\)

Thus, the solution can be given as:

\(\begin{aligned}{c}x(t) = 2.5\left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right){e^{4t}} - 1.5\left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right){e^{2t}}\\ = \left( {\begin{aligned}{{20}{c}}{ - 7.5{e^{4t}} + 1.5{e^{2t}}}\\{2.5{e^{4t}} - 1.5{e^{2t}}}\end{aligned}} \right)\end{aligned}\)

Hence, this is the required position.

Most popular questions for Math Textbooks

For the matrices A find real closed formulas for the trajectory $\vec{x} (t + 1) = A \vec{x} (t)$ where $\vec{x} (0) = [\begin{matrix} 0 \\ 1 \end{matrix}]$ . Draw a rough sketch $A = [\begin{matrix} 1 & - 3 \\ 1.2 & - 2.6 \end{matrix}]$

For the matrices A in Exercises 1 through 12, find closed formulas for , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.

$1 . A = [\begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix}]$

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

17. \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\- 5&1&0&0&0\\3&8&0&0&0\\0&- 7&2&1&0\\- 4&1&9&- 2&3\end{array}} \right]\)

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

14. \(\left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right)\)

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\).

Reduce \(A\) to echelon form \(U\) with no row scaling, and use \(U\) in formula (1) (before Example 2) to compute \(\det A\). (If \(A\) happens to be singular, start over with a new random matrix.)
Compute the eigenvalues of \(A\) and the product of these eigenvalues (as accurately as possible).
List the matrix \(A\), and, to four decimal places, list the pivots in \(U\) and the eigenvalues of \(A\). Compute \(\det A\) with your matrix program, and compare it with the products you found in (a) and (b).

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