Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 5E from Chapter 5.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
In Exercises 3–6, solve the initial value problem
Step-by-Step Solution
Given Information
We have to find the position of the particle at time $t ,$ when the particle moving in a planar force field has the position vector $x$ that satisfies $x ^ { \prime } = A x$ with the condition, \[ \mathbf { x } ( 0 ) = \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] \]
Step-1:
The characteristic polynomial \[\begin{array}{l} det (A - \lambda I) = \left| {\begin{array}{*{20}{c}} {7 - \lambda }&{ - 1}\\ 3&{3 - \lambda } \end{array}} \right|\\ = (7 - \lambda )(3 - \lambda ) + 3\\ = {\lambda ^2} - 10\lambda + 24\\ = {\lambda ^2} - 6\lambda - 4\lambda + 24\\ = \lambda (\lambda - 6) - 4(\lambda - 6)\\ = (\lambda - 6)(\lambda - 4) \end{array}\] Thus, the eigenvalues are 4 and 6
Step-2: The eigenvector for eigenvalue 4
\[ \begin{aligned} ( A - 4 I ) \mathbf { v } _ { 1 } & = \mathbf { 0 } \\ \left[ \begin{array} { c c } { 7 - 4 } & { - 1 } \\ { 3 } & { 3 - 4 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = \left[ \begin{array} { c } { 0 } \\ { 0 } \end{array} \right] \\ \left[ \begin{array} { c c } { 3 } & { - 1 } \\ { 3 } & { - 1 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = \left[ \begin{array} { c } { 0 } \\ { 0 } \end{array} \right] \end{aligned} \] The general solution is: \[ \begin{aligned} \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = \left[ \begin{array} { l } { x _ { 1 } } \\ { 3 x _ { 1 } } \end{array} \right] \\ & = \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] x _ { 1 } \end{aligned} \] Hence, the eigenvector is: \[v_1 = \left[ {\begin{array}{*{20}{r}} { 1}\\ 1 \end{array}} \right]\]
Step-3: The eigenvector for eigenvalue 6
\[ \begin{aligned} ( A - 6 I ) \mathbf { v } _ { 2 } & = \mathbf { 0 } \\ \left[ \begin{array} { c c } { 7 - 6 } & { - 1 } \\ { 3 } & { 3 - 6 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] & = \left[ \begin{array} { l } { 0 } \\ { 0 } \end{array} \right] \\ \left[ \begin{array} { c c } { 1 } & { - 1 } \\ { 3 } & { - 3 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] & = \left[ \begin{array} { c } { 0 } \\ { 0 } \end{array} \right] \end{aligned} \] The general solution is: \[ \begin{aligned} \left[ \begin{array} { l } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] & = \left[ \begin{array} { l } { y _ { 1 } } \\ { y _ { 1 } } \end{array} \right] \\ & = \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] y _ { 1 } \end{aligned} \] Hence, the eigenvector is: \[v_2 = \left[ {\begin{array}{*{20}{r}} { 1}\\ 1 \end{array}} \right]\]
Step-4: The general solution
Using the eigenvalues and eigenvectors: \[ \mathbf { x } ( t ) = c _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] e ^ { 4 t } + c _ { 2 } \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] e ^ { 6 t } \]
Step-5:
Apply the initial condition: \[ \begin{array} { l } { \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] = c _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] e ^ { 4 ( 0 ) } + c _ { 2 } \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] e ^ { \phi ( 0 ) } } \\ { c _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] + c _ { 2 } \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] = \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] } \\ { \left[ \begin{array} { l } { c _ { 1 } + c _ { 2 } } \\ { 3 c _ { 1 } + c _ { 2 } } \end{array} \right] = \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] } \end{array} \] Thus, the equations are: \[ \begin{aligned} c _ { 1 } + c _ { 2 } & = 3 \\ 3 c _ { 1 } + c _ { 2 } & = 2 \end{aligned} \] Which can be solved to get: \[ c _ { 1 } = - \dfrac { 1 } { 2 } \text { and } c _ { 2 } = \dfrac { 7 } { 2 } \]
Step-6:
Since both the eigenvalues are greater than 1, so, origin is the repellor of the dynamical system.
Since the Eigen value 6 has the greatest magnitude. Therefore, the direction of greatest repulsion is the line through $v _ { 2 } = \left[ \begin{array} { l l } { 1 } \\ { 1 } \end{array} \right]$ and the origin.
We have to find the position of the particle at time $t ,$ when the particle moving in a planar force field has the position vector $x$ that satisfies $x ^ { \prime } = A x$ with the condition, \[ \mathbf { x } ( 0 ) = \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] \]
Step-1:
The characteristic polynomial \[\begin{array}{l} det (A - \lambda I) = \left| {\begin{array}{*{20}{c}} {7 - \lambda }&{ - 1}\\ 3&{3 - \lambda } \end{array}} \right|\\ = (7 - \lambda )(3 - \lambda ) + 3\\ = {\lambda ^2} - 10\lambda + 24\\ = {\lambda ^2} - 6\lambda - 4\lambda + 24\\ = \lambda (\lambda - 6) - 4(\lambda - 6)\\ = (\lambda - 6)(\lambda - 4) \end{array}\] Thus, the eigenvalues are 4 and 6
Step-2: The eigenvector for eigenvalue 4
\[ \begin{aligned} ( A - 4 I ) \mathbf { v } _ { 1 } & = \mathbf { 0 } \\ \left[ \begin{array} { c c } { 7 - 4 } & { - 1 } \\ { 3 } & { 3 - 4 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = \left[ \begin{array} { c } { 0 } \\ { 0 } \end{array} \right] \\ \left[ \begin{array} { c c } { 3 } & { - 1 } \\ { 3 } & { - 1 } \end{array} \right] \left[ \begin{array} { c } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = \left[ \begin{array} { c } { 0 } \\ { 0 } \end{array} \right] \end{aligned} \] The general solution is: \[ \begin{aligned} \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \end{array} \right] & = \left[ \begin{array} { l } { x _ { 1 } } \\ { 3 x _ { 1 } } \end{array} \right] \\ & = \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] x _ { 1 } \end{aligned} \] Hence, the eigenvector is: \[v_1 = \left[ {\begin{array}{*{20}{r}} { 1}\\ 1 \end{array}} \right]\]
Step-3: The eigenvector for eigenvalue 6
\[ \begin{aligned} ( A - 6 I ) \mathbf { v } _ { 2 } & = \mathbf { 0 } \\ \left[ \begin{array} { c c } { 7 - 6 } & { - 1 } \\ { 3 } & { 3 - 6 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] & = \left[ \begin{array} { l } { 0 } \\ { 0 } \end{array} \right] \\ \left[ \begin{array} { c c } { 1 } & { - 1 } \\ { 3 } & { - 3 } \end{array} \right] \left[ \begin{array} { c } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] & = \left[ \begin{array} { c } { 0 } \\ { 0 } \end{array} \right] \end{aligned} \] The general solution is: \[ \begin{aligned} \left[ \begin{array} { l } { y _ { 1 } } \\ { y _ { 2 } } \end{array} \right] & = \left[ \begin{array} { l } { y _ { 1 } } \\ { y _ { 1 } } \end{array} \right] \\ & = \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] y _ { 1 } \end{aligned} \] Hence, the eigenvector is: \[v_2 = \left[ {\begin{array}{*{20}{r}} { 1}\\ 1 \end{array}} \right]\]
Step-4: The general solution
Using the eigenvalues and eigenvectors: \[ \mathbf { x } ( t ) = c _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] e ^ { 4 t } + c _ { 2 } \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] e ^ { 6 t } \]
Step-5:
Apply the initial condition: \[ \begin{array} { l } { \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] = c _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] e ^ { 4 ( 0 ) } + c _ { 2 } \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] e ^ { \phi ( 0 ) } } \\ { c _ { 1 } \left[ \begin{array} { l } { 1 } \\ { 3 } \end{array} \right] + c _ { 2 } \left[ \begin{array} { l } { 1 } \\ { 1 } \end{array} \right] = \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] } \\ { \left[ \begin{array} { l } { c _ { 1 } + c _ { 2 } } \\ { 3 c _ { 1 } + c _ { 2 } } \end{array} \right] = \left[ \begin{array} { l } { 3 } \\ { 2 } \end{array} \right] } \end{array} \] Thus, the equations are: \[ \begin{aligned} c _ { 1 } + c _ { 2 } & = 3 \\ 3 c _ { 1 } + c _ { 2 } & = 2 \end{aligned} \] Which can be solved to get: \[ c _ { 1 } = - \dfrac { 1 } { 2 } \text { and } c _ { 2 } = \dfrac { 7 } { 2 } \]
Step-6:
Since both the eigenvalues are greater than 1, so, origin is the repellor of the dynamical system.
Since the Eigen value 6 has the greatest magnitude. Therefore, the direction of greatest repulsion is the line through $v _ { 2 } = \left[ \begin{array} { l l } { 1 } \\ { 1 } \end{array} \right]$ and the origin.