/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q15E In Exercises 9-18, construct the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 9-18, construct the general solution of \(x' = Ax\) involving complex eigenfunctions and then obtain the general real solution. Describe the shapes of typical trajectories.

15. (M) \(A = \left( {\begin{aligned}{ {20}{c}}{ - 8}&{ - 12}&{ - 6}\\2&1&2\\7&{12}&5\end{aligned}} \right)\)

Short Answer

Expert verified

The general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 4}\\1\\4\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - 6}\\1\\5\end{aligned}} \right){e^{ - t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\0\\1\end{aligned}} \right){e^{ - 2t}}\). It is observed that the origin is a saddle point. A solution of \({c_1} = 0\) attracts the origin but a solution of \({c_2} = {c_3} = 0\) repels.

Step by step solution

01

Determine the eigenvalues and eigenvector of the matrix

The given matrix is \(A = \left( {\begin{aligned}{ {20}{c}}{ - 8}&{ - 12}&{ - 6}\\2&1&2\\7&{12}&5\end{aligned}} \right)\).

Use the MATLAB code to compute the eigenvalues of the matrix \(A\) as shown below:

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( { - 8\,\,\, - 12\,\,\,\, - 6;\,2\,\,\,\,1\,\,\,\,2;\,7\,\,\,12\,\,\,5} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

\({\mathop{\rm ev}\nolimits} = \left( {\begin{aligned}{ {20}{c}}{1.0000}\\{ - 1.0000}\\{ - 2.0000}\end{aligned}} \right)\)

Use the MATLAB code to compute the eigenvector of the matrix A as shown below:

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 1 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{ - 1.0000}\\{0.2500}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}{ - 4}\\1\\4\end{aligned}} \right)\).

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 2 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{ - 1.2000}\\{0.2000}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}{ - 6}\\1\\5\end{aligned}} \right)\).

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 3 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{ - 1.0000}\\{0.0000}\\{ - 1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _3} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\0\\1\end{aligned}} \right)\).

02

Construct the general solution of \(x' = Ax\)

Therefore, the general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 4}\\1\\4\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{ - 6}\\1\\5\end{aligned}} \right){e^{ - t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\0\\1\end{aligned}} \right){e^{ - 2t}}\), with \({c_1}\) , and \({c_2}\) are arbitrary complex numbers.

03

Describe the shape of the trajectories

It is observed that the origin is a saddle point. A solution of \({c_1} = 0\) attracts the origin but a solution of \({c_2} = {c_3} = 0\) repels the origin.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

21. Use mathematical induction to prove that for \(n \ge {\bf{2}}\),\(\begin{aligned}{c}det\left( {{C_p} - \lambda I} \right) = {\left( { - {\bf{1}}} \right)^n}\left( {{a_{\bf{0}}} + {a_{\bf{1}}}\lambda + ... + {a_{n - {\bf{1}}}}{\lambda ^{n - {\bf{1}}}} + {\lambda ^n}} \right)\\ = {\left( { - {\bf{1}}} \right)^n}p\left( \lambda \right)\end{aligned}\)

(Hint: Expanding by cofactors down the first column, show that \(det\left( {{C_p} - \lambda I} \right)\) has the form \(\left( { - \lambda B} \right) + {\left( { - {\bf{1}}} \right)^n}{a_{\bf{0}}}\) where \(B\) is a certain polynomial (by the induction assumption).)

Let\(\varepsilon = \left\{ {{{\bf{e}}_1},{{\bf{e}}_2},{{\bf{e}}_3}} \right\}\) be the standard basis for \({\mathbb{R}^3}\),\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space \(V\) and\(T:{\mathbb{R}^3} \to V\) be a linear transformation with the property that

\(T\left( {{x_1},{x_2},{x_3}} \right) = \left( {{x_3} - {x_2}} \right){{\bf{b}}_1} - \left( {{x_1} - {x_3}} \right){{\bf{b}}_2} + \left( {{x_1} - {x_2}} \right){{\bf{b}}_3}\)

  1. Compute\(T\left( {{{\bf{e}}_1}} \right)\), \(T\left( {{{\bf{e}}_2}} \right)\) and \(T\left( {{{\bf{e}}_3}} \right)\).
  2. Compute \({\left( {T\left( {{{\bf{e}}_1}} \right)} \right)_B}\), \({\left( {T\left( {{{\bf{e}}_2}} \right)} \right)_B}\) and \({\left( {T\left( {{{\bf{e}}_3}} \right)} \right)_B}\).
  3. Find the matrix for \(T\) relative to \(\varepsilon \), and\(B\).

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

8. \(\left[ {\begin{array}{*{20}{c}}7&- 2\\2&3\end{array}} \right]\)

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

8. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{1}}\\{\bf{0}}&{\bf{5}}\end{array}} \right)\)

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.