Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 5: Q6E (page 267)

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) an eigenvector of\(\left){\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\)? If so, find the eigenvalue.

Short Answer

Expert verified

Yes, \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\) , and the eigenvalue is \( - 2\).

Step by step solution

01

Definition of eigenvector

If there exists a non-zero vector \({\bf{x}}\) that satisfies \(A{\bf{x}} = \lambda {\bf{x}}\) for some scaler \(\lambda \), then \({\bf{x}}\) be the eigenvector of an \(n \times n\) matrix \(A\), and if \(A{\bf{x}} = \lambda {\bf{x}}\) exists, then scaler \(\lambda \) is the eigenvalue of the matrix.

02

Determine whether \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the given matrix

Denote the given matrix by \(A\) and the given vector by \({\bf{x}}\).

\(A = \left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\)

According to the definition of eigenvalue, \({\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the matrix \(A\), if \(A{\bf{x}} = \lambda {\bf{x}}\).

Find the product of \(A\), and \({\bf{x}}\).

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( 1 \right) + 6\left( { - 2} \right) + 7\left( 1 \right)}\\{3\left( 1 \right) + 3\left( { - 2} \right) + 7\left( 1 \right)}\\{5\left( 1 \right) + 6\left( { - 2} \right) + 5\left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3 - 12 + 7}\\{3 - 6 + 7}\\{5 - 12 + 5}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 2}\\4\\{ - 2}\end{array}} \right)\end{array}\)

The obtained matrix in the form of the given vector can be written as:

\(A{\bf{x}} = \left( { - 2} \right)\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\)

So, the vector \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\).

03

Determine the eigenvalue 

As the given vector satisfies the condition \(A{\bf{x}} = \lambda {\bf{x}}\). Which implies that \(\lambda \) is the eigenvalue of the given matrix. So, \(\lambda = - 2\).

So, \( - 2\) is the eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).

Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

6. \(\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{0}}&{{\bf{ - 2}}}\\{\bf{2}}&{\bf{5}}&{\bf{4}}\\{\bf{0}}&{\bf{0}}&{\bf{5}}\end{array}} \right){\bf{ = }}\left( {\begin{array}{*{20}{c}}{{\bf{ - 2}}}&{\bf{0}}&{{\bf{ - 1}}}\\{\bf{0}}&{\bf{1}}&{\bf{2}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{5}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\bf{0}}&{\bf{0}}&{\bf{1}}\\{\bf{2}}&{\bf{1}}&{\bf{4}}\\{{\bf{ - 1}}}&{\bf{0}}&{{\bf{ - 2}}}\end{array}} \right)\)

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

For the Matrices A find real closed formulas for the trajectory x→(t+1)=Ax→(t)wherex→(0)=[01]A=[2-332]

For the matrix A, find real closed formulas for the trajectoryx→(t+1)=Ax¯(t)where x→=[01]. Draw a rough sketch

A=[15-27]
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation
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.

Sign-up for free