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 1E from Chapter 5.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
Find the characteristic polynomial and the real eigenvalues of the matrices in Exercises 1-8...
Step-by-Step Solution
Step 1
Given matrix: \[A = \left[ {\begin{array}{*{20}{c}}2&7\\7&2\end{array}} \right]\]We have to find the characteristic polynomial and the corresponding eigenvalues of the matrix given above
Step 2: Write the equation for characteristic polynomial
\[\begin{array}{l}\det \left( {A - \lambda I} \right) = \det \left( {\left[ {\begin{array}{*{20}{c}}2&7\\7&2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}\lambda &0\\0&\lambda \end{array}} \right]} \right)\\ = \det \left( {\left[ {\begin{array}{*{20}{c}}{2 - \lambda }&7\\7&{2 - \lambda }\end{array}} \right]} \right)\\ = \left( {2 - \lambda } \right) \times \left( {2 - \lambda } \right) - 7 \times 7\\ = {\left( {2 - \lambda } \right)^2} - {7^2}\\ = 4 - 4\lambda + {\lambda ^2} - 49\\ = {\lambda ^2} - 4\lambda - 45\end{array}\]
Step 3: Solve the equation for eigenvalues
\[\begin{array}{l}\det \left( {A - \lambda I} \right) = 0\\{\lambda ^2} - 4\lambda - 45 = 0\\{\lambda ^2} - 9\lambda + 5\lambda - 45 = 0\\\lambda \left( {\lambda - 9} \right) + 5\left( {\lambda - 9} \right) = 0\\\left( {\lambda + 5} \right)\left( {\lambda - 9} \right) = 0\\\lambda = - 5,\,\,9\end{array}\]
ANSWER
The characteristic polynomial \[P\left( \lambda \right) = {\lambda ^2} - 4\lambda - 45\]The eigenvalues: \[\lambda = - 5,\,\,9\]
Given matrix: \[A = \left[ {\begin{array}{*{20}{c}}2&7\\7&2\end{array}} \right]\]We have to find the characteristic polynomial and the corresponding eigenvalues of the matrix given above
Step 2: Write the equation for characteristic polynomial
\[\begin{array}{l}\det \left( {A - \lambda I} \right) = \det \left( {\left[ {\begin{array}{*{20}{c}}2&7\\7&2\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}\lambda &0\\0&\lambda \end{array}} \right]} \right)\\ = \det \left( {\left[ {\begin{array}{*{20}{c}}{2 - \lambda }&7\\7&{2 - \lambda }\end{array}} \right]} \right)\\ = \left( {2 - \lambda } \right) \times \left( {2 - \lambda } \right) - 7 \times 7\\ = {\left( {2 - \lambda } \right)^2} - {7^2}\\ = 4 - 4\lambda + {\lambda ^2} - 49\\ = {\lambda ^2} - 4\lambda - 45\end{array}\]
Step 3: Solve the equation for eigenvalues
\[\begin{array}{l}\det \left( {A - \lambda I} \right) = 0\\{\lambda ^2} - 4\lambda - 45 = 0\\{\lambda ^2} - 9\lambda + 5\lambda - 45 = 0\\\lambda \left( {\lambda - 9} \right) + 5\left( {\lambda - 9} \right) = 0\\\left( {\lambda + 5} \right)\left( {\lambda - 9} \right) = 0\\\lambda = - 5,\,\,9\end{array}\]
ANSWER
The characteristic polynomial \[P\left( \lambda \right) = {\lambda ^2} - 4\lambda - 45\]The eigenvalues: \[\lambda = - 5,\,\,9\]