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Q5.2-13E

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Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

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

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)

The characteristic polynomial of the matrix is \( - {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of the characteristic polynomial

The eigenvalue of an \(n \times n\) matrix \(A\) is a scalar \(\lambda \) such that \(\lambda \) satisfies the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

When \(A\) is an \(n \times n\) matrix, \(\det \left( {A - \lambda I} \right)\) is the characteristic polynomial of \(A\), which is the polynomial of degree \(n\).

Step 2: Determine the characteristic polynomial of the matrix

Use the cofactor expression down the third column to obtain the characteristic polynomial of the matrix, as shown below.

\[\begin{array}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{6 - \lambda }&{ - 2}&0\\{ - 2}&{9 - \lambda }&0\\5&8&{3 - \lambda }\end{array}} \right]\\ = \left( {3 - \lambda } \right)\det \left[ {\begin{array}{*{20}{c}}{6 - \lambda }&{ - 2}\\{ - 2}&{9 - \lambda }\end{array}} \right]\\ = \left( {3 - \lambda } \right)\left[ {\left( {6 - \lambda } \right)\left( {9 - \lambda } \right) - \left( { - 2} \right)\left( { - 2} \right)} \right]\\ = \left( {3 - \lambda } \right)\left( {{\lambda ^2} + - 15\lambda + 50} \right)\\ = - {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150\\ = \left( {3 - \lambda } \right)\left( {\lambda - 5} \right)\left( {\lambda - 10} \right)\end{array}\]

Thus, the characteristic polynomial of the matrix is \( - {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150\).

Most popular questions for Math Textbooks

Suppose \({\bf{x}}\) is an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \).

a. Show that \(x\) is an eigenvector of \(5I - A\). What is the corresponding eigenvalue?

b. Show that \(x\) is an eigenvector of \(5I - 3A + {A^2}\). What is the corresponding eigenvalue?

[M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

20. \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{3}}&{\bf{2}}\\{\bf{2}}&{{\bf{12}}}&{{\bf{13}}}&{{\bf{11}}}\\{{\bf{ - 2}}}&{\bf{3}}&{\bf{0}}&{\bf{2}}\\{\bf{4}}&{\bf{5}}&{\bf{7}}&{\bf{2}}\end{array}} \right]\]

Let \(G = \left( {\begin{aligned}{*{20}{c}}A&X\\{\bf{0}}&B\end{aligned}} \right)\). Use formula \(\left( {\bf{1}} \right)\) for the determinant in section \({\bf{5}}{\bf{.2}}\) to explain why \(\det G = \left( {\det A} \right)\left( {\det B} \right)\). From this, deduce that the characteristic polynomial of \(G\) is the product of the characteristic polynomials of \(A\) and \(B\).

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

5. \(\left[ {\begin{array}{*{20}{c}}2&1\\-1&4\end{array}} \right]\)

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)
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