• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q5.2-13E

Expert-verified Found in: Page 267 ### 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 # 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 $$- {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150$$.

See the step by step solution

## 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 $$- {\lambda ^3} + 18{\lambda ^2} - 95\lambda + 150$$. ### Want to see more solutions like these? 