Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 5 - Eigenvalues and Eigenvectors - 5.2 Exercises - Page 282: 11

Answer

$24-26\lambda+9\lambda^2-\lambda^3$

Work Step by Step

$A=\begin{bmatrix} 4-\lambda&0&0\\5&3-\lambda&2\\-2&0&2-\lambda \end{bmatrix}$ $det(A)=(4-\lambda)\left|\begin{matrix} 3-\lambda&2\\0&2-\lambda \end{matrix}\right|=(4-\lambda)(3-\lambda)(2-\lambda)=0$ $(12-7\lambda+\lambda^2)(2-\lambda)=24-26\lambda+9\lambda^2-\lambda^3$
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