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: 9

Answer

P= −1(λ^3)+4(λ^2)−9λ−6

Work Step by Step

Finding Characteristic equation: Given: matrix A= 1 0 -1 2 3 -1 0 6 0 Step 1: A−λI A−λI = (1-λ) 0 -1 2 (3-λ) -1 0 6 (0-λ) Step 2: Now take determinant of the above matrix det (A-λI) = det of : (1-λ) 0 -1 2 (3-λ) -1 0 6 (0-λ) |A-λI| = (1-λ) 0 -1 2 (3-λ) -1 0 6 (0-λ) =(1−λ)[−3λ+(λ^2)+6]−0−1[12−0] =(1−λ)[(λ^2)−3λ+6]−12 =(λ^2)−3λ+6−(λ^3)+3(λ^2)−6λ−12 =−λ^3+4(λ^2)−9λ−6 Hence, the characteristic polynomial is: The characteristic polynomial is: P= −1(λ^3)+4(λ^2)−9λ−6
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