{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# quiz9 - Solutions for Quiz on March 29 We need to...

This preview shows pages 1–3. Sign up to view the full content.

Solutions for Quiz on March 29 We need to diagonalize the matrix, A = 1 0 0 - 1 0 1 1 1 0 . First we find the eigenvalues. We need to solve fl fl fl fl 1 - λ 0 0 - 1 0 - λ 1 1 1 0 - λ fl fl fl fl = 0 . We expand the determinant in cofactors of the first row. This row has only one nonzero element, so we have (1 - λ )( - 1) 1+1 M 13 = (1 - λ ) fl fl fl fl - λ 1 1 - λ fl fl fl fl = 0 . The minor is given by λ 2 - 1, so the cherecteristic equation becomes (1 - λ )( λ 2 - 1) = 0 . This has roots λ 1 = 1, λ 2 = 1 and λ 3 = - 1. Now we need to find the eigenvectors. We start from the eigenvector corresponding to λ 3 . We have, 2 0 0 - 1 1 1 1 1 1 x 1 x 2 x 3 = 0 . From the first equation, we have 2 x 1 = 0, which means x 1 = 0. From the second and third equations, we have, 0+ x 2 + x 3 = 0, which means x 2 = - x 3 . Therefore, the corresponding solution is α 0 - 1 1 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Remark. This equation can also be solved by row reduction, if you like.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}