This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 43, Fall 2007 B. Dodson Week 9: Monday: Finish Suggested Hw8, start on Hw9 material 1. Determinants 2. Properties of Dets 3. Eigenvalues and Eigenvectors (2by2 in 4.1; then nbyn in 4.3) ——– We compute det 2 1 5 4 2 3 9 5 1 using the (first) row expansion (by minors): det ( A ) = 2 fl fl fl fl 2 3 5 1 fl fl fl fl 1 fl fl fl fl 4 3 9 1 fl fl fl fl + 5 fl fl fl fl 4 2 9 5 fl fl fl fl = 2(2 15) (4 27) + 5(20 18) = 2( 13) ( 23) + 5(2) = 26 + 33 = 7 . ———– Problem Reduce A = 2 1 3 5 3 1 2 4 1 4 3 5 2 5 3 to an upper triagular matrix and use the reduction to find det ( A ) . Solution: A →  1 1 2 3 3 1 2 4 1 4 3 5 2 5 3 ( r 1 → r 1 r 2 ) →  1 1 2 3 3 7 11 5 12 15 7 15 18 ( r 2 → r 2 + 3 r 1 , r 3 → r 3 + 4 r 1 , r 4 → r 4 + 5 r 1 ) 2 →  1 1 2 3 3 7 11 5 12 15 1 1 4 ( r 4 → r 4 2 r 2 ) ——– →  1 1 2 3 1 1 4 5 12 15 3 7 11 ( r 2 ↔ r 4 ) →  1 1 2 3 1 1 4 7 35 4 23 ( r 3 → r 3 5 r 2 , r 4 → r 4 3 r 2 ) →  1 1 2 3 1 1 4 1 11 4 23 ( r 3 → r 3 2 r 4 ) →  1 1 2 3 1 1 4 1 11 21 = U ( r 4 → r 4 + 4 r 3 ). Now we have det U = ( 1)(1)( 1)( 21) = 21 , and det A = ( 1) det U = 21 , since all of the 3rd elementary operations change the determinant by a factor of (1) – so no change  there are no 2nd EROs and exactly one 1st ERO, with each 1st ERO changing the determinant by a factor of ( 1) . Notice that, for the problem of finding the value of the determinant, we do not need a reduced row echelon or 1’s on the diagonal, and stop at a triangular form. 3 3. Eigenvalues and Eigenvectors ————— A vector v = 0 in R n is an eigenvector with eigenvalue λ of an nby n matrix A if Av = λv....
View
Full Document
 Spring '08
 Dodson
 Linear Algebra, Determinant, Eigenvectors, Vectors, Eigenvalue, eigenvector and eigenspace, linearly independent eigenvectors, Dets

Click to edit the document details