HW5 - p.301 problem 6 1 a 1 a 1 1 a 1 1 1 a 1 → 0 a − 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: p.301, problem 6 1 a 1 a 1 1 a 1 1 1 a 1 → 0 a − 1 1 − a → 0 a − 1 1− a a 1 1 0 1 − a 1 − a 2 0 0 2 − a − a2 2 2 − a − a = 0 ⇒ a = 1,−2 Rank = 3 if a≠ 1or -2. Rank = 2 if a=-2. Rank = 1 if a=1. Page 301, problem 11 4 8 16 4 8 16 2 4 8 16 2 2 16 8 4 2 0 − 24 − 60 − 126 0 − 24 − 60 − 126 → → 4 8 16 2 0 0 0 0 − 30 0 0 − 30 0 − 12 0 − 30 − 75 2 16 8 4 0 12 0 4 8 16 2 0 − 24 − 60 − 126 → 0 0 − 30 − 75 0 0 − 30 0 Rank = 4. Page 301, problem 15 6 0 3 0 −1 2 12 3 0 6 0 → 0 − 1 0 0 1 4 2 3 1 4 2 6 0 7 0 5 → 0 − 1 2 7 0 5 0 3 − 6 − 21 0 − 15 − 19 8 − 11 3 1 4 2 2 7 0 5 0 0 0 0 Therefore linearly dependent. Page 314, problem 8 14 2 5 2 0 8 = 14( 0 − 8 ⋅ 8) − 2( − 2 ⋅ 2 − 8 ⋅ 5) + 5( 2 ⋅ 8 − 0) = −728 5 8 −2 Page 322, problem 4 2 / 3 1/ 3 2 / 3 1 0 0 2 / 3 1/ 3 2 / 3 1 1 1 1 − 2 / 3 2 / 3 1/ 3 0 1 0 → 0 1/ 3 2 / 3 − 2 / 3 0 0 1 0 1/ 2 − 1 − 1/ 2 2 / 3 1/ 3 2 / 3 1 0 0 1 1/ 2 1 3 / 2 → 0 1 11 1 0 → 0 1 1 1 0 0 − 3 / 2 − 1 − 1/ 2 1 0 0 1 2 / 3 0 0 1 0 0 1 0 0 1 0 1/ 3 − 2 / 3 1 1/ 2 0 5 / 6 − 1/ 3 2 / 3 1 0 0 2 / 3 − 2 / 3 1/ 3 → 0 1 0 1/ 3 2 / 3 2 / 3 → 0 1 0 1/ 3 2 / 3 2/3 0 0 1 2 / 3 1/ 3 − 2 / 3 0 0 1 2 / 3 1/ 3 − 2 / 3 2 / 3 1/ 3 2 / 3 ∴ − 2 / 3 2 / 3 1/ 3 1 / 3 2 / 3 − 2 / 3 −1 2 / 3 − 2 / 3 1/ 3 = 1/ 3 2 / 3 2/3 2 / 3 1 / 3 − 2 / 3 Page 322, problem 5 3 −1 1 1 0 0 3 −1 1 1 1 0 5 − 15 6 − 5 0 1 0 → 0 5 0 − 1/ 3 1/ 3 − 5 / 3 − 2 2 0 0 1 3 − 1 1 1 0 0 1 − 1/ 3 1/ 3 1/ 3 → 0 1 0 5 1 0 → 0 1 05 0 0 1/ 3 0 1/ 3 1 0 0 10 0 0 1 0 0 1 0 0 1 0 1 3 1 − 1 / 3 0 1 / 3 − 1 / 3 − 1 1 0 0 2 0 − 1 → 0 1 05 1 0 → 0 1 0 5 1 0 0 0 10 1 3 0 0 1 0 1 3 −1 −1 1 3 2 0 − 1 ∴ − 15 6 − 5 = 5 1 0 5 0 1 3 −2 2 Page 338, problem 19: 13 − λ 5 2 3 2 7 − λ − 8 = −λ3 + 27λ2 − 243λ + 729 = −( λ − 9 ) = 0 ⇒ λ = 9,9,9. 5 4 7−λ 2 x1 0 4 5 2 − 2 − 8 x 2 = 0 5 4 − 2 x 0 3 Let x3 = 1 ⇒ 4 x1 + 5 x 2 = −2, 2 x1 − 2 x 2 = 8 ⇒ x1 = 2, x 2 = −2 2 ⇒ − 2 1 is an eigenvector. Page 338, problem 25: −1− λ 0 0 0 0 −1− λ 0 0 12 0 −1− λ −4 0 12 2 2 = ( − 1 − λ ) λ 2 + 2λ − 15 = ( λ + 1) ( λ + 5) ( λ − 3) = 0 −4 −1− λ ( ) ⇒ λ = −1, − 1, − 5, 3 λ = −1 : 0 0 0 0 0 12 0 x 1 0 0 12 x 2 0 0 − 4 x 3 0 − 4 0 x 4 λ = −5 : 0 1 0 0 0 1 = 0 ⇒ x 3 = 0, x 2 = 0, x 1 and x 2 arbitrary ⇒ 0 and 0 eigenvectors 0 0 0 0 12 0 x 1 4 0 12 x 2 0 4 − 4 x 3 0 − 4 4 x 4 λ =3: 0 − 3 0 − 3 = 0 Let x 3 = 1 ⇒ x 1 = −3, x 4 = 1, x 2 = −3 ⇒ 1 is an eigenvector 1 0 4 0 0 0 − 4 0 12 0 x 1 0 − 4 0 12 x 2 0 0 − 4 − 4 x 3 0 0 − 4 − 4 x 4 0 3 0 − 3 = 0 Let x 3 = 1 ⇒ x 1 = 3, x 4 = −1, x 2 = −3 ⇒ 1 is an eigenvector − 1 0 ...
View Full Document

This note was uploaded on 09/05/2011 for the course EGM 4313 taught by Professor Mei during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online