hw09sols - AAE 250 Homework 7 Solutions...

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: AAE 250 Homework 7 Solutions http://nsgsun.aae.uiuc.edu/AAE250 1. Solution: (a) We calculate K- q = 3 1 0 1 3 1 0 1 2 1- 2 1 = 1- 4 - q T K = h 1- 2 1 i 3 1 0 1 3 1 0 1 2 = h 1- 4 0 i- q T K T = h 1- 2 1 i 3 1 0 1 3 1 0 1 2 = h 1- 4 0 i 1 2- q T K- q = 1 2 h 1- 2 1 i 3 1 0 1 3 1 0 1 2 1- 2 1 = 1 2 h 1- 4 0 i 1- 2 1 = 9 2 (b) The matrix K is positive definite if all of its principal minors are positive definite. These minors are given by 1 = | k 11 | = | 3 | > 2 = 3 1 1 3 = 8 > 3 = 3 1 0 1 3 1 0 1 2 = 13 > So, since all of the principal leading minors are positive, the matrix K is positive definite. (c) The matrix K- 2 I is given by K- 2 I = 1 1 0 1 1 1 0 1 0 1 The principal leading minors are given by 1 = | k 11 | = 1 > 2 = 1 1 1 1 = 0 3 = 1 1 0...
View Full Document

Page1 / 5

hw09sols - AAE 250 Homework 7 Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online