Week5 - h 6 = 0 and we will find the inverse We must row...

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

View Full Document Right Arrow Icon
Solutions for Assignment 5 Applied Linear Algebra MATH 232 (Fall 2008) Section 2.3 Section 2.6
Background image of page 1

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

View Full DocumentRight Arrow Icon
Assignment 5 Applied Linear Algebra Math 232(Fall 2008) Additional question: For which values of h is the matrix 1 1 1 0 h 2 h 0 1 1 invertible? Find the inverse when it exists. A1. We create the 3 × 6 matrix 1 1 1 | 1 0 0 0 h 2 h | 0 1 0 0 1 1 | 0 0 1 . The first column is OK. Let’s switch rows two and three to get a one as a pivot in the second row: 1 1 1 | 1 0 0 0 1 1 | 0 0 1 0 h 2 h | 0 1 0 . Now we take row 3 and subtract h times row 2 from it: 1 1 1 | 1 0 0 0 1 1 | 0 0 1 0 0 h | 0 1 - h . The matrix on the left is in echelon form. We have a pivot in every row and column if and only if h is nonzero. So the answer to the first part is: the matrix is invertible for every h except for h = 0. For the remainder, we’ll assume
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: h 6 = 0 and we will find the inverse. We must row reduce to the identity, so let’s divide the third row by h . (This is OK, because we are assuming h is nonzero.) We get: 1 1 1 | 1 0 1 1 | 1 0 0 1 | 0 1 /h-1 . Next we take row 3 and subtract it from both rows 2 and 1: 1 1 0 | 1-1 /h 1 0 1 0 |-1 /h 2 0 0 1 | 1 /h-1 . 1 To finish, let’s take row 2 and subtract it from row 1: 1 0 0 | 1-1 0 1 0 |-1 /h 2 0 0 1 | 1 /h-1 . This shows that the inverse (when h is nonzero) is the matrix 1-1-1 /h 2 1 /h-1 . 2...
View Full Document

Page1 / 3

Week5 - h 6 = 0 and we will find the inverse We must row...

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