{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# week08 - Math 43 Fall 2007 B Dodson Week 8 Problem 3.2.32...

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

Math 43, Fall 2007 B. Dodson Week 8: Problem 3.2.32 Find the inverse of the elementary matrix E = 1 2 0 1 . Solution: Left multiplication by E does the 3rd elementary row operation R 1 R 1 + 2 R 2 . (check) To undo this row operation, we subtract 2 R 2 from the first row, so E - 1 = 1 - 2 0 1 . (why?) Problem 3.2.40 Write the invertible matrix A = 2 4 1 1 as a product of elementary row matrices. Solution. We reduce A to the identity matrix I, keeping track of the row operations. We have A = 2 4 1 1 1 1 2 4 1 1 0 2 1 1 0 1 1 0 0 1 = I, which in terms of elementary matrices looks like E 4 E 3 E 2 E 1 A = I, and gives A = E - 1 1 E - 1 2 E - 1 3 E - 1 4 . Tracing through the row operations E 1 = E - 1 1 = 0 1 1 0 ; E 2 = 1 0 - 2 1 , so E - 1 2 = 1 0 2 1 , E 3 = 1 0 0 1 2 , so E - 1 3 = 1 0 0 2 ; and E 4 = 1 - 1 0 1 , so E - 1 4 = 1 1 0 1 . We get A = 0 1 1 0 ‚ • 1 0 2 1 ‚ • 1 0 0 2 ‚ • 1 1 0 1 .

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

View Full Document
2 A collection S of vectors in R n is a Subspace of R n if it satisfies the two subspace conditions, called closure rules: (1) for every u, v S, check u + v S (closure under vector +), and (2) for every u S,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

week08 - Math 43 Fall 2007 B Dodson Week 8 Problem 3.2.32...

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

View Full Document
Ask a homework question - tutors are online