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Unformatted text preview: Math 43, Fall 2007 B. Dodson Week 8: Problem 3.2.32 Find the inverse of the elementary matrix E = 1 2 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 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 2 1 1 1 1 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 = 1 1 ; E 2 = 1 2 1 , so E 1 2 = 1 2 1 , E 3 = 1 1 2 , so E 1 3 = 1 2 ; and E 4 = 1 1 1 , so E 1 4 = 1 1 1 . We get A = 1 1 1 2 1 1 2 1 1 1 . 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...
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This note was uploaded on 02/29/2008 for the course MATH 43 taught by Professor Dodson during the Spring '08 term at Lehigh University .
 Spring '08
 Dodson
 Multiplication

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