# 2.6 - Math 2940 Solutions, Fall 2011 Section 2.6 5)...

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Math 2940 Solutions, Fall 2011 Section 2 . 6 5 ) Multiplying A by the E 31 matrix below gives 1 0 0 0 1 0 - 3 0 1 2 1 0 0 4 2 6 3 5 = 2 1 0 0 4 2 0 0 5 so L = 1 0 0 0 1 0 - 3 0 1 - 1 = 1 0 0 0 1 0 3 0 1 and 2 1 0 0 4 2 6 3 5 = 1 0 0 0 1 0 3 0 1 2 1 0 0 4 2 0 0 5 . 6 ) We have A = 1 1 1 2 4 5 0 4 0 . We ﬁrst use E 21 as below to get 1 0 0 - 2 1 0 0 0 1 1 1 1 2 4 5 0 4 0 = 1 1 1 0 2 3 0 4 0 . We then multiply by the E 32 matrix to get 1 0 0 0 1 0 0 - 2 1 1 1 1 0 2 3 0 4 0 = 1 1 1 0 2 3 0 0 - 6 . so A = 1 0 0 2 1 0 0 0 1 1 0 0 0 1 0 0 2 1 1 1 1 0 2 3 0 0 - 6 = 1 0 0 2 1 0 0 2 1 1 1 1 0 2 3 0 0 - 6 = LU. 13

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## This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell.

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2.6 - Math 2940 Solutions, Fall 2011 Section 2.6 5)...

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