# Mata04y 5 a we may carry out a sequence of row

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Unformatted text preview: uence of row operations 2 4 2 4 2 4 2 4 page 6 321 731 ?3 2 2 3 5 3 5 3 5 3 5 R 1 \$ R2 ?! 2 4 2 4 2 4 2 4 731 321 ?3 2 2 1 ?1 ?1 054 0 ?1 ?1 1 ?1 ?1 011 054 1 ?1 ?1 011 001 3 5 3 5 3 5 3 5 R1 ! R1 ? 2R2 ?! 1 ?1 ?1 321 ?3 2 2 1 ?1 ?1 0 ?1 ?1 054 1 ?1 ?1 011 0 0 ?1 2 4 ?! R2 ! R2 ? 3R1 R3 ! R3 + 3R1 R2 ! ?R2 ?! R2 \$ R 3 R3 ! R3 ? 5R2 ?! ?! ?! R3 ! ?R3 R2 ! R 2 ? R 3 ?! R1 ! R 1 + R2 100 011 001 3 5 ?! 2 4 100 010 001 3 5 ? ?? Thus E10 E9 E1A = I , or, A = E1 1 E9 1E101 , where, in general, the elementary matrix E corresponding to the elementary row operation R is obtained for the identity matrix I by I ? R!E : ??? MATA04Y ? From this we calculate the Ek and the Ek 1, for k = 1 : : : 10 as page 7 010 ? 4 1 0 0 5 = E1 1 E1 = 001 1 ?2 0 E2 = 4 0 1 0 0 01 100 4 ?3 1 0 E3 = 001 100 40 1 0 E4 = 301 2 2 3 5 3 2 2 3 5 3 5 2 3 120 ; E2?1 = 4 0 1 0 001 100 ?1 = 4 3 1 0 ; E3 001 2 2 2 3 5 3 5 3 5 100 ?1 = 4 0 1 0 ; E4 ?3 0 1 100 ? 4 0 0 1 5 = E5 1 E5 = 010 1 00 ? 4 0 ?1 0 5 = E6 1 E6 = 0 01 1 00 E7 = 4 0 1 0 0 ?5 1 2 2 3 5 3 2 3 100 ; E7?1 = 4 0 1 0 051 2 3 5 10 0 ? 40 1 0 5 = E8 1 E8 = 0 0 ?1 110 E9 = 4 0 1 0 001 2 2 3 5 1 ?1 0 ; E9?1 = 4 0 1 0 0 01 3 5 2 3 5 3 5 10 0 4 0 1 ?1 E10 = 00 1 100 ?1 = 4 0 1 1 ; E10 001 2 : MATA04Y page 8 (b) The two matrices will be row equivalent if and only if their reduced row echelon forms are the same. The rst of these is calculated to be the identity in part (a) and the second is, by the TI85, 38 2 ?5 100 4 13 3 5 ?! 4 0 1 0 0 A= ?3 2 2 001 2 3 2 3 5 which is also the identity. Hence A and B are row equivalent....
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## This note was uploaded on 04/13/2010 for the course ECON 330 taught by Professor Minetti during the Fall '08 term at Michigan State University.

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