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Unformatted text preview: MATH 235 (Algebra 1) Solutions to Assignment 3 October 1, 2007 Solution of 1. Let F be a field. Lets check that the ring axioms are satisfied for the set M 2 ( F ) of 2 2 matrices over F endowed with matrix addition and multiplication. It is a bit tedious but its the kind of things that one has to do at some point. Closure for +: Let A = a 1 b 1 c 1 d 1 and B = a 2 b 2 c 2 d 2 be any two elements of M 2 ( F ). Then clearly A + B = a 1 + a 2 b 1 + b 2 c 1 + c 2 d 1 + d 2 M 2 ( F ) . Associativity of +: Let A = a 1 b 1 c 1 d 1 , B = a 2 b 2 c 2 d 2 , and C = a 3 b 3 c 3 d 3 be any three elements of M 2 ( F ). Using the associativity of + in F , we can compute ( A + B ) + C = a 1 + a 2 b 1 + b 2 c 1 + c 2 d 1 + d 2 + a 3 b 3 c 3 d 3 = ( a 1 + a 2 ) + a 3 ( b 1 + b 2 ) + b 3 ( c 1 + c 2 ) + c 3 ( d 1 + d 2 ) + d 3 = a 1 + ( a 2 + a 3 ) b 1 + ( b 2 + b 3 ) c 1 + ( c 2 + c 3 ) d 1 + ( d 2 + d 3 ) = a 1 b 1 c 1 d 1 + a 2 + a 3 b 2 + b 3 c 2 + c 3 d 2 + d 3 = A + ( B + C ) . Commutativity of +: Let A = a 1 b 1 c 1 d 1 and B = a 2 b 2 c 2 d 2 be any two elements of M 2 ( F ). Then using the commutativity of + in F we check that A + B = a 1 + a 2 b 1 + b 2 c 1 + c 2 d 1 + d 2 = a 2 + a 1 b 2 + b 1 c 2 + c 1 d 2 + d 1 = B + A. Zero element: Let = 0 0 0 0 M 2 ( F ). We see that is indeed a neutral element for +, since for every A = a b c d M 2 ( F ) we have A + = a b c d + 0 0 0 0 = a + 0 b + 0 c + 0 d + 0 = a b c d = A and + A = 0 0 0 0 + a b c d = 0 + a 0 + b 0 + c 0 + d = a b c d = A. 1 Existence of additive inverse: For A = a b c d M 2 ( F ), we see that A + a b c d = a a b b c c d d = 0 0 0 0 = hence A has an additive inverse, namely a b c d . Closure for : Let A = a 1 b 1 c 1 d 1 and B = a 2 b 2 c 2 d 2 be any two elements of M 2 ( F ). Then clearly A B = a 1 a 2 + b 1 c 2 a 1 b 2 + b 1 d 2 c 1 a 2 + d 1 c 2 c 1 b 2 + d 1 d 2 M 2 ( F ) ....
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 Fall '07
 Goren
 Algebra, Addition, Multiplication, Matrices

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