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HW3_solutions 167

# HW3_solutions 167 - Math 167 homework 3 solutions 2.1.22...

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Math 167 homework 3 solutions October 20, 2010 2.1.22: For which right-hand sides (find a condition on b 1 , b 2 , b 3 ) are these systems solvable? (a) 1 4 2 2 8 4 - 1 - 4 - 2 x 1 x 2 x 3 = b 1 b 2 b 3 r 2 - 2 r 1 and r 3 + r 1 give the conditions b 2 - 2 b 1 = 0 and b 3 + b 1 = 0 for solvability. These conditions are equivalent to b 2 = 2 b 1 and b 3 = - b 1 . (b) 1 4 2 9 - 1 - 4 x 1 x 2 x 3 = b 1 b 2 b 3 r 2 - 2 r 1 and r 1 + r 3 = 0 give the condition b 3 + b 1 = 0 for solvability. This condition is equivalent to b 3 = - b 1 . 2.1.26 The columns of AB are linear combinations of the columns of A . This means: The column space of AB is contained in (possibly equal to) the column space of A . Give an example where the column spaces of A and AB are not equal. Let A = 1 0 0 0 1 0 0 0 1 and AB = B = 1 0 0 0 8 0 0 0 0 . 2.1.30 If the 9 by 12 system Ax = b is solvable for every b , then C ( A ) = R 9 . 1

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2.2.24 Every column of AB is a combination of the columns of A . Then the dimension of the column space gives rank( AB ) rank( A ). Prove also that rank( AB ) rank( B ). Note that rank( AB ) = n - dim N ( AB ) , where n is the number of columns of AB and dim N ( AB ) is dimension of the nullspace of AB . It is true that dim N ( AB ) dim N ( B ) , since x ∈ N ( B ) implies x ∈ N ( AB ). Also, note that rank( B ) = n - dim N ( B ) .
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HW3_solutions 167 - Math 167 homework 3 solutions 2.1.22...

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