Quiz8Solutions

Quiz8Solutions - a and b such that a + b-3 a-b 2 b + 1 = ....

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Newberger Math 247 Spring 03 Quiz 4.1-4.2 name: 1. (12 points) Show that the following set is not a subspace of R 3 . H = a + b - 3 a - b 2 b + 1 a,b R We will show that the zero vector 0 = 0 0 0 is not in H . To see if 0 is in H , we will try to find a and b such that a + b - 3 a - b 2 b + 1 = 0 0 0 . To solve this system using linear algebra, we must move all the constants to the right hand side of the equation. We get a + b a - b 2 b = 3 0 - 1 . This becomes 1 1 1 - 1 0 2 a b = 3 0 - 1 . Reducing, we get 1 1 3 1 - 1 0 0 2 - 1 1 1 3 0 - 2 - 3 0 2 - 1 1 1 3 0 - 2 - 3 0 0 - 4 which shows that the system is inconsistent. Thus there are no real numbers
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Unformatted text preview: a and b such that a + b-3 a-b 2 b + 1 = . This means is not in H and H is not a subspace. 2. (12 points) Show that the following set is not a subspace of R 2 . W = ( x y -x y x ) Here is a graph of the region W . W u-u Note that the vector u = (1 , 0) is in W , since-1 1 . However-u = (-1 , 0) is not in W , since 1 -1 is false. Thus W is not closed under scalar multiplication....
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This note was uploaded on 01/12/2010 for the course MATH 247 taught by Professor F,newberger during the Spring '03 term at Stanford.

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Quiz8Solutions - a and b such that a + b-3 a-b 2 b + 1 = ....

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