Vjeko2s - Math 33A, Midterm 2 solutions 1. We have to solve...

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Unformatted text preview: Math 33A, Midterm 2 solutions 1. We have to solve the linear system of 1 equation in 3 variables. Its augmented coefficient matrix is ( 2- 1 1 | ) , and after dividing by 2 we get its RREF ( 1- 1 2 1 2 | ) . Alternatively, we can say that we are finding the kernel of the matrix ( 2- 1 1 ) . The solution is x 1 x 2 x 3 = s 1 2 1 + t - 1 2 1 where s,t are arbitrary parameters, i.e. the subspace (the plane) can be written as span { 1 2 1 , - 1 2 1 } . The above two vectors are linearly independent (which is always the case whenever we compute the kernel from the RREF), and thus 1 2 1 , - 1 2 1 is a basis for the given subspace. 2. Denote the matrices by A and B respectively. image( A ) = span { 1 1 1 , 2 1 , - 2- 1 } = span { 1 1 1 , 2 1 } , because - 2- 1 =- 2 1 1 1 + 2 1 . Also image( B ) = span { 3 1 2 ,...
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This note was uploaded on 02/06/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

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Vjeko2s - Math 33A, Midterm 2 solutions 1. We have to solve...

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