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Unformatted text preview: MATH 102 SOLUTIONS TO HW #4 Section 3.1, problem 6. First we find a basis for the space of vectors orthogonal to both a = (1 , 1 , 1) and b = (1 , 1 , 0). this space is nothing other that the nullspace of the matrix: A = 1 1 1 1 1 . To compute it, after the following Gaussian elimination steps: (1) 1 × { Row 1 } + { Row 2 } ⇒ { Row 2 } (2) 1 2 × { Row 2 } + { Row 1 } ⇒ { Row 1 } (3) 1 2 × { Row 2 } ⇒ { Row 2 } we are left with the reduces matrix: R = 1 1 2 1 1 2 . From this, a basis for the nullspace of A is easily computed to be: c =  1 2 1 2 1 . To normalize these vectors, we just divide through by their lengths. Doing this yields the orthonormal set: ˆ a = 1 √ 3 1 √ 3 1 √ 3 , ˆ b = 1 √ 2 1 √ 2 , ˆ c =  1 √ 6 1 √ 6 2 √ 6 . Section 3.1, problem 12. This problem again asks to compute the nullspace of A . Recall that N ( A ) = [ R ( A )] ⊥ . After performing the reduction step: (1) 1 × { Row 1 } + { Row 2 } ⇒ { Row 2 } we have the reduced matrix: R = 1 2 1 2...
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This note was uploaded on 08/02/2011 for the course MATH 102 taught by Professor Szypowsk during the Fall '08 term at UCSD.
 Fall '08
 SZYPOWSK
 Linear Algebra, Algebra, Vectors

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