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Unformatted text preview: Math 330 Homework 4.3 (Pages 243,244,245) For problems (4) and (6) decide if the collection is (i) linearly independent and/or (ii) spans R 3 and/or (iii) is a basis for R 3 (4) 2 2 1 , 1 3 2 ,  7 5 4 SOLUTION: Since there are 3 vectors, they will span R 3 iff they are linearly independent iff they form a basis for R 3 . To see if they do, we make them the columns of a single matrix which we can row reduce: A = 2 1 7 2 3 5 1 2 4 R 1 ↔ R 3 = ⇒ 1 2 4 2 3 5 2 1 7 ( R 2 ,R 3) ← ( R 2+2 R 1 ,R 3 2 R 1) = ⇒ 1 2 4 1 13 3 15 R 3 ← R 3+3 R 2 = ⇒ 1 2 4 0 1 13 0 0 24 This is Echelon form. We have three pivots and so the vectors are linearly independent, they span  bfR 3 , and they form a basis for R 3 . (6) 1 2 3 ,  4 5 6 SOLUTION: Since there are only two vectors we know that they cannot span R 3 (they can only span a 2dimensional plane within R 3 ) and therefore they cannot be a basis for R 3 . Furthermore, it is always easy to see if two vectors are linearly independent  we just need to see if they are multiples of each other. They are not multiples of each other and hence they are linearly independent....
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.
 Spring '08
 Johnson,J
 Math, Linear Algebra, Algebra, Vectors

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