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# sol34 - Partial Solution Set Leon 3.4 3.4.3 Given the...

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Partial Solution Set, Leon § 3.4 3.4.3 Given the vectors x 1 = (2 , 1) T , x 2 = (4 , 3) T , and x 3 = (7 , - 3) T , (a) Show that x 1 and x 2 form a basis for R 2 . (b) Why must x 1 , x 2 , and x 3 be linearly dependent? (c) What is the dimension of Span( x 1 , x 2 , x 3 )? Solution : (a) This follows because they are (by inspection) linearly independent in R 2 . Since the dimension of R 2 is 2, x 1 and x 2 form a basis for R 2 . (b) Since R 2 is 2-dimensional, any collection of more than two vectors from R 2 must be linearly dependent. (c) Since x 1 and x 2 form a basis for R 2 , the dimension of Span( x 1 , x 2 , x 3 ) is 2, i.e., Span( x 1 , x 2 , x 3 ) = R 2 . 3.4.4 Given the vectors, x 1 = (3 , - 2 , 4) T , x 2 = ( - 3 , 2 , - 4) T , and x 3 = ( - 6 , 4 , - 8) T , what is the dimension of Span( x 1 , x 2 , x 3 )? Solution : Both x 2 and x 3 are scalar multiples of x 1 , so the dimension of Span( x 1 , x 2 , x 3 ) is 1. 3.4.5 Given the vectors, x 1 = (2 , 1 , 3) T , x 2 = (3 , - 1 , 4) T , and x 3 = (2 , 6 , 4) T , (a) Show that x 1 , x 2 , and x 3 are linearly dependent. (b) Show that x 1 and x 2 are linearly independent. (c) What is the dimension of Span( x 1 , x 2 , x 3 )? (d) Give a geometric description of Span( x 1 , x 2 , x 3 ). Solution : (a) Letting A = x 1 x 2 x 3 , we consider the solutions to A x = 0 . (Yes, this is one of those situations in which the matrix turns out to be square, so the determinant is a possibility. But it is not recommended. Use Gaussian elimination instead.)

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