Quiz4PracticeProblemsSolutions.pdf - Extra Practice Solutions 1 Describe geometrically the following sets and find a simplified vector equation for each

Quiz4PracticeProblemsSolutions.pdf - Extra Practice...

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Extra Practice Solutions 1. Describe geometrically the following sets and find a simplified vector equation for each. (a) S 1 = Span ⇢ 1 2 (b) S 2 = Span 8 < : 2 4 1 2 3 3 5 , 2 4 2 3 4 3 5 , 2 4 1 1 1 3 5 9 = ; (c) S 3 = Span 8 > > < > > : 2 6 6 4 1 1 0 0 3 7 7 5 , 2 6 6 4 0 0 1 1 3 7 7 5 9 > > = > > ; Solution. (a) A vector equation for S 1 is ~ x = c 1 1 2 , c 1 2 R which represents a line through the origin in R 2 . (b) Since 2 4 1 1 1 3 5 = 2 4 2 3 4 3 5 - 2 4 1 2 3 3 5 we have that S 1 = Span 8 < : 2 4 1 2 3 3 5 , 2 4 2 3 4 3 5 , 2 4 1 1 1 3 5 9 = ; = Span 8 < : 2 4 1 2 3 3 5 , 2 4 2 3 4 3 5 9 = ; Since 2 4 1 2 3 3 5 and 2 4 2 3 4 3 5 are not scalar multiples of one another, a simplified vector equation for S 2 is ~ x = c 1 2 4 1 2 3 3 5 + c 2 2 4 2 3 4 3 5 , c 1 , c 2 2 R which represents a plane through the origin in R 3 . (c) Since the set 8 > > < > > : 2 6 6 4 1 1 0 0 3 7 7 5 , 2 6 6 4 0 0 1 1 3 7 7 5 9 > > = > > ; contains two vectors that are not scalar multiples of one another, a simplified vector equation for S 3 is ~ x = c 1 2 6 6 4 1 1 0 0 3 7 7 5 + c 2 2 6 6 4 0 0 1 1 3 7 7 5 , c 1 , c 2 2 R which represents a plane through the origin in R 4 . Page 1
2. Determine which of the following sets is linearly independent. If it is linearly dependent, write one of the vectors as a linear combination of the others. (a) ⇢ 1 2 , 1 3 , 1 4 (b) 8 < : 2 4 1 2 - 3 3 5 , 2 4 6 1 2 3 5 , 2 4 0 0 0 3 5 9 = ; (c) 8 < : 2 4 1 1 0 3 5 , 2 4 1 0 1 3 5 , 2 4 0 1 1 3 5 9 = ; Solution. (a) For c 1 , c 2 , c 3 2 R , consider c 1 1 2 + c 2 1 3 + c 3 1 4 = 0 0 We obtain the system of equations c 1 + c 2 + c 3 = 0 2 c 1 + 3 c 2 + 4 c 3 = 0 We see from the first equation that c 3 = - c 1 - c 2 and thus the second equation becomes 2 c 1 + 3 c 2 + 4( - c 1 - c 2 ) = 0, that is, - 2 c 1 - c 2 = 0. There are infinitely many solutions to this equation. In particular, if we let c 2 = 2, then c 1 = - 1 and it follows that c 3 = - ( - 1) - 2 = - 1.

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