18 11 points previous answers holtlinalg1 23049

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18. 1/1 points | Previous Answers HoltLinAlg1 2.3.049. Determine if the statement is true or false, and justify your answer. If is a linear combination of then is linearly independent. u 4 { u 1 , u 2 , u 3 }, { u 1 , u 2 , u 3 , False. If u 4 = x 1 u 1 + x 2 u 2 + x 3 u 3 , then x 1 u 1 + x 2 u 2 + x 3 u 3 u 4 = 0 , and since the coefficient of u 4 is 1, { u 1 , u 2 , u 3 , u 4 } is linearly dependent. u 4 }
Solution or Explanation False. If then and since the coefficient of is 1, is linearly dependent. u 4 = x 1 u 1 + x 2 u 2 + x 3 u 3 , x 1 u 1 + x 2 u 2 + x 3 u 3 u 4 = 0 , u 4 { u 1 , u 2 , u 3 , u 4 }
False. Consider u 1 = (1, 0, 0, 0), u 2 = (0, 1, 0, 0), u 3 = (0, 0, 1, 0), u 4 = (1, 1, 1, 1).
19. 1/1 points | Previous Answers HoltLinAlg1 2.3.051. Determine if the statement is true or false, and justify your answer. If is not a linear combination of then is linearly independent. u 4 { u 1 , u 2 , u 3 }, { u 1 , u 2 , u 3 , u 4 }
Solution or Explanation False. Consider True. If { u 1 , u 2 , u 3 , u 4 } is linearly dependent, then u 4 = x 1 u 1 + x 2 u 2 + x 3 u 3 , which is a contradiction.
False. Consider u 1 = (1, 0, 0), u 2 = (0, 1, 0), u 3 = (0, 0, 1), u 4 = (0, 1, 0).
20. 1/1 points | Previous Answers HoltLinAlg1 2.3.052. Determine if the statement is true or false, and justify your answer. If is not a linear combination of then is linearly dependent. u 4 } False. Consider u 1 = (1, 0, 0, 0), u 2 = (0, 1, 0, 0), u 3 = (0, 0, 1, 0), u 4 = (0, 0, 0, 1). True. The echelon form of the augmented matrix [ u 1 u 2 u 3 u 4 ] will have at least one row of zeroes at the bottom, which means the vectors are linearly dependent.