# A x 0 true for a x 0 to have no nontrivial solutions

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A x = True. For A x = 0 to have no nontrivial solutions, the columns of A could not span R n . False. Consider A = [0]. False. Consider A = [1]. 0 .
A x = b
17. 1/1 points | Previous Answers HoltLinAlg1 2.2.057. Determine if the statement is true or false, and justify your answer. If { u 1 , u 2 , u 3 } spans R 3 , then so does { u 1 , u 2 , u 3 , True, the span of a set of vectors can only increase (with respect to set containment) when adding a vector to the set. True, since the span of { u 1 , u 2 , u 3 , u 4 } is a subset of the span of { u 1 , u 2 , u False. Consider u 4 = u 1 u 4 }. 3 }. . .
18. 1/1 points | Previous Answers HoltLinAlg1 2.2.058. Determine if the statement is true or false, and justify your answer. If { u 1 , u 2 , u 3 } does not span R 3 , then neither does { u 1 , u 2 , u 3 , u 4 }. Solution or Explanation False. Consider and }. 3 }.
19. 1/1 points | Previous Answers HoltLinAlg1 2.2.059. Determine if the statement is true or false, and justify your answer. If { u 1 , u 2 , u 3 , u 4 } spans R 3 , then so does { u 1 , u 2 , u 3 }. }. 3 }.
20. 1/1 points | Previous Answers HoltLinAlg1 2.2.060. Determine if the statement is true or false, and justify your answer. If { u 1 , u 2 , u 3 , u 4 } does not span R 3 , then neither does { u 1 , u 2 , u 3 }. 1).
Solution or Explanation True. The span of { u 1 , u 2 , u 3 } will be a subset of the span of { u 1 , u 2 , u 3 , u 4 }. True. The span of { u 1 , u 2 , u 3 , u 4 } will be a subset of the span of { u 1 , u 2 , u 3 }. True. The span of { u 1 , u 2 , u 3 } will be a subset of the span of { u 1 , u 2 , u 3 , u 4 }. False. Consider u 1 = (1, 0, 0), u 2 = (0, 1, 0), u 3 = (0, 0, 1), and u 4 = (0, 0, 0). False. Consider u 1 = (0, 0, 0), u 2 = (1, 0, 0), u 3 = (0, 1, 0), and u 4 = (1, 1, 0).