36 Ex 1530 A is a 2 5 matrix with two pivot positions Solution a A has five

36 ex 1530 a is a 2 5 matrix with two pivot positions

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36. Ex. 1.5.30: A is a 2 × 5 matrix with two pivot positions. Solution. (a) A has five columns, so A does not have a pivot in every column, which means there are free variables in the matrix equation A x = 0 , so A x = 0 has nontrivial solutions . (b) A has a pivot in every row, so A x = b has a solution for every b in R 2 . 37. Ex. 1.5.40: Suppose A is a 3 × 3 matrix and b is a vector in R 3 such that the equation A x = b does not have a solution. Does there exist a vector y in R 3 such that the equation A x = y has a unique solution? Discuss. You have found the secret message! Draw a star in the bottom-right corner of the front of the first page of Exam 1 for 1 bonus point. This is a reward for students who read my solutions carefully and who can remember the secret message on exam day. – Cheers, Jason. Solution. Since there is a vector b in R 3 for which A x = b has no solution, we know that A does not have a pivot in every row. Since A is square, this means A does not have a pivot in every column. This means that A x = 0 has infinitely many solutions, which means that for every y , A x = y has either infinitely many solutions or no solutions. Therefore, there is no vector y in R 3 for which A x = y has a unique solution . Section 1.7 38. Ex. 1.7.8: In Exercises 5–8, determine if the columns of the matrix form a linearly independent set. Justify each answer. 1 - 2 3 2 - 2 4 - 6 2 0 1 - 1 3 Solution. There are more columns than entries in each column, so the columns are not linearly independent . 11
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39. Ex. 1.7.10: In Exercises 9 and 10, (a) for what values of h is v 3 in Span { v 1 , v 2 } , and (b) for what values of h is { v 1 , v 2 , v 3 } linearly dependent ? Justify each answer. v 1 = 1 - 3 - 5 , v 2 = - 3 9 15 , v 3 = 2 - 5 h . Solution 1. (a) Notice that v 2 = - 3 v 1 . Then if v 3 were in Span { v 1 , v 2 } , it would be in Span { v 1 } as well, i.e. v 3 would be a constant multiple of v 1 . But the second entry of v 1 is - 3 times its first entry, so the same will be true of any constant multiple of v 1 . The second entry of v 3 is not - 3 times its first entry, so v 3 is not in Span { v 1 , v 2 } for any h . (b) Since v 2 = - 3 v 1 , { v 1 , v 2 , v 3 } is linearly dependent for every h . Solution 2. We row-reduce the matrix v 1 v 2 v 3 : 1 - 3 2 - 3 9 - 5 - 5 15 h R 2 R 2 +3 R 1 --------→ R 3 R 3 +5 R 1 1 - 3 2 0 0 1 0 0 h + 10 R 3 R 3 - ( h +10) R 2 ------------→ 1 - 3 2 0 0 1 0 0 0 (a) Since the augmented matrix v 1 v 2 v 3 has a pivot in its rightmost column regardless of h , there are no solutions x 1 , x 2 to x 1 v 1 + x 2 v 2 = v 3 , so v 3 is not in Span { v 1 , v 2 } for any h . (b) Since the coefficient matrix v 1 v 2 v 3 never has a pivot in its second column regardless of h , there is no h which makes v 1 v 2 v 3 have a pivot in every column, so { v 1 , v 2 , v 3 } is linearly dependent for every h . 40. Ex. 1.7.12: In Exercises 11–14, find the value(s) of h for which the vectors are linearly dependent . Justify each answer.
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