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HWSolutions2.1&amp;2.2&amp;2.3

# HWSolutions2.1&amp;2.2&amp;2.3 - Newberger Math 247...

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Newberger Math 247 Spring 03 Homework solutions: Section 2.1 #18,20-24, Section 2.2 #14, 16, 18, 20 and Section 2.3 #16, 18, 22 Section 2.1 #18,20-24. 18. Suppose the first two columns, b 1 and b 2 , of B are equal. What can you say about the columns of AB (if AB is defined)? Why? We start with the assumption (the ”Suppose” part) that b 1 = b 2 . So A b 1 = A b 2 . Now we know AB = [ A b 1 , A b 2 , . . . , A b n ] , so we see that the first two columns of AB are also equal. 20. Suppose the second column of B is all zeros. What can you say about the second column of AB ? We start with the assumption that b 2 = 0 . So A b 2 = A 0 = 0 . Now we know AB = [ A b 1 , A b 2 , . . . , A b n ] = [ A b 1 , 0 , . . . , A b n ] , so we see that the second column of AB is all zeros as well. 21. Suppose the last column of AB is entirely zero but B itself has no column of zeros. What can you say about the columns of A ? Let B = [ b 1 , . . . , b n ] . Then AB = [ A b 1 , A b 2 , . . . , A b n ] . We start with the assumption that the last column of AB is the zero vector. So we get that A b n = 0 . Thus x = b n is a solution to the equation A x = 0 . Since B has no columns that are entirely zero, we know b n 6 = 0 , so x = b n is a non-trivial solution to the equation A x = 0 . Since the homogeneous equation A x = 0 has non-trivial solutions, the columns of A are linearly dependent. 22. Show that if the columns of B are linearly dependent, then so are the columns of AB . You might want to reword this problem into the “Suppose..., Explain why...” form that you are used to. In that form the statement would be: Suppose that the columns of B are linearly dependent. Explain why the columns of AB are also independent. Now we begin with the assumption (the ”Suppose” part) that the columns of B are linearly dependent. One way for the columns of B to be linearly dependent is for one column to be a multiple of another column. However we cannot assume that that is the case, because there are other relationships between the columns of B that would cause the columns to be linearly dependent. Theorem 7 on page 68 gives us a characterization of linearly dependent sets that is always true.

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HWSolutions2.1&amp;2.2&amp;2.3 - Newberger Math 247...

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