Solutions 2 - Math 601 Solutions to Homework 2 1 Consider...

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Math 601 Solutions to Homework 2 1. Consider the following matrix: A = 1 2 1 2 5 4 1 - 1 - 4 (a) Compute A - 1 . (b) Use A - 1 to solve A x = 2 - 1 2 . Answer: (a) First we compute A - 1 : we create the augmented matrix ( A | I ) and row reduce. 1 2 1 1 0 0 2 5 4 0 1 0 1 - 1 - 4 0 0 1 -→ 1 0 0 - 16 7 3 0 1 0 12 - 5 - 2 0 0 1 - 7 3 1 Thus, A - 1 = - 16 7 3 12 - 5 - 2 - 7 3 1 (b) We can solve for x by multiplying both sides of the matrix equation by A - 1 . We get x = A - 1 2 - 1 2 = - 16 7 3 12 - 5 - 2 - 7 3 1 2 - 1 2 = - 33 25 15 1
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2. Consider the following vectors: u = 3 1 0 2 v = 1 1 2 - 3 w = 2 0 - 2 5 x = 6 2 0 4 y = 1 0 0 4 (a) Find a basis for the subspace S = Span { u , v , w , x , y } . (b) Is the vector 1 0 2 - 1 in S ? Answer: (a) Here are two methods to find a basis for S . Method 1: We create a matrix with the vectors as columns, and then row reduce: 3 1 2 6 1 1 1 0 2 0 0 2 - 2 0 0 2 - 3 5 4 4 -→ 1 0 1 2 0 0 1 - 1 0 0 0 0 0 0 1 0 0 0 0 0 In the row reduced matrix, we see that the columns 1, 2, and 5 are linearly independent (these are the columns containing piv- ots), and that adding any other column to this list would create a dependency. Since row operations preserve the dependencies among the columns, we know that columns 1, 2, and 5 from the
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