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Unformatted text preview: 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 2. Consider the following vectors: u = 3 1 2 v = 1 1 2 3 w = 2 2 5 x = 6 2 4 y = 1 4 (a) Find a basis for the subspace S = Span { u , v , w , x , y } . (b) Is the vector 1 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 2 2 2 2 3 5 4 4 → 1 1 2 1 1 1 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...
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This homework help was uploaded on 04/01/2008 for the course MATH 601 taught by Professor Alndy during the Spring '08 term at A.T. Still University.
 Spring '08
 alndy
 Math

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