Final415practice_sol - Math 415 Final practice solutions...

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Math 415 — Final practice solutions Problem 1: Let’s start by writing the augmented matrix 1 1 - 2 1 2 1 0 0 2 1 1 2 - 1 0 1 1 1 1 1 0 We now use Gauss-Jordan reduction to get 1 0 0 2 1 0 1 0 - 1 - 1 / 3 0 0 1 0 - 2 / 3 0 0 0 0 0 Therefore w is a free variable and the solution is x = 1 - 2 w ; y = - 1 3 + w ; z = - 2 3 ; w R Problem 2: Lets’ start by setting v 1 = w 1 . Then according to the Gram- Schmidt process v 2 = w 2 - < w 2 , v 1 > || v 1 || 2 v 1 = 1 / 2 - 1 / 2 1 0 v 3 = w 3 - < w 3 , v 1 > || v 1 || 2 v 1 - < w 3 , v 2 > || v 2 || 2 v 2 = 0 0 0 1 Now we just need to normalize these vectors u 1 = 1 2 1 1 0 0 ; u 2 = 1 6 1 - 1 2 0 ; u 3 = 0 0 0 1 1
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Problem 3: If we put the vectors v 1 , v 2 , and v 3 in a matrix as columns we have A = 1 1 1 0 2 - 2 1 2 0 and using Gaussian elimination we have U = 1 1 1 0 2 - 2 0 0 0 Since we have only two pivots, only two vectors are linearly independent ( v 1 and v 2 , given the position of the pivots) but we need three vectors for a basis of R 3 . Let’s look for a vector w that is orthogonal to v 1 and v 2 . Starting from w = ( x,y,z ) T we impose orthogonality with v 1 and v 2 < v 1 , w > = x + z = 0; < v 2 , w > = x + 2 y + 2 z = 0 and solving these we get x = 2 y and z = - 2 y . So we can choose y = 1 to get the vector w = (2 , 1 , - 2) T . The vectors
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This note was uploaded on 02/10/2009 for the course MATH 415 taught by Professor Bertrandguillou during the Spring '08 term at University of Illinois at Urbana–Champaign.

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Final415practice_sol - Math 415 Final practice solutions...

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