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# solution12 - M341 Homework Solution 12 Section 4.6 1(c...

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M341 Homework Solution 12 Section 4.6: 1(c): Using the Simpli ed Span Method, 0 1 1 0 6 2 1 0 2 1 1 2 1 1 2 3 2 0 2 3 1 1 1 1 4 2 1 1 1 3 is reduced to 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 Therefore { [1 , 0 , 0 , 0 , 0] , [0 , 1 , 0 , 0 , 0] , [0 , 0 , 1 , 0 , 0] , [0 , 0 , 0 , 1 , 0] , [0 , 0 , 0 , 0 , 1] } is a basis for V . 4(f): We can use Independence Test Method to nd a basis. 3 2 0 0 6 1 1 1 0 5 2 5 0 7 0 21 0 7 is reduced to 1 0 0 2 2 0 0 1 0 3 0 0 0 0 0 0 0 1 The pivot columns are 1st, 2nd and 6th columns. Therefore B = { [3 , 1 , 0] , [2 , 1 , 7] , [1 , 5 , 7] } is a basis for V by independence test method. 4(g): Consider B = { [1 , 0 , 0] , [0 , 0 , 1] } . I will show that B is a basis for span ( S ) = V . It is obvious that B is a subset of S and B is linearly independent. It remains to show that span ( B ) = V . First we have span ( B ) span ( S ) = V , since B S . For any vector v in S , v is of the form v = [ a, 0 , b ] for some a, b R . Hence we have v = a [1 , 0 , 0] + b [0 , 0 , 1] and therefore v span ( B ) . It follows S span ( B ) and V = span ( S ) span ( B ) . This shows span ( B ) = V and thus B is a basis for V . 5(f): Consider B = { 8 x 3 + 1 , 8 x 3 + x, 8 x 3 + x 2 , 8 x 3 } . It is obviously that B is a subset of S . To see B is a basis for V , we check if B is linearly independent rst. Suppose there exists a 1 , a 2 , a 3 , a 4 R such that a 1 (8 x 3 + 1) + a 2 (8 x 3 + x ) + a 3 (8 x 3 + x

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## This note was uploaded on 12/05/2011 for the course M 341 taught by Professor Hietmann during the Spring '08 term at University of Texas.

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solution12 - M341 Homework Solution 12 Section 4.6 1(c...

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