PS #5 SOLUTION - 1 ]8 ,3 ,4 ,1[ ,]1 ,0 ,1 ,2[ ,]1 ,1 ,2...

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Math 3013 Problem Set 5 Problems from § 2.1 (pgs. 134-136 of text): 1,3,11,12,13,16,23 Problems from § 2.2 (pgs. 140-141 of text): 1,3,5,7,11 Problems from § 2.3 (pgs. 152-154 of text): 1,2,3,4,5,7,13,15,19,29 1. (Problem 2.1.1 in text). Give a geometric criterion for a set of two distinct nonzero vectors in R 2 to be dependent. If two vectors v 1 and v 2 are linearly dependent, then there must exist a solution of c 1 v 1 + c 2 v 2 = 0 with at least one of the coefficients c 1 ,c 2 not zero. Suppose (without loss of generality) that c 2 =0 . Then c 1 can not equal zero either (otherwise we’d have c 2 v 2 = 0 with neither c 2 or v 2 zero). Then we can multiply both sides of this equation by 1 /c 2 to obtain c 1 c 2 v 1 + v 2 = 0 v 2 = - c 1 c 2 v 1 So v 2 must be a non-zero scalar multiple of v 1 . But then, this implies that v 2 is either parallel (or anti-parallel) to v 1 . 2. (Problem 2.1.3 in text). Give a geometric criterion for a set of two distinct nonzero vectors in R 3 to be dependent. By exactly the same reasoning we used in Problem 1, we can conclude that if two distinct non-zero vectors in R 3 are dependent then they must be parallel (or anti-parallel). 3. (Problem 2.1.11 in text). Find a basis for the subspace spanned by the vectors [1 , 2 , 1 , 1] , [2 , 1 , 0 , - 1] , [ - 1 , 4 , 3 , 8] , [0 , 3 , 2 , 5 R 4 . First we form a 4 × 4matr ix A whose columns correspond to the above set of vectors. A = 12 - 10 21 4 3 10 3 2 1 - 18 5 Now we row-reduce A to row-echelon form. R 2 R 2 - 2 R 1 R 3 R 3 - R 1 R 4 R 4 - R 1 -------------→ - 0 - 36 3 0 - 24 2 0 - 39 5 R 2 →- 1 3 R 2 R 3 R 3 - 2 3 R 2 R 4 R 4 - R 2 - 01 - 2 - 1 000 0 003 2 R 3 R 4 ------→ - - 2 - 1 The pivots of the final matrix (a row-echelon form of A ) are in the first three columns. Hence, the first three columns { [1 , 2 , 1 , 1] , [2 , 1 , 0 , - 1] , [ - 1 , 4 , 3 , 8] } 1
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2 of A will form a basis for the column space ColSp ( A )= span ([1 , 2 , 1 , 1] , [2 , 1 , 0 , - 1] , [ - 1 , 4 , 3 , 8] , [0 , 3 , 2 , 5]) 4. (Problem 2.1.12 in text). Find a basis for the column space of the matrix A = 23 1 52 1 17 2 6 - 20 We’ll apply the same technique used in Problem 3. 6 - 6 - 17 2 0 - 33 - 9 0 - 11 - 3 0 - 44 - 12 01 1 - 3 00 0 The pivots in the row-echelon form of A are in the first two columns. Therefore, the first two columns of A { [2 , 5 , 1 , 6] , [3 , 2 , 7 , - 2] } will form a basis for the column space of A . 5. (Problem 2.1.13 in text). Find a basis for the row space of the matrix A = 1357 2042 3287 The row space of A is the span of the row vectors { [1 , 3 , 5 , 7] , [2 . 0 , 4 , 2] , [3 , 2 , 8 , 7] } of A To find a basis for the span of these vectors we arrange them as the columns of a new matrix A A = 123 302 548 727 which happens to be the transpose of our original matrix A . We now row-reduce A . 12 3 0 - 6 - 7 0 - 6 - 7 0 - 12 - 14 067 000 = H The pivots of H are contained in the first two columns, therefore the first two columns of A form a basis for the column space of A
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This note was uploaded on 07/04/2011 for the course MATH 3013 taught by Professor Staff during the Spring '08 term at Oklahoma State.

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PS #5 SOLUTION - 1 ]8 ,3 ,4 ,1[ ,]1 ,0 ,1 ,2[ ,]1 ,1 ,2...

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