08spr-m1sols

08spr-m1sols - Math 51 Exam 1 Solutions April 22, 2008 1....

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Unformatted text preview: Math 51 Exam 1 Solutions April 22, 2008 1. (10 points) Compute, showing all steps, the reduced row echelon form of the matrix 0 2- 1 1 1 1 2 4 1 1 3 5 1 1 1 3 . 0 2- 1 1 1 1 2 4 1 1 3 5 1 1 1 3 swap swap 1 1 2 4 0 2- 1 1 1 1 3 5 1 1 1 3 - I- I 1 1 2 4 0 2- 1 1 0 0 1 1 0 0- 1- 1 ( 1 2 ) 1 1 2 4 0 1- 1 / 2 1 / 2 0 0 1 1 0 0- 1- 1 - II 1 0 5 / 2 7 / 2 0 1- 1 / 2 1 / 2 0 0 1 1 0 0- 1- 1 - 5 2 III + 1 2 III + III 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 Math 51, Spring 2008 Exam 1 Solutions April 22, 2008 Page 2 of 10 2. (15 points) Consider the matrix A below, and its reduced row echelon form (which you do not have to verify!): A = 1 0 1 3 1 0 2 1 0- 1 1 1 0 1 3 , rref ( A ) = 1 0 0 2 0 0 1 1 0 0 0 0 0 0 0 0 . (a) Find a basis for the null space of A . You do not need to prove that your collection is a basis. (5 points; more than one answer is possible, but this is the most direct solution.) From rref ( A ), we see that the system A x = has been reduced to: x 1 + 2 x 4 = 0 , x 3 + x 4 = 0 . Thus a solution vector x can be expressed: x = x 1 x 2 x 3 x 4 = - 2 x 4 x 2- x 4 x 4 = x 4 - 2- 1 1 + x 2 1 , so that N ( A ) is the span of the vectors - 2- 1 1 and 1 . As is always true for vectors obtained through this method, these vectors are linearly independent, which can be seen directly by con- sidering the second and fourth components. Thus, a basis for N ( A ) is - 2- 1 1 , 1 . Notes: Many people incorrectly took x 2 = 0 because there is no x 2 in any of the initial equations; this means they did not get 1 in the null space. (b) Are the columns of A linearly independent? If so, explain why; if not, write a non-trivial linear relation they satisfy. (2 points) No, the columns of A are linearly dependent, as seen by the presence of columns in rref ( A ) that do not contain pivots. If we denote the columns of A by v 1 , v 2 , v 3 , v 4 , then two relations that we can easily write down are: v 4 = 2 v 1 + v 3 , or v 1 + 1 v 2 + 0 v 3 + 0 v 4 = . Math 51, Spring 2008 Exam 1 Solutions April 22, 2008 Page 3 of 10 (ctd. from previous page) A = 1 0 1 3 1 0 2 1 0- 1 1 1 0 1 3 , rref ( A ) = 1 0 0 2 0 0 1 1 0 0 0 0 0 0 0 0 (c) Find a basis for the column space of A . You do not need to prove that your collection is a basis....
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08spr-m1sols - Math 51 Exam 1 Solutions April 22, 2008 1....

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