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Unformatted text preview: Mathematics 54W Professor A. Ogus Spring 2005 Final Solution SetMay 18, 2005 Work each problem on a separate sheet of paper. Be sure to put your name, your section number, and your GSIs name on each sheet of paper. Also, at the top of the page, in the center, write the problem number, and be sure to put the pages in order. Write clearlyexplanations (with complete sentences when appropriate) will help us understand what you are doing. Math 49 students taking the linear algebra portion should work problems 15; Math 49 students taking the differential equation should work problems 610. 1. Let A := 1- 2 1 1- 2- 1 2- 1 1 1- 2 1- 1 . (2 points each) (a) Find a matrix B which is in reduced row echelon form and which is row equivalent to A . B = 1- 2 1 0- 1 0 1- 1 0 0 (b) Find a basis for the null space of A . 1 1 1 , - 1 1 , 2 1 . (c) Find a basis for the column space of A from among the columns of A . 1- 1 1 , 1- 1 . (d) Find all X such that AX = 2- 1 . X = 1 1 + X , where X is any element of the null space of A . (e) Find at least one X such that A T AX = A T B , where B = 3- 2 1 . Hint: you do not need to calculate A T A to do this. Recall that the above equation says that AX is the orthogonal projection B of B on the column space W of A . Luckily we have above an orthogonal basis w 1 ,w 2 for W , and so B = ( B | w 1 ) ( w 1 | w 1 ) w 1 + ( B | w 2 ) ( w 2 | w 2 ) w 2 = 2 1- 1 1 + 1- 1 Hence X = 2 1 plus any element of the null space of A will do. 2. Write the definition of each of the following concepts. Use complete sentences and be as precise as you can. (2 points each) (a) The inverse of a matrix. Let A be an n n matrix. Then the inverse of A is a matrix B such that BA = AB = I ; such a matrix is unique if it exists. It can be proved that either of these conditions implies the other, but only for n n matrices. (b) A linearly independent sequence in a vector space V . A sequence ( v 1 ,v 2 ,...v n ) is linearly independent if for every se- quence of numbers ( c 1 ,c 2 ,...c n ) such that c 1 v 1 + c n v n = 0, each c i = 0....
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