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Unformatted text preview: 18.06 Professor Strang Final Exam May 20, 2008 Grading 1 2 3 4 5 6 7 8 9 10 Your PRINTED name is: Please circle your recitation: 1) M 2 2131 A. Ritter 2085 21192 afr 2) M 2 4149 A. Tievsky 2492 34093 tievsky 3) M 3 2131 A. Ritter 2085 21192 afr 4) M 3 2132 A. Tievsky 2492 34093 tievsky 5) T 11 2132 J. Yin 2333 37826 jbyin 6) T 11 8205 A. Pires 2251 37566 arita 7) T 12 2132 J. Yin 2333 37826 jbyin 8) T 12 8205 A. Pires 2251 37566 arita 9) T 12 26142 P. Buchak 2093 31198 pmb 10) T 1 2132 B. Lehmann 2089 31195 lehmann 11) T 1 26142 P. Buchak 2093 31198 pmb 12) T 1 26168 P. McNamara 2314 41459 petermc 13) T 2 2132 B. Lehmann 2089 21195 lehmann 14) T 2 26168 P. McNamara 2314 41459 petermc Thank you for taking 18.06. If you liked it, you might enjoy 18.085 this fall. Have a great summer. GS 1 (10 pts.) The matrix A and the vector b are A = 1 1 2 1 4 b = 3 1 (a) The complete solution to Ax = b is x = . (b) A T y = c can be solved for which column vectors c = ( c 1 ,c 2 ,c 3 ,c 4 ) ? (Asking for conditions on the c ’s, not just c in C ( A T ) .) (c) How do those vectors c relate to the special solutions you found in part (a)? Solution (10 points) a) The complete solution is a particular solution x p plus any vector in the nullspace x n . Since the matrix A is already reduced, we can just read the special solutions off: [ 1 , 1 , , 0] T and [ 2 , , 4 , 1] T . To find a particular solution to Ax = b , we put any numbers (we may as well choose ) in for the free variables. This yields the two equations x 1 = 3 and x 3 = 1 , so x p = [3 , , 1 , 0] T . In the end we get x comp = 3 1 + c 1  1 1 + c 2  2 4 1 (1) b) You can do this computation by hand by augmenting A T with the column ( c 1 ,c 2 ,c 3 ,c 4 ) and row reducing. The solution is given by the equations that correspond to rows in the reduced matrix. A quicker way is to note that A T y = c has a solution whenever c is in the column space C ( A T ) , i.e. the row space of A . This is perpendicular to the nullspace. Thus, we can find the equations by taking a basis for the nullspace and using the components as coefficients in our equations. We find equations c 1 + c 2 = 0 and 2 c 1 4 c 3 + c 4 = 0 . c) Because these c are in the row space, they are perpendicular to vectors in the nullspace of A , and in particular are perpendicular to the special solutions. 2 2 (8 pts.) (a) Suppose q 1 = (1 , 1 , 1 , 1) / 2 is the first column of Q . How could you find three more columns q 2 ,q 3 ,q 4 of Q to make an orthonormal basis?...
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This note was uploaded on 08/10/2008 for the course MATH 250 taught by Professor Chanillo during the Spring '08 term at Rutgers.
 Spring '08
 CHANILLO

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