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q1sols18085f05

# q1sols18085f05 - MIT OpenCourseWare http/ocw.mit.edu 18.085...

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MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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SOLUTIONS 18.085 Quiz 1 Fall 2005 1) (a) The incidence matrix A is 12 by 9. Its 4th row comes from edge 4: Row 4 of A = [ 0 0 0 0 1 1 0 0 0 ] (node 5 to node 6) (b) The 5th column of A indicates edges 3, 4, 9, 10 in and out of node 5: Column 5 of A = [ 0 0 1 1 0 0 0 0 1 1 0 0 ] The (5 , 5) entry in A T A is (column 5) T (column 5) = 4. The 5th row of A T A indicates nodes 2, 4, 6, 8 that are connected to node 5: Row 5 of A T A (also column 5) = [ 0 1 0 1 4 1 0 1 0 ] . (c) There are 4 independent solutions to A T w = 0 ( n r = 12 8 = 4 = number of loops). The lower left loop uses edges 1, 9, back on 3, back on 7: w loop = [ 1 0 1 0 0 0 1 0 1 0 0 0 ] (d) A T A is not positive definite because Au = 0 for u = [ 1 1 . . . 1 ] = ones (9 , 1).
2) (a) The equation u �� = ( x a ) with u (0) = 0 and u (1) = 0 is solved by x for x a u ( x ) = a for x a The slope drops from 1 to 0. The graph shows linear displacement

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