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pset3-s10-soln

# pset3-s10-soln - 18.06 PSET 3 SOLUTIONS Problem 1(3.2#18...

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18.06 PSET 3 SOLUTIONS FEBRUARY 22, 2010 Problem 1. ( § 3.2, #18) The plane x 3 y z = 12 is parallel to the plane x 3 y z = 0 in Problem 17. One particular point on this plane is (12 , 0 , 0). All points on the plane have the form (fill in the first components) x y z = 0 0 + y 1 0 + z 0 1 . Solution. (4 points) The equation x = 12 + 3 y + z says it all: x y z = 12 + 3 y + z y z = 12 0 0 + y 3 1 0 + z 1 0 1 . square Problem 2. ( § 3.2, #24) (If possible. . . ) Construct a matrix whose column space contains (1 , 1 , 0) and (0 , 1 , 1) and whose nullspace contains (1 , 0 , 1) and (0 , 0 , 1). Solution. (4 points) Not possible : Such a matrix A must be 3 × 3. Since the nullspace is supposed to contain two independent vectors, A can have at most 3 2 = 1 pivots. Since the column space is supposed to contain two independent vectors, A must have at least 2 pivots. These conditions cannot both be met! square Problem 3. ( § 3.2, #36) How is the nullspace N ( C ) related to the spaces N ( A ) and N ( B ), if C = bracketleftbigg A B bracketrightbigg ? Solution. (12 points) N ( C ) = N ( A ) N ( B ) just the intersection: Indeed, C x = bracketleftbigg A x B x bracketrightbigg so that C x = 0 if and only if A x = 0 and B x = 0. (...and as a nitpick, it wouldn’t be quite sloppy instead write “if and only if A x = B x = 0”—those are zero vectors of potentially different length, hardly equal). square Problem 4. ( § 3.2, #37) Kirchoff’s Law says that current in = current out at every node. This network has six currents y 1 ,...,y 6 (the arrows show the positive direction, each y i could be positive or negative). Find the four equations A y = 0 for Kirchoff’s Law at the four nodes. Find three special solutions in the nullspace of A . Solution. (12 points) The four equations are, in order by node, y 1 y 3 + y 4 = 0 y 1 + y 2 + y 5 = 0 y 2 + y 3 + y 6 = 0 y 4 y 5 y 6 = 0 1

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or in matrix form A y = 0 for A = 1 0 1 1 0 0 1 1 0 0 1 0 0 1 1 0 0 1 0 0 0 1 1 1 Adding the last three rows to the first eliminates it, and shows that we have three “pivot variables” y 1 ,y 2 ,y 4 and three “free variables” y 3 ,y 5 ,y 6 . We find the special solutions by back-substitution from ( y 3 ,y 5 ,y 6 ) = (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1): y 1 y 2 y 3 y 4 y 5 y 6 = 1 1 1 0 0 0 , 1 0 0 1 1 0 , 1 1 0 1 0 1
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