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Unformatted text preview: bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg 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). All points on the plane have the form (fill in the first components) x y = + y 1 + z . z 1 Solution. (4 points) The equation x = 12 + 3 y + z says it all: x 12 + 3 y + z 12 3 1 y = y = 0 + y 1 + z 0 . z z 1 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 , , 1) and (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! A Problem 3. ( § 3.2, #36) How is the nullspace N ( C ) related to the spaces N ( A ) and N ( B ), if C = ? B Solution. (12 points) N ( C ) = N ( A ) ∩ N ( B ) just the intersection: Indeed, A x C x = B x 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). 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 bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg or in matrix form A y = 0 for 1 − 1 1 0 0 − 1 1 0 1 0 A = 0 − 1 1 0 0 1 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 ....
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This note was uploaded on 08/22/2011 for the course MATH 1806 taught by Professor Strang during the Fall '10 term at MIT.
 Fall '10
 Strang
 Linear Algebra, Algebra

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