final_sol_1806-S05

# Final_sol_1806-S05 - 18.06 Final Exam Monday May 16th 2005 solutions 1(a We want the coordinates(ai1 ain to satisfy the equation c1 x1 cn xn = 1

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18.06 - Final Exam, Monday May 16th, 2005 solutions 1. (a) We want the coordinates ( a i 1 ,...,a in ) to satisfy the equation c 1 x 1 + ... + c n x n =1; thus the system of equations we want to solve is Ac = ones : c 1 a 11 + c 2 a 12 + + c n a 1 n =1 c 1 a 21 + c 2 a 22 + + c n a 2 n c 1 a n 1 + c 2 a n 2 + + c n a nn (b) There is no plane of the given form, if one of the points P i is the origin. In R 3 an example is given by the points (0 , 0 , 0), (1 , 0 , 0) and (0 , 1 , 0): they lie in the (unique) plane x 3 =0 and this plane does not have the required form. More than one plane contains the P i ’s if the three points are on a line not through the origin. (c) There is not a unique solution precisely det A = 0. This means geometrically that the points P i lie in an ( n 1) dimensional subspace of R n . 2. (a) Subtracting the ﬁrst row from the second, we ﬁnd the matrix 12345 U = 00001 . 00000 (In the row reduced echelon form R , the 5 changes to 0.) The pivot variables are the ﬁrst and the last, while the remaining ones are the free variables. Thus the “special solutions” to Ax are 2 3 4 1 0 0 0 , 1 , 0 . 0 1 0 0 0 0 (b) and (c) need to prove that these three vectors are linearly independent and they span the nullspace . By considering the second, third and fourth coordinates, a combination of the vectors adding to zero must have zero coeﬃcients. The vectors span the nullspace, since the dimension of the nullspace is three (note that the rank of the matrix A is 2). 3. (a) The condition that Ax = b has no solution means that the column space of A has dimension strictly smaller than m . In particular, the rank is r< m .S i n c e A T y = c has exactly one solution, the columns of A T are independent. This means that the rank of A T is r = m . This contradiction proves that we cannot ﬁnd A , b and c .

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## This note was uploaded on 01/14/2011 for the course EECS 18.06 taught by Professor Strang during the Spring '05 term at University of Michigan.

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Final_sol_1806-S05 - 18.06 Final Exam Monday May 16th 2005 solutions 1(a We want the coordinates(ai1 ain to satisfy the equation c1 x1 cn xn = 1

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