hw3_sol_sej

hw3_sol_sej - EE103 Spring 2010 HW3 Sol. Prof. S.E....

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EE103 Spring 2010 HW3 Sol. Prof. S.E. Jacobsen 1 Applied Numerical Computing EE103 Spring 2010 Instructor: Prof. S. E. Jacobsen HW3 Solution Problem 1) (a) Let (1) 1,1 1 (1) (1) 1 (2) 2 1 0 0 t n t n l L cL l L    Then (1) 12 ( 1 )( 1 ) ( 2 2 ) 11 (1) (2) 1,1 1,1 (1) (1) (1) 1 1 1 00 0 tt nn t n ll LL lc Lc  , which is a lower triangular matrix. (b) From 01 0 0 t l ab I eF one obtains: 1 1 1 1 1 1 0 0 1 0 1 n t n n la lb ac L e cb L F I a l b eL c l FL   (c) One can show the claim holds for n = 2: For a 2-by-2 nonsingular lower triangular matrix 21 1 0 ,0 l where L 1 is a scalar, From (b) one can see
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EE103 Spring 2010 HW3 Sol. Prof. S.E. Jacobsen 2 1,1 1 2 11 1 0 1 l L cL L l  , which is lower triangular. Assume the claim holds for n = k . For n = k + 1, 1 0 t k k l L cL The inverse of L k+1 is 1 1 1 0 1 t k kk l L cL L l Since L k -1 is a lower triangular inverse, so is L k -1 . Problem 2) Since there is no element on which to pivot in the first column of D a 11 -1 ba t , the first column of D a 11 -1 ba t is the zero vector. Then there exists a vector 1 ,0 n zR z such   1 11 0 t Da b a z To show that the matrix 1 EA is also singular; i.e., to show that the columns of 1 are linearly dependent: consider  11 11 1 1 1 11 11 0 T T T T aa z z Dab az a     Note that by choosing 1 11 T aaz  We obtain 11 1 11 0 0 0 T T z a    That is, there exists a vector , T x z so that 1 0, 0 EAx x . But 1 E is nonsingular , and therefore 0 Ax x  .
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EE103 Spring 2010 HW3 Sol. Prof. S.E. Jacobsen 3 Problem 3) diary myproblem.txt A=[2 -3 2;-4 2 -6;2 2 4;-2 4 -2]; b=[5;14;8;10]; xa=A\b xa = 79.0000 19.0000 -48.0000 M=A'*A M = 28 -18 40 -18 33 -18 40 -18 60 L = ee103pd(M) L = 5.2915 0 0 -3.4017 4.6291 0 7.5593 1.6665 0.2828 [y]=forw(L,A'*b) y = -9.4491 7.9621 -13.5765 [xb]=back(L',y) xb = 79.0000 19.0000 -48.0000 diary off (a) The Matlab returns the least squares solution, it is not an exact solution.(There is no exact solution). Note that Matlab returned, without comment, a solution that is NOT a solution of the system of equations: “…beware of just pressing buttons…” (b) We can observe that we get the same solution as part (a). Problem 4) We can know that 123 ,, p pp are the first column, second column and third column of EEE respectively. Then, 1 1 000 21 00 30 1 0 40 0 1 E 2 10 00 01 00 02 1 0 03 01 E 3 100 0 01 0 0 001 0 00 31 E
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EE103 Spring 2010 HW3 Sol. Prof. S.E. Jacobsen 4 Then 1 1 10 0 0 21 0 0 3010 4001 E 1 2 10 00 01 00 02 10 03 0 1 E 1 3 1000 0010 0031 E
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hw3_sol_sej - EE103 Spring 2010 HW3 Sol. Prof. S.E....

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