hwsolB10 - (iii) First, solve L*y = b = [1 2 -2 -1].' y1 =...

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Sheet1 Page 1 HW 10 Solution ______________ (i) To solve U*x = b for arbitrary x = [x1 x2 x3 x4].' and b = [b1 b2 b3 b4].': x4 = b4 -2*x4 + x3 = b3 therefore x3 = 2*b4 + b3 x2 - 2*x3 + x4 = b2 therefore x2 = -b4 + 2*(2*b4+b3) + b2 = b2 + 2*b3 + 3*b4 x1 - 2*x2 + x3 = b1 therefore x1 = b1 + 2*(b2 + 2*b3+ 3*b4) - (2*b4+b3) = b1 + 2*b2 + 3*b3 + 4*b4 inv(U) = 1 2 3 4 0 1 2 3 0 0 1 2 0 0 0 1 (ii) A = LU, L = A*inv(U) Ans: L = U^T = 1 0 0 0 -2 1 0 0 1 -2 1 0 0 1 -2 1 inv(L) = inv(U)^T Note that L is lower triangular and inv(L) is also lower triangular
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Unformatted text preview: (iii) First, solve L*y = b = [1 2 -2 -1].' y1 = 1 y y2 = 2*1 + 2 = 4 y3 = -y1 + 2*y2 - 2 = 5 y4 = -y2 + 2*y3 - 1 = -4 + 10 -1 = 5 Then solve U*x = y: From part (i) we have x4 = y4 = 5 x3 = 2*y4 + b3 = 15 x2 = y2 + 2*y3 + 3*y4 = 4 + 10 + 15 = 29 x1 = y1 + 2*y2 + 3*y3 + 4*y4 = 44 Sheet1 Page 2 (iv) G = D*P*L implies that the rows of L are first permuted, the scaled. P = 0 0 0 1 0 0 1 0 1 0 1 0 0 D = 4 0 0 0 3 0 0 0 2 0 0 0 1...
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This note was uploaded on 03/21/2010 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

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hwsolB10 - (iii) First, solve L*y = b = [1 2 -2 -1].' y1 =...

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