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A^Tx=b given LU factorization for A
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[QUOTE="dbkats, post: 3661472, member: 378413"] [h2]Homework Statement [/h2] Suppose you are given the LU factorization for some nxn square matrix A. Assume A is non-singular. This factorization is a result of partial pivoting. Can you use this factorization to solve A^Tx=b for x (given A and b).[h2]Homework Equations[/h2] A^T is the transpose of matrix A. PA = LU is the assumed factorization of A with partial pivoting Since P is a permutation matrix, P^T=P^-1 [h2]The Attempt at a Solution[/h2] Haha...I figured it out... PA=LU A = (P^T)LU A^T = (U^T)(L^T)P A^Tx = (U^T)(L^T)Px = b Then let Px = y A^Tx = (U^T)(L^T)y = b U^T is then lower-triangular, L^T is unit-upper-triangular. Therefore I can solve for y in the usual way, and then figure out what x is based on the permutation matrix P. [/QUOTE]
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A^Tx=b given LU factorization for A
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