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By Mariesa L. Crow

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-- IEEE strength & power journal

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2. A quick check to verify the correctness of the solution is to substitute the solution vector x back into the linear system Ax = b. 1 LU Factorization with Partial Pivoting The LU factorization process presented assumes that the diagonal element is non-zero. Not only must the diagonal element be non-zero, it must be on the same order of magnitude as the other non-zero elements. 30) By inspection, the solution to this linear system is x1 ≈ 2 x2 ≈ 1 The LU factors for A are L= 0 10−10 2 1 − 2 × 1010 U = 1 1010 0 1 Applying forward elimination to solve for the dummy vector y yields: y1 = 1010 5 − 2 × 1010 ≈1 y2 = (1 − 2 × 1010 ) Back substituting y into U x = y yields x2 = y2 ≈ 1 x1 = 1010 − 1010 x2 ≈ 0 The solution for x2 is correct, but the solution for x1 is considerably off.

In this approach, two permutation matrices are developed: one for row exchange as in partial pivoting, and a second matrix for column exchange. 33) Therefore to solve the linear system of equations Ax = b requires that a slightly different approach be used. 36) can be solved using forward and backward substitution for z. 35). In complete pivoting, both rows and columns may be interchanged to place the largest element (in magnitude) on the diagonal at each step in the LU factorization process. The pivot element is chosen from the remaining elements below and to the right of the diagonal.

N where θ= π n+1 (b) Let a = 2. i. Will the Jacobi iteration converge for this matrix? ii. Will the Gauss-Seidel iteration converge for this matrix? 12. An alternative conjugate gradient algorithm for solving Ax = b may be based on the error functional Ek xk = xk − x, xk − x where · denotes inner product. The solution is given as xk+1 = xk + αk σk Using σ1 = −AT r0 and σk+1 = −AT rk + βk σk , derive this conjugate gradient algorithm. 7 using this conjugate gradient algorithm. 13. Write a subroutine with two inputs (A, flag) that will generate for any non-singular matrix A, the outputs (Q, P ) such that if • flag=0, A = LU, P + I • flag=1, P A = LU where ⎡ l11 ⎢ l21 ⎢ ⎢ L = ⎢ l31 ⎢ ..

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