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We shall solve the system Stage l: An LU decomposition of the system is Stage 2: Set y = U1x and then solve the system L1y = b, i.e., Using forward substitution, we obtain: Stage 3: Solve Back-substitution now yields: Thus the solution of Ax = b is: which may be checked, using the original equations. We turn now to the problem of finding an LU decomposition of a given square matrix A. Example
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