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Then we may solve the linear system by the following procedure: Stage l: Write Ax = LUx = b. Stage 2: Set y = Ux, so that Ax = Ly = b. Use forward substitution on Ly = b to find y1, y2, . . , yn in that order, i.e., suppose the augmented matrix for the system Ly = b is: Then forward substitution yields y1 = b1/l11, and, subsequently, Note that the value of yi depends on the values y1, y2, . . , yi-1, which have already been calculated. Stage 3: Finally, use back-substitution on Ux = y to find xn, . . . , x1x in that order. Later on we shall outline a general method for finding LU decompositions of square matrices. The following example shows this method in action, involving the matrix A = L1U1 above. If we wish to solve Ax = b for a number of different bs, then this method is more efficient than Gauss elimination. Once we have found an LU decomposition of A, we need only do forward and backward substitutions to solve the system for any b. Procedure
Suppose we have to solve a linear system Ax = b, and that we can express the coefficient matrix A in the form of the socalled LU decomposition A = LU.