Math 2.0

concept

Back substitution

Summary

Back substitution is a method to solve a system Ux=bUx = b, where UU is an n×nn \times n upper triangular matrix with nonzero diagonal entries. The last equation gives xn=bn/unnx_n = b_n / u_{nn}. Then, for i=n−1,…,1i = n-1, \dots, 1 in this order, equation ii gives xi=1uii(bi−∑j=i+1nuijxj),x_i = \frac{1}{u_{ii}} \left( b_i - \sum_{j = i+1}^{n} u_{ij} x_j \right), which uses only the unknowns that are already known.

Remarks

Back substitution uses on the order of n2n^2 arithmetic operations. Forward substitution solves Lx=bLx = b for a lower triangular matrix LL in the same way, from the first equation down. After Gaussian elimination, back substitution completes the solution of the system. With an LU decomposition A=LUA = LU, the system Ax=bAx = b is solved by forward substitution for Ly=bLy = b and then back substitution for Ux=yUx = y.

Explanations

L1 human

In a triangular system, the last equation has only one unknown, so we solve it at once. We put that value into the equation above it, which now has only one new unknown. We continue upwards until we know all the unknowns.

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Sources

  1. Web https://en.wikipedia.org/wiki/Triangular_matrix, Forward and back substitution