Page 25 - Elementary_Linear_Algebra_with_Applications_Anton__9_edition

Interchanging the third and fourth rows and then multiplying the third row of the resulting matrix by gives the row-echelon form

Adding −3 times the third row to the second row and then adding 2 times the second row of the resulting matrix to the first row
yields the reduced row-echelon form

The corresponding system of equations is

(We have discarded the last equation, , since it will be satisfied automatically by the

solutions of the remaining equations.) Solving for the leading variables, we obtain

If we assign the free variables , , and arbitrary values r, s, and t, respectively, the general solution is given by the formulas

Back-Substitution

It is sometimes preferable to solve a system of linear equations by using Gaussian elimination to bring the augmented matrix into
row-echelon form without continuing all the way to the reduced row-echelon form. When this is done, the corresponding system of
equations can be solved by a technique called back-substitution. The next example illustrates the idea.

EXAMPLE 5 Example 4 Solved by Back-Substitution
From the computations in Example 4, a row-echelon form of the augmented matrix is

To solve the corresponding system of equations

we proceed as follows:
Step 1. Solve the equations for the leading variables.