Linear Algebra

46 Chapter One. Linear SystemsIIIReduced Echelon FormAfter developing the mechanics of Gauss’ method, we observed that it can bedone in more than one way. One example is that we sometimes have to swaprows and there can be more than one row to choose from. Another example isthat from this matrix (2)24 3Gauss’ method could derive any of these echelon form matrices.(2)2(1)1(2)00 −1 0 −1 0 −1The first results from −2ρ 1 + ρ 2 . The second comes from following (1/2)ρ 1 with−4ρ 1 + ρ 2 . The third comes from −2ρ 1 + ρ 2 followed by 2ρ 2 + ρ 1 (after the firstpivot the matrix is already in echelon form so the second one is extra work butit is nonetheless a legal row operation).The fact that the echelon form outcome of Gauss’ method is not uniqueleaves us with some questions. Will any two echelon form versions of a systemhave the same number of free variables? Will they in fact have exactly the samevariables free? In this section we will answer both questions “yes”. We willdo more than answer the questions. We will give a way to decide if one linearsystem can be derived from another by row operations. The answers to the twoquestions will follow from this larger result.III.1 Gauss-Jordan ReductionGaussian elimination coupled with back-substitution solves linear systems, butit’s not the only method possible. Here is an extension of Gauss’ method thathas some advantages.1.1 Example To solvex + y − 2z = −2y + 3z = 7x − z = −1we can start by going to echelon form as usual.⎛⎞ ⎛⎞1 1 −2 −2 1 1 −2 −2−ρ 1 +ρ 3−→ ⎝0 1 3 7 ⎠ ρ 2+ρ 3−→ ⎝0 1 3 7 ⎠0 −1 1 1 0 0 4 8We can keep going to a second stage by making the leading entries into ones⎛⎞1 1 −2 −2(1/4)ρ 3−→ ⎝0 1 3 7 ⎠0 0 1 2