Linear Algebra

Section I. Solving Linear Systems 31.4 Example To solve this system3x 3 = 9x 1 + 5x 2 − 2x 3 = 213 x 1 + 2x 2 = 3we transform it, step by step, until it is in a form that we can easily solve.The first transformation rewrites the system by interchanging the first andthird row.swap row 1 with row 3−→13 x 1 + 2x 2 = 3x 1 + 5x 2 − 2x 3 = 23x 3 = 9The second transformation rescales the first row by multiplying both sides ofthe equation by 3.multiply row 1 by 3−→x 1 + 6x 2 = 9x 1 + 5x 2 − 2x 3 = 23x 3 = 9The third transformation is the only nontrivial one in this example. We mentallymultiply both sides of the first row by −1, mentally add that to the second row,and write the result in as the new second row.add −1 times row 1 to row 2−→x 1 + 6x 2 = 9−x 2 − 2x 3 = −73x 3 = 9The point of these steps is that we’ve brought the system to a form where we caneasily find the value of each variable. The bottom equation shows that x 3 = 3.Substituting 3 for x 3 in the middle equation shows that x 2 = 1. Substitutingthose two into the top equation gives that x 1 = 3. Thus the system has a uniquesolution; the solution set is { (3, 1, 3) }.Most of this subsection and the next one consists of examples of solvinglinear systems by Gauss’s Method. We will use it throughout the book. It is fastand easy. But before we do those examples we will first show that this Methodis also safe in that it never loses solutions or picks up extraneous solutions.1.5 Theorem (Gauss’s Method) If a linear system is changed to another by one ofthese operations(1) an equation is swapped with another(2) an equation has both sides multiplied by a nonzero constant(3) an equation is replaced by the sum of itself and a multiple of anotherthen the two systems have the same set of solutions.