Matrices & Linear Algebra
Matrices & Linear Algebra MATH 2164
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Date Created: 10/25/15
Invariance of the Solution Set of a Linear System Under Elementary Row Operations Suppose we have the following linear system of m equations in 71 variables C11 612 Cin n d1 C21 622 62nn d2 Cm i Cm z 01min dm Recall that a solution is a list of numbers 5152 5 that when substi tuted for 1 2 xn makes each of the above equations a true statement and that the solution set is the set of all solutions this set contains none exactly one or in nitely many solutions as we discussed in class Question Why do the three elementary row operations row interchange multiplying a row by a nonzero constant and adding a multiple of one row into another not affect the solution set of a linear system Answer The book gives a perfunctory answer mentioning something about row operations being reversible but doesnt really give a satisfac tory answer l7m going to give you a more detailed explanation actually a proof but dont let that word scare you You dont have to know this proof for any exams so if this handout bores or confuses you then just skip it l7m providing this for those of you that want to understand this better Suppose we have two sets A and B We call A a subset of B denoted A C B if whenever an element x belongs to A denoted z E A then it also belongs to B That is if z E A implies z E B then A C B We say two sets A and B to be equal if they contain exactly the same elements Formally we de ne A B iff meaning if and only if each set is a subset of the other That is A B iff A C B and B C A If you ponder this formal de nition for a minute you7ll see that it agrees with your common sense So if we want to prove A B then according to the de nition we must show both A C B and B C A With these ideas in mind let us establish that performing elementary row operations does not change the solution set of a linear system of m equations in 71 variables In all of what follows7 let S be the solution set of the original linear system We will consider each of the three elementary row operations separately The easiest is the row interchange operation Suppose 517 52 75 is any solution to the initial linear system Then by de nition it must make each equation true when substituted for 1727 7x Now suppose we swap any two rows actually7 swap any number of rows jumble them all up7 and let S1 be the solution set of this new linear system Clearly the same solution 517 52 7 5 still satis es each equation7 because the equations themselves haven7t changedl Thus any solution of the original system is also a solution of the new system ie S C S1 On the other hand7 if 5152 75 is a solution to the new linear system with the rows jumbled up7 then we can just swap the equations back into their original order and the same reasoning still applies7 whence S1 C S By our de nition of equal sets7 S S17 so interchanging rows leaves the solution set the same Now suppose we multiply any row7 say the jth one7 by the nonzero con stant a 31 0 So our new linear system is the same as the original one7 except now the jth equation reads acjlzl acjgzg acjnzn adj Call the solution set of the new linear system S2 Well7 if 515275n makes the equation cjlzl cjgzg Bins dj true7 then it also makes the above equation true To see this7 just factor out and cancel the a from both sides ofthe modi ed jth equation This is allowed because we said a 31 0 otherwise we7d be dividing by zero Thus S C S2 Similarly7 if we started with the new system above7 then we could multiply the jth equation by i to get back the original linear system7 and by the same logic7 517 52 7 5 would solve that system Hence S2 C S and so S S2 Finally7 consider adding a nonzero multiple of one row into another row say we add 1 times rowj into row k Then the kth equation now reads czle cum 1072 6mm acm 61mm ad dk and the rest of the linear system is unchanged Call the solution set of this new linear system S3 lf 517 52 75 is any solution to the original system7 then substituting it for 17 2 7x must have made the following two equations true Cj i Cj z 6min dj Ck i Ck z Ckn n dk Well we argued in the previous paragraph this would imply that 51 52 5n solves the equation acjlzl acjgzg acjnzn adj so if we add this equation to the second equation above we get exactly 1071 Cki i 0072 Ckz z mm Ckn n adj dk Therefore S C Sg To see that it works the other way just add 7a tirnes equationj into equation k to recover the original kth equation Then Sg C S and so Sg S We7ve show that if you perform any of the three elernentary row opera tions then the solution set remains unchanged