WHY ROW REPLACEMENT DOESN’T CHANGE THE SOLUTION SET
A couple of you have asked about a rigorous explanation of why this is true. Here’s how
I would explain it.
Suppose we are given a system of linear equations in nvariables and mequations,
a11x1+. . . +a1nxn=b1
a21x1+. . . +a2nxn=b2
. . .
am1x1+. . . +amn xn=bm
For shorthand, lets call this system of equations S. We want to show that it has the same
solution as the system obtained by adding ctimes the ith row of Sto the jth row of S. (i
and jare arbitrary integers between 1 and m). We call this new, row-replaced system S0.
There are two things to show. Every solution of Sis a solution of S0AND every solution
of S0is a solution of S.
To start with, let t= (t1, . . . , tn) be a solution of S. We want to show it is also a solution
of S0. Note that
(1) ak1t1+. . . +akn tn=bk
holds for every value of k, and so it is easy to see that tis a solution for every equation of
S0(except possibly the jth one, ie. the modified one). The jth equation of S0is
c(ai1x1+. . . +ain xn) + aj1x1+. . . +aj nxn=cbi+bj.
However, if we substitute the t’s into the x’s, using (1), we get that
cbi+bj=cbi+bj,
which is certainly true. Therefore, every solution of Sis also a solution of S0.
To show the converse (that is, to show that every solution of S0is a solution of S) we note
the following fact. One can do a single row replacement operation to S0to re-obtain S(add
−ctimes row ito row j). Thus we can perform the same argument as above (with the roles
of Sand S0reversed) and see that every solution of S0is also a solution of S.
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