44 CHAPTER 2. SYSTEMS OF LINEAR EQUATIONS

As illustrated above, such a system of linear equations may have a unique solution, nosolution, or infinitely many solutions and these are the only three cases which can occur forany linear system. Furthermore, you do exactly the same things to solve any linear system.You write the augmented matrix and do row operations until you get a simpler system inwhich it is possible to see the solution, usually obtaining a matrix in echelon or reducedechelon form. All is based on the observation that the row operations do not change thesolution set. You can have more equations than variables, fewer equations than variables,etc. It doesn’t matter. You always set up the augmented matrix and go to work on it.

Definition 2.2.8 A system of linear equations is called consistent if there exists a solution.It is called inconsistent if there is no solution.

These are reasonable words to describe the situations of having or not having a solution.If you think of each equation as a condition which must be satisfied by the variables, con-sistent would mean there is some choice of variables which can satisfy all the conditions.Inconsistent would mean there is no choice of the variables which can satisfy each of theconditions.

2.3 When are Two Polynomials Relatively Prime?Suppose you have two polynomials having coefficients in a field of scalars F. How canyou tell if they are relatively prime? One way is outlined in an earlier excercise. Here isanother. By the method of partial fractions if p(x) ,q(x) are relatively prime polynomialsof degree at least 1, then it follows from the partial fractions theorem, Proposition 1.14.2that there is a partial fractions expansion of the following form.

1p(x)q(x)

=a(x)p(x)

+b(x)q(x)

where the degree of a(x) is smaller than the degree of p(x) and the degree of b(x) issmaller than the degree of q(x). Conversely, if there is such a partial fractions expansion,then 1 = a(x)q(x)+b(x) p(x) and these two polynomials are relatively prime. Checkingthe existence of such a partial fractions expansion is a simple example of finding a solutionto a linear system of equations. Thus this is a question which can be resolved withouthaving to factor the polynomials and instead uses the method of row operations to resolvethe question.

Example 2.3.1 Consider x2− 3x+ 2 and x4− 4x2 + 4. Then these are relatively prime ifand only if there is such a partial fractions expansion just described. We would need

1(x2−3x+2)(x4−4x2 +4)

=ax+b

x2−3x+2+

c+dx+ ex2 + f x3

x4−4x2 +4

Now multiply and write

1 = (ax+b)(x4−4x2 +4

)+(c+dx+ ex2 + f x3)(x2−3x+2

)

= (a+ f )x5 +(b−3 f + e)x4 +(d−4a+2 f −3e)x3

+(c−4b−3d +2e)x2 +(4a−3c+2d)x+(4b+2c)