Linear Algebra

Section I. Solving Linear Systems 25
The next lemma finishes the proof of Theorem 3.1 by considering the particular
solution part of the solution set’s description.
3.8 Lemma For a linear system, where �p is any particular solution, the solution
set equals this set.
{�p + � h � � � h satisfies the associated homogeneous system}
Proof. We will show mutual set inclusion, that any solution to the system is
in the above set and that anything in the set is a solution to the system. ∗
For set inclusion the first way, that if a vector solves the system then it is in
the set described above, assume that �s solves the system. Then �s − �p solves the
associated homogeneous system since for each equation index i between 1 and
n,
ai,1(s1 − p1)+···+ ai,n(sn − pn) =(ai,1s1 + ···+ ai,nsn)
− (ai,1p1 + ···+ ai,npn)
= di − di
=0
where pj and sj are the j-th components of �p and �s. We can write �s − �p as � h,
where � h solves the associated homogeneous system, to express �s in the required
�p + � h form.
For set inclusion the other way, take a vector of the form �p + � h, where �p
solves the system and � h solves the associated homogeneous system, and note
that it solves the given system: for any equation index i,
ai,1(p1 + h1)+···+ ai,n(pn + hn) =(ai,1p1 + ···+ ai,npn)
+(ai,1h1 + ···+ ai,nhn)
= di +0
= di
where hj is the j-th component of � h. QED
The two lemmas above together establish Theorem 3.1. We remember that
theorem with the slogan “General = Particular + Homogeneous”.
3.9 Example This system illustrates Theorem 3.1.
Gauss’ method
−2ρ1+ρ2
−→
x +2y − z =1
2x +4y =2
y − 3z =0
x +2y − z =1
2z =0
y − 3z =0
ρ2↔ρ3
−→
∗ More information on equality of sets is in the appendix.
x +2y − z =1
y − 3z =0
2z =0