A First Course in Linear Algebra, 2017a

30 Systems of Equations
Example 1.31: Basic Solutions of a Homogeneous System
Consider the following homogeneous system of equations.
Find the basic solutions to this system.
x + 4y + 3z = 0
3x + 12y + 9z = 0
Solution. The augmented matrix of this system and the resulting reduced row-echelon form are
[ ] [ ]
1 4 3 0
1 4 3 0
→···→
3 12 9 0
0 0 0 0
When written in equations, this system is given by
x + 4y + 3z = 0
Notice that only x corresponds to a pivot column. In this case, we will have two parameters, one for y and
one for z. Lety = s and z = t for any numbers s and t. Then, our solution becomes
which can be written as
⎡
⎣
x
y
z
⎤
⎡
⎦ = ⎣
0
0
0
x = −4s − 3t
y = s
z = t
⎤
⎡
⎦ + s⎣
−4
1
0
⎤
⎡
⎦ +t ⎣
You can see here that we have two columns of coefficients corresponding to parameters, specifically one
for s and one for t. Therefore, this system has two basic solutions! These are
⎡ ⎤
−4
⎡ ⎤
−3
X 1 = ⎣ 1 ⎦,X 2 = ⎣
0
0 ⎦
1
We now present a new definition.
Definition 1.32: Linear Combination
Let X 1 ,···,X n ,V be column matrices. Then V is said to be a linear combination of the columns
X 1 ,···,X n if there exist scalars, a 1 ,···,a n such that
V = a 1 X 1 + ···+ a n X n
−3
0
1
⎤
⎦
♠
A remarkable result of this section is that a linear combination of the basic solutions is again a solution
to the system. Even more remarkable is that every solution can be written as a linear combination of these