A First Course in Linear Algebra, 2017a

26 Systems of Equations
Example 1.24: Basic and Free Variables
Find the basic and free variables in the system
x + 2y − z + w = 3
x + y − z + w = 1
x + 3y − z + w = 5
Solution. Recall from the solution of Example 1.23 that the row-echelon form of the augmented matrix of
this system is given by
⎡
⎤
1 2 −1 1 3
⎣ 0 1 0 0 2 ⎦
0 0 0 0 0
You can see that columns 1 and 2 are pivot columns. These columns correspond to variables x and y,
making these the basic variables. Columns 3 and 4 are not pivot columns, which means that z and w are
free variables.
We can write the solution to this system as
x = −1 + s −t
y = 2
z = s
w = t
Here the free variables are written as parameters, and the basic variables are given by linear functions
of these parameters.
♠
In general, all solutions can be written in terms of the free variables. In such a description, the free
variables can take any values (they become parameters), while the basic variables become simple linear
functions of these parameters. Indeed, a basic variable x i is a linear function of only those free variables
x j with j > i. This leads to the following observation.
Proposition 1.25: Basic and Free Variables
If x i is a basic variable of a homogeneous system of linear equations, then any solution of the system
with x j = 0 for all those free variables x j with j > i must also have x i = 0.
Using this proposition, we prove a lemma which will be used in the proof of the main result of this
section below.
Lemma 1.26: Solutions and the Reduced Row-Echelon Form of a Matrix
Let A and B be two distinct augmented matrices for two homogeneous systems of m equations in n
variables, such that A and B are each in reduced row-echelon form. Then, the two systems do not
have exactly the same solutions.
Proof. With respect to the linear systems associated with the matrices A and B, there are two cases to
consider: