Fundamentals of Matrix Algebra, 2011a

Chapter 1
Systems of Linear Equaons
is chosen, the value of x 1 is determined. We have infinite choices for the value of x 2 ,
so therefore we have infinite soluons.
For example, if we set x 2 = 0, then x 1 = 1; if we set x 2 = 5, then x 1 = −4. .
Let’s try another example, one that uses more variables.
. Example 9 Find the soluon to the linear system
x 2 − x 3 = 3
x 1 + 2x 3 = 2
−3x 2 + 3x 3 = −9
.
S To find the soluon, put the corresponding matrix into reduced
row echelon form.
⎡
⎤
⎡
⎤
⎣ 0 1 −1 3
1 0 2 2 ⎦
−→
rref ⎣ 1 0 2 2
0 1 −1 3 ⎦
0 −3 3 −9
0 0 0 0
Now convert this reduced matrix back into equaons. We have
or, equivalently,
x 1 + 2x 3 = 2
x 2 − x 3 = 3
x 1 = 2 − 2x 3
x 2 = 3 + x 3
x 3 is free.
These two equaons tell us that the values of x 1 and x 2 depend on what x 3 is. As
we saw before, there is no restricon on what x 3 must be; it is “free” to take on the
value of any real number. Once x 3 is chosen, we have a soluon. Since we have infinite
choices for the value of x 3 , we have infinite soluons.
As examples, x 1 = 2, x 2 = 3, x 3 = 0 is one soluon; x 1 = −2, x 2 = 5, x 3 = 2 is
another soluon. Try plugging these values back into the original equaons to verify
that these indeed are soluons. (By the way, since infinite soluons exist, this system
of equaons is consistent.) .
In the two previous examples we have used the word “free” to describe certain
variables. What exactly is a free variable? How do we recognize which variables are
free and which are not?
Look back to the reduced matrix in Example 8. Noce that there is only one leading
1 in that matrix, and that leading 1 corresponded to the x 1 variable. That told us that
x 1 was not a free variable; since x 2 did not correspond to a leading 1, it was a free
variable.
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