Lecture 4
Andrei Antonenko
February 05, 2003
1
Analysis of the system
Now, we can formulate the main theoretical result about the system. This theoretical result
follows directly from our practical method of solving them.
Case 1
If during our procedures we got an equation of the form 0 =
b
, where
b
6
= 0, then we
should stop and deduce that the system has no solution.
Case 2
If in the row echelon form we have no equation of the form 0 =
b
, where
b
6
= 0, and we
have free variables, than the system has infinitely many solutions, since we can assign to
them any (real) values.
Case 3
If in the row echelon form we have no equation of the form 0 =
b
, where
b
6
= 0, and we don’t
have free variables, than the solution can be determined uniquely, so the corresponding
system has only 1 solution.
Theorem 1.1.
The system of linear equations can have 0, 1 or infinitely many solutions.
Last lecture we provided an example of a linear system which had free variables in its REF.
Now we will provide an example of a system without free variables, so it would have a unique
numerical solution.
Example 1.2.
Let’s reduce the following system to the row echelon form and then solve it.
x
1
+
x
2
+
x
3
=
4
x
1
+
2
x
2

x
3
=
1
2
x
1

x
2
+
x
3
=
3
(1)
First variable with nonzero coefficient is
x
1
, and the first equation has a nonzero coefficient
before it too.
So, we don’t have to interchange the equations.
Now, we’ll subtract the first
1
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equation from the second one, then multiply it by 2 and subtract from the third one. We’ll get
the following system (we should not omit the first equation!!!):
x
1
+
x
2
+
x
3
=
4
x
2

2
x
3
=

3

3
x
2

x
3
=

5
(2)
Now, we’ll apply the same steps to equations from the 2nd to the 3rd.
So, we multiply the
second equation by 3 and add it to the third one (again, do not omit any equations!!!):
x
1
+
x
2
+
x
3
=
4
x
2

2
x
3
=

3

7
x
3
=

14
(3)
Now we have a system in a row echelon form. Now it possible to solve it using the back substitu
tion process. From the last equation we can determine the value for
x
3
:
x
3
= (

14)
/
(

7) = 2
.
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 Spring '03
 Andant
 Gaussian Elimination, Row echelon form

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