Math 141 Lecture Notes for Section 2.3
Section 2.3 -
1
Systems of Linear Equations: Other Cases
Previously, we used augmented matrices in order to obtain the unique solution to a system of linear
equations in 3 variables.
Example 2.3.1:
Consider the system of equations
2x + 4y = 5
and
2y + x = 3.
Using matrices, obtain the row-reduced form of the augmented matrix for this system and express
whether it has 0, 1 or infinitely many solutions.
First we obtain the augmented matrix associated with the two equations:
2 4 5
1 2 3
Then we use Gauss-Jordan to obtain the row-reduced form of the matrix:
1 2 0
0 0 1
Since we observed before that the first two columns relate to x and y coefficients, we can relate this
matrix to the system of equations
x + 2y = 0
and
0=1
Since the last row corresponds to an equation that is false under all possible cases, this system must
have no solutions.
From this example we can observe a simple fact. A system of equations has no possible solutions if the
associated row-reduced form of the augmented matrix has a row of zeros followed by a nonzero entry
after the |.
Suppose we have a system of m linear equations in n variables. In order to solve for all of the n
variables, then we need at least n equations.
Example 2.3.2:
Consider the following matrices:
(a)

1 0 0
 0 1 0
0 0 1
This has a single solution of x = 3, y = −1, z = 2.

3
−1 
2
Math 141 Lecture Notes for Section 2.3
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(b)

1 0 0
 0 1 0

 0 0 1
0 0 0

4
5 

3 
0
This has a single solution of x = 4, y = 5, z = 3. Since it is in 3 variables with 4 equations, we
call this an over-determined system of equations.
(c)

0 1 0
 0 0 1

 0 0 0
0 0 0
1
−2
0
0

3
4 

0 
0
This has infinitely many solutions in 4 variables, x, y, z and w. Because the first column is 0 we
know that x is completely independent in the set of solutions, and therefore we must assign a
parameter r for x. Similarly since we have no leading 1 in the 4th column, we know that w must
also be assigned a parameter s. Then we have
z − 2w = 4
y+w =3
Then we substitute the parameter s for w and we have y = 3 − s and z = 4 + 2s. Therefore our
solutions are (r, 3 − s, 4 + 2s, s)
Example 2.3.3:
Solve the following systems of equations using Gauss-Jordan elimination:
(a)
x +2y
2x −3y
x −4y
=3
= −8
= −9


3
−8 
−9
The matrix form of this system is
1
 2
1
which has row reduced form:

2
−3
−4
1 0
 0 1
0 0

−1
2 
0
Which has the sole solution x = −1, y = 2. This system is over-determined since there are three
equations in two variables..
Math 141 Lecture Notes for Section 2.3
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(b)
3x −y
x −y
5x −2y
+2z
+2z
+4z
=5
=1
= 12
The matrix form of this system is

3
 1
5
−1 2
−1 2
−2 4

5
1 
12

0
1
0

0
0 
1
which has row reduced form:
1
 0
0
0
−2
0
By inspecting the last row of the matrix, it is clear that the system has no solutions, since the last
row is equivalent the the equation 0 = 1.
(c)
2x1
x1
3x1
+6x2
+3x2
+9x2
−5x3
+x3
−x3
+7x4
+13x4
=5
= −1
=1
The matrix form of this system is

2 6
 1 3
3 9
which has row-reduced form:
−5 0
1 7
−1 13

1 3 0 5
 0 0 1 2
0 0 0 0

5
−1 
1

0
−1 
0
This system is under-determined since it has three equations in four variables. Thus this system
must have infinitely many or no solutions. Since the row-reduced matrix has two nonzero rows (all
of which have at least one nonzero entry to the left of the |) , and there are four variables, we
know that we must have precisely 2 parameters. One possible parametrization of the solutions of
this system is:
(−3s − 5t, s, −1 − 2t, t)
or
x = −3s − 5t,
y = s, z = −1 − 2t, w = t.
(d)
3x
x
4x
8x
−2y
+3y
+y
+2y
−z
+7z
+6z
+12z
=4
=5
=9
= 18
Math 141 Lecture Notes for Section 2.3
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The matrix for this system of equations is:

3 −2 −1
 1
3
7

 4
1
6
8
2 12
which has row-reduced form:

1 0 1
 0 1 2

 0 0 0
0 0 0

4
5 

9 
18

2
1 

0 
0
This system is over-determined since it has four equations in three variables. Since the row-reduced
matrix has only two nonzero rows (all of which have at least one nonzero entry to the left of the |)
and there are three variables, we know that there must be infinitely many solutions and precisely
one parameter. A possible parametrization of the solutions of this system is:
(2 − t, 1 − 2t, t)
or
x = 2 − t,
Suggested Homework Problems: 1,3,5,7,9,17,24,31,35,37.
y = 1 − 2t,
z = t.