Question 4: Do all systems of linear equations have unique solutions?
Earlier we examined two systems where the numbers of variables was equal to the
number of equations. Often there are fewer equations than variables or more equations
than variables. Even in systems where the number of variables initially equals the
number of equations, one of the rows may become all zeros, meaning that the equation
was unnecessary. In all of these cases, we can still use the strategy from Question 3 to
solve the problem.
Example 6
Solve a System with More Variables Than Equations
Solve the system of equations
x1  x2 
x3  70
2 x1  3 x2  3.5 x3  10
by transforming its augmented matrix to reduced row echelon form.
Solution This system does not use x, y, and z like earlier examples.
However, if we let the first column in the augmented matrix correspond
to x1 , the second column to x2 , and the third column to x3 , we can write
the augmented matrix for this system as
1 1 1 70 
 2 3 3.5 10 


To put this matrix into reduced row echelon form, we must transform it
using row operations to
1 0 ? ? 
0 1 ? ?


Since there is no third row, we have no hope of solving for a solution
where x1 , x2 , and x3 are each a unique value. Instead, we’ll be able to
solve for x1 and x2 in terms of x3 .
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The original augmented matrix already has a 1 in the first column and
row, so there is no need to put a pivot there. To put a 0 below the pivot,
multiply the first row by -2, add it to the second row, and put the result in
place of the second row:
2 R1 : 2 2 2 140
 R2 : 2 3 3.5 10
0
1
2 R1  R2
becomes
R2
1.5 130
70 
1 1 1
 0 1 1.5 130 


The previous row operation not only put a 0 in the first column, but also
placed the pivot in the second row and column (very lucky!). To put this
matrix into reduced row echelon form, we need to put a zero above the
pivot in the second column. Multiply the second row by -1, add it to the
first row, and place the result in the first row:
1R2 : 0 1 1.5 130
 R1 : 1 1
1
70
1
0
1 0 0.5 200 
 0 1 1.5 130 


1R2  R1
becomes
R1
 0.5 200
This matrix is in reduced row echelon form, but we can’t read the
solution off as we have in earlier examples. If we convert this back to a
system of equations, we get
1 0 0.5 200 
 0 1 1.5 130 



x1  0.5 x3  200
x2  1.5 x3  130
Notice that each of these equations contains x3 and it is easy to solve
for x1 in the first equation and x2 in the second equation. If we solve for
x1 and x2 we get
x1  0.5 x3  200
x2  1.5 x3  130
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Although we don’t have specific numbers for each variable, we do have
a recipe for finding values. If we choose any value for x3 , we can find
the corresponding values for x1 and x2 . For instance, if we choose
x3  100 , then
x1  0.5 100   200  250
x2  1.5 100   130  280
If x3  0 , then
x1  0.5  0   200  200
x2  1.5  0   130  130
Since x3 can be any number, there are an infinite number of solutions
to the system. However, not just any combination of numbers works.
Once a value for x3 is chosen, the equations above must be used to
calculate corresponding values for x1 and x2 . This can be summarized
by writing
x3 can be any real number
x1  0.5 x3  200
x2  1.5 x3  130
In this case, we have more variables than equations so we would
expect to be able to solve for only some of the variables explicitly. In
each row we can solve for a variable explicitly as long as the row is not
entirely zeros. Any variables beyond the number of nonzero rows, in
this case one, will be values that we can pick. These variables are
called parameters. Parameters are variables whose values are arbitrary
and can be picked to be anything that is reasonable for the system of
equations.
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Example 7
Solve a System with More Equations Than Variables
Solve the system of equations
x1  2 x2  2
2 x1  8 x2  16
x1  x2  5
by transforming its augmented matrix to reduced row echelon form.
Solution The augmented matrix for this system is
 1 2 2 
 2 8 16 


 1 1 5 
To put this matrix into reduced row echelon form, we need to place
pivots in the first and second columns and zeros in the rest of these
columns.
The entry in the first column and row is a 1 so the pivot is already in
place. To put zeros below the pivot,
2 R1 : 2 4 4
 R2 : 2 8 16
2 R1  R2
becomes
R2
0 12 12
1R1 : 1 2 2
 R3 : 1 1 5
0
3
1R1  R3
becomes
R3
3
1 2 2 
 0 12 12 


 0 3 3 
In the second column and row, we need to transform the 12 to a 1 by
multiplying the row by
1
12
:
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1 2 2 
0 1 1 


 0 3 3 
0 12 12

0
1
1
12
1
12
R2
becomes
R2
1
Now place zeros in the rest of the column,
2 R2 : 0 2 2
 R1 : 1 2 2
1
0
4
3R2 : 0 3 3
 R3 : 0 3 3
0
0
0
2 R2  R1
becomes
R1
1 0 4 
0 1 1 


 0 0 0 
3R2  R3
becomes
R3
If we convert this matrix back to a system of equations, we find that
x1  4 and x2  1 . Notice that the last row of the matrix is all zeros and
does not contribute to the solution. This means that there are two
variables and two nonzero rows so no parameters are needed in the
solution.
We can check the solution by substituting  4,1 into each equation:
?
First equation: 4  2 1  2
2  2 TRUE
?
Second equation: 2  4   8 1 16
16  16 TRUE
?
Third equation: 4  1  5
5  5 TRUE
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Since  x1 , x2    4,1 satisfies each equation in the system, it is the
solution to the original system of equations.
Example 8
Solve a System with More Equations Than Variables
Solve
10 x1  x2  44
 12 x1  x2  2
x1  2 x2  8
by transforming its augmented matrix to reduced row echelon form.
Solution Like the last example, we’ll let the first column in the
augmented matrix correspond to x1 and the second column to x2 . The
augmented matrix for this system is
 10
 1
 2
 1
1 44 
1 2 
2 8 
To create a pivot in the first row and column, we have two possibilities.
We could multiply the first row by
1
10
or interchange the first and third
row. Since interchanging rows does not introduce and fractions into the
matrix, we’ll do that to yield
2 8 
1
 1 1
2 
 2
10 1 44 
Now we’ll use row operations to fill the rest of the first column with
zeros:
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1
2
1
2
R1 :
 R2 : 
0
1 4
1
2
1 2
R1  R2
becomes
R2
2 6
1
2
10 R1 : 10 20 80
 R3 : 10
1 44
0
10 R1  R3
becomes
R3
 21 36
8 
1 2
0 2
6 

 0 21 36 
To create a pivot in the second row and column, multiply the second
row by 12 :
8 
1 2
0 1
3 

 0 21 36 
0 2 6

0 1
1
2
1
2
R2
becomes
R2
3
Now use row operations to place zeros above and below the pivot in the
second column:
2 R2 : 0 2 6
 R1 : 1 2 8
1
0
2
21R2 : 0 21 63
 R3 : 0 21 36
0
0
27
2 R2  R1
becomes
R1
1 0 2 
0 1 3 


 0 0 27 
21R2  R3
becomes
R3
The first two rows in the reduced row echelon form suggest a solution,
but the third is problematic. If we write this row as an equation, we get
0 x1  0 x2  27
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The left side is simply 0. Since 0 cannot equal 27, this implies that there
are no solutions to this system. This system is an inconsistent system.
These examples illustrate what may happen when there are more equations than
variables or when there are more variables than equations. It is possible for the system
to have a unique solution as in Example 5. The system may be a system with infinitely
many solutions like Example 6 or have no solutions like the inconsistent system in
Example 8.
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