3.6
Existence of solutions to linear equations
So far most of the systems of equations that we have encountered have had solutions.
Now we want to look at situations where the system of equations does not have a solution.
In particular, we want to give a systematic algorithm for determining if a system of
equations does or does not have a solution. A related question is the following.
Given a set of linear equations where the coefficients are fixed, but the numbers on
the right hand side of the equations might change from time to time, does the set of
linear equations have a solution for every set of numbers on the right hand side, i.e.
is there always a solution?
Here is set of linear equations that does not have a solution.
Example 1.
x + 2y = 3
(1)
2x + 4y = 1
The reason this does not have a solution is because if we multiply the first equation by 2
we get 2x + 4y = 6 which contradicts the second equation which says 2x + 4y = 1.
The fact that this system does not have a solution has a nice interpretation in terms of
superpositions. If we write the system in vector form we have
 x + 2y 
3
=
 2x + 4y 
1
or
 x 
 2y 
3
+
=
 2x 
 4y 
1
or
1
2
3
x2 + y4 = 1
or
xv1 + yv2 = b
where
1
v1 =  2 ,
2
v2 =  4 ,
3
b = 1
So we want to write b as a superposition of v1 and v2. Note that v2 = 2v1, so if
xv1 + yv2 = b, then xv1 + y2v1 = b or (x+2y)v1 = b or tv1 = b where t = x+2y. So, if there
were a solution, then b would be a scalar multiple of v1, but this is not the case since the
only vectors that are scalar mulitiples of v1 are those whose second components are twice
their first.
3.6 - 1
How do we tell if a set of equations has a solution in terms of the elimination procedure
we have been using. Let’s look at the above system from the point of view of the
elimination procedure, both in terms of the equations and also in terms of the augmented
matrix.
Equations
Augmented matrix
 1 2 | 3
 2 4 | 1
x + 2y = 3
2x + 4y = 1
In the equations, subtract 2E1 from E2. In the augmented matrix, subtract 2R1 from R2.
 1 2 | 3
 0 0 | -5
x + 2y = 3
0 = -5
In the equations, we have an equation (the last) that has the form
0 = (some non-zero number). In the augmented matrix, we have a row which is all zeros
to the left of the vertical bar, and which is non-zero to the right of the vertical bar. In
general, this is how we tell that a set of equations does not have a solution. Such a set of
equations is often called inconsistent.
Example 2. If we change the right hand side of the second equation in (1) to 6, so that
the set is
x + 2y = 3
(2)
2x + 4y = 6
then the second equation is twice the first, so that any solution to the first is a solution to
the second. So, if x = 3 – 2y, then this will be a solution. For example, x = 3, y = 0 is a
solution. In fact, there are infinitely many solutions and we can describe all the solutions
as follows.
x
 3 - 2y 
3
-2
y =  y  = 0 + y 1 
x
Thus the solutions can be viewed as those vectors  y  that can be written as the vector
3
-2
plus
any
multiple
of
the
vector
0
 1 .
If we look at the system (2) in terms of superpositions we want
1
2
3
x2 + y4 = 6
3
1
Since  6  = 3 2 , one solution is x = 3, y = 0, as we have already noted.
What does this look like in terms of the elimination procedure.
3.6 - 2
Equations
Augmented matrix
 1 2 | 3
 2 4 | 6
x + 2y = 3
2x + 4y = 6
Subtract 2E1 from E2 or subtract 2R1 from R2.
 1 2 | 3
 0 0 | 0
x + 2y = 3
0 = 0
In the equations, we have an equation (the last) that has the form 0 = 0. In the augmented
matrix, we have a row which is all zeros both to the left and to the right of the vertical bar.
Such an equation or row does not affect the solutions and can be ignored.
Example 3. To generalize Examples 1 and 2, consider the system
x + 2y = b1
2x + 4y = b2
 b1 
3
where b =  b  can have different sets of values. In the first example (1), where b =  1 
2
3
there was no solution, but in the second example (2) where b =  6  there was a solution.
To summarize, there is a solution for some, but not all b. In fact there is a solution
precisely if b2 = 2b1.
Here is an example with three equations and three unknowns.
Example 4.
(3)
x + 2y + 3z = 6
2x + 4y + 8z = 14
3x + 6y + 12z = 22
We go through the solution algorithm with the augmented matrix.
 1 2 3 | 6  R2 - 2R1  R2
 2 4 8 | 14  R - 3R  R
1
3

 3

 3 6 12 | 22 
1 2
0 0

0 0
3 | 6  R1 - 3R2  R1
1 | 1  R3 - 3R2  R3


3 | 4
1 2
0 0

0 0
1 2
0 0

0 0
3 | 6
2 | 2  -R2/2  R2


3 | 4
0 | 3
1 | 1

0 | 1
The final matrix is in reduced row echelon form. Interpreting it as equations we have
3.6 - 3
x + 2y
= 3
z = 1
0 = 1
Since the final equation is 0 = 1, there is no solution.
Let's generalize the above examples. We have a system of linear equations.
a11 x1 + a12 x2 +  + a1n xn = b1
a21 x1 + a22 x2 +  + a2n xn = b2


am1 x1 + am2 x2 +  + amm xn = bm
(4)
The number of equations may be different from the number of unknowns. When written
in vector-matrix form it would be
 a11
 a21
…
 am1
a12  a1n
a22  a2n
am2 
  x1 
  x2 
…
amn  xm
 bb12 
…
 bm 
=
which has the form Au = b. The augmented matrix is
M =
 a11
 a21
…
 am1
a12  a1n
a22  a2n
am2
| b1 
| b2 

 amn | bm
Using the row operations we transform M to reduced row echelon form. Some possible
reduced row echelon form matrices one might get are the following
M1 =
1 0
0 1
…
0 0
 0 | c1
 0 | c2
 1 | cm




M2 =
3.6 - 4
1 0
0 1
…
0 0
 0 d1 | c1
 0 d2 | c2
 1 dm | cm




M3 =




1 0
0 1
…
0 0
0 0
 0 | c1
 0 | c2
 1 | cm-1
 0 | 0




M4 =




1 0
0 1
…
0 0
0 0
 0 | c1
 0 | c2
 1 | cm-1
 0 | cm




In the first three case the original equations have a solution. In the first case n = m and
the solution is
x1 = c1
x2 = c2

xn = cn
In the second case n = m + 1 and the solution is
x1 = c1 – d1
x2 = c2 – d2

xn = cn – dn
In the third case n = m – 1 and the solution is the same as in the first case. In the last case
n = m – 1. The final equation is 0 = cm. If cm  0 there is no solution.
In general, we have the following rule to tell if the system (4) has a solution? We
transfrom the augmented matrix to reduced row echelon form. Suppose the leading 1 in
some row is in the column to the right of the vertical line, i.e. it is all zero's to the left of
the vertical line and a 1 to the right of the vertical line. Then the corresponding equation
is 0 = 1 and there is no solution. If such a row is not present in an augmented matrix in
reduced row echelon form then there is a solution. To summarize
Theorem 1. Consider a system of equations and suppose we transform the original
augmented matrix M to reduced row echelon form using elementary row operations. Call
the reduced row echelon form matrix M'. If no row of M' has a leading 1 in the column to
the right of the vertical line, then there is a solution. If some row of M' has a leading 1 in
the column to the right of the vertical line, then there is no solution.
Now we look at a more general question. Consider a system of equations of the form (4).
Suppose the coefficients aij are fixed and we consider the system for different choices of
the numbers bi on the right side. How can we tell if the system has a solution for every
choice of numbers bi on the right side? Let's look at an example.
Example 5. Does the following system of equations have a solution for every choice of
b1, b2 and b3?
3.6 - 5
(5)
-x
3x +
+ 2z - w = b1
y - z + 4w = b2
6y + 5z + w = b3
We go through the solution algorithm with the augmented matrix.
 1 2 3 | 6  R2 - 2R1  R2
 2 4 8 | 14  R - 3R  R
1
3

 3

 3 6 12 | 21 
1 2
0 0

0 0
3 | 6  R1 - 3R2  R1
1 | 1  R3 - 3R2  R3


3 | 4
1 2
0 0

0 0
1 2
0 0

0 0
3 | 6
2 | 2  -R2/2  R2


3 | 4
0 | 3
1 | 1

0 | 1
Problem 1. Show that the following system of equations has a solution for every choice of b1 and b2.
3x + y = b1
x + 2y = b2
Example 6. Does the following system of equations have a solution?
Equations
Augmented matrix
0 1 -4 | -2
M =  1 - 1 3 | 4 
3 1 -7 | 9 
y - 4z = - 2
x - y + 3z = 4
3x + y - 7z = 9
0 1 -4x
-2




Remark. These equations are equivalent to  1 - 1 3   y  =  4  or Au = b where
3 1 -7z
 9
0 1 -4
x
-2
A =  1 - 1 3 , u =  y  and b =  4 . They are also equivalent to
3 1 -7
z
 9
 0
 1
-4 -2
 0








1
1
1
3
4
x +y
+z
=
 or xv1 + yv2 + zv3 = b where v1 =  ,
3
 1
-7  9
3
1
4




v2 =  - 1  and v3 =  3 . To solve, we interchange the first and second equations.
 1
-7
Equations
(9)
Augmented matrix
1 -1 3 | 4 
M1 =  0 1 - 4 | - 2 
3 1 -7 | 9
x - y + 3z = 4
y - 4z = - 2
3x + y - 7z = 9
3.6 - 6
Subtract 3 times the first equation from the third giving
Equations
Augmented matrix
x - y + 3z = 4
y - 4z = - 2
4y - 16z = - 3
M2
1 -1 3 | 4 
=  0 1 -4 | - 2 
 0 4 -16 | - 3 
Subtract 4 times the second equation from the third giving
Equations
x - y + 3z = 4
y - 4z = - 2
0 = 5
Augmented matrix
M3
1 -1 3 | 4 
=  0 1 -4 | - 2 
0 0 0 | 5 
Since the last equation is 0 = 5, there is no solution.
3.6 - 7