Discrete Structures Need done 3/26
1
Part 2: Systems of Equations Which Do Not Have A Unique Solution
On the previous pages we learned how to solve systems of equations using Gaussian elimination. In each of the examples and exercises of part 1(except for exercise 1 parts d and e) the systems of equations had a unique solution. That is, a single value for each of the variables.
In example 3 we found the solution to be 7 23 3, . This means that the graphs of the two lines in example 3 intersect at this unique point. In 2-space, the xy-plane, we have the geometric bonus of being able to draw a picture of the solutions to a system of two equations two unknowns. Clearly, if we were asked to draw the graphs of two lines in the xy-plane we have 3 basic choices/cases:
1. Draw the two lines so they intersect. This point of intersection can only happen once for a given pair of lines. That is, the two lines intersect in a unique point. There is a unique common solution to the system of equations. Discussed in part 1.
2. Draw the two lines so that one is on "top of" the other. In this case there are an infinite number of common points, an infinite number of solutions to the given system. Discussed in part 2.
3. Draw two parallel lines. In this case there are no points common to both lines. There is no solution to the system of equations that describe the lines. Discussed in part 2.
The 3 cases above apply to any system of equations.
Theorem 1. For any system of m equations with n unknowns (m < n) one of the following cases applies:
1. There is a unique solution to the system.
2. There is an infinite number of solutions to the system.
3. There are no solutions to the system.
Again, in this section of the notes we will illustrate cases 2 and 3. To solve systems of equations where these cases apply we use the matrix procedure developed previously.
Example 6. Solve the system
x + 2y = 1
2x + 4y = 2
2
It is probably already clear to the reader that the second equation is really the first in disguise. (Simply divide both sides of the second equation by 2 to obtain the first). So if we were to draw the graph of both we would obtain the same line, hence have an infinite number of points common to both lines, an infinite number of solutions. However it would be helpful in solving other systems where the solutions may not be so apparent to do the problem algebraically, using matrices. The matrix of the system with its simplification follows. Recall,
we try to express the matrix 1 2 1
2 4 2
in the form 1 2
1 0
0 1
b
b
from which we can read off the
solution. However after one step we note that
1 2 1
2 4 2
1 22 R R 1 2 1
0 0 0
. It should be clear to the reader that no matter what further
elementary row operations we perform on the matrix 1 2 1
0 0 0
we cannot change it to the form
we hoped for, namely, 1 2
1 0
0 1
b
b
. To understand what our result means simply write the system
of equations that the matrix 1 2 1
0 0 0
represents, that is,
x + 2y = 1
0x + 0y = 0.
The first equation tells us that x = 1 - 2y, or equivalently that y = 1 2 (1 - x), so that (1,0),
(-1,1), and (-5,3) are three of the infinite number of possible solutions of the first equation. The second equation places no restrictions on what values x and y can assume, hence there are an infinite number of solutions to both equations, to the system. Any pair of real numbers of the form (1 - 2y,y) where y can be any real number is a solution to the given system of equations. The solutions of the system can also be expressed in the form (x, 1 2 (1 - x) where x can be any
real number.
Warning. When there are an infinite number of solutions to a system there are frequently several “different-looking” ways to describe the solutions, as in the above example.
3
Example 7. Determine the solutions of the system of equations whose matrix is row equivalent
to 1 0 0 1
0 1 1 0
0 0 0 0
. Give three examples of the solutions. If we use the variables x1, x2, and x3 the
system of equations which is represented by this matrix is: 1x
1 + 0x
2 + 0x
3 = 1 This equation simply says that x1 = 1.
0x 1 + 1x
2 + -1x
3 = 0 This equation simply tells us that x3 = x2.
0x 1 + 0x
2 + 0x
3 = 0 All this equation says is that 0 = 0.
So equation 1 indicates that there is a restriction on x1, namely it must be 1. Equation 2 gives us restrictions on x2 and x3, namely, they must equal each other. Equation 3 does not place any restrictions on any of the variables.
There are an infinite number of solutions to this system of equations. Any triple of the form (1, x2, x2) where x2 can be any real number is a solution. So (1,1,1), (1,3,3) and (1,-1,-1) are three examples of solutions to this system of equations.
Example 8. Solve the system of equations whose matrix is 2 1 0 1
1 2 1 0
0 3 2 1
. Give three
examples of the solutions.
2 1 0 1
1 2 1 0
0 3 2 1
1
interchange rows 1 and 2 and then -1R
1 2 1 0
2 1 0 1
0 3 2 1
1 2 -2R R 1 2 1 0
0 3 2 1
0 3 2 1
2 3 R R 1 2 1 0
0 3 2 1
0 0 0 0
.
At this point we could row-reduce the last matrix further by but this is really not necessary. If we call the variables x
1 , x
2 and x
3 the system of equations that this last matrix
represents is: x1 - 2x2 + x3 = 0
3x 2 - 2x
3 = 1.
From the latter equation we can say x 3 = ½ (3x
2 - 1). If we substitute this expression for
x 3 in the first equation we obtain x1 - 2x2 + ½ (3x2 - 1) = 0 or x1 = 2x2 - ½ (3x2 - 1) which can be
simplified to x 1 = ½(x
2 + 1). If we replace x
2 by 0 then one solution is x
1 = 1/2, x
2 = 0 and
x 3 = -1/2. Another solution is (3/2, 2, 5/2). Why? We ask the reader to substitute these
solutions into the original system to verify that they are solutions, and to find two more solutions.
4
Another situation that one encounters in solving systems of equations is that when the number of unknowns is greater than the number of equations. For example:
x1 + x2 - 3x3 = -1
x2 - x3 = 0.
Here we have only two equations but three unknowns
But upon closer inspection this is simply another form of the above examples as we show in example 9.
Example 9. Solve the system of equations
x1 + x2 - 3x3 = -1
x2 - x3 = 0.
This system is the same as the system
x1 + x2 - 3x3 = -1
0x1 + x2 - x3 = 0
0x1 + 0x2 + 0x3 = 0,
So we can represent the above system by the matrix 1 1 3 1
0 1 1 0
or the matrix 1 1 3 1
0 1 1 0
0 0 0 0
.
Clearly no matter what elementary row operations we perform on this matrix we cannot change
it to the form 1
2
3
1 0 0
0 1 0
0 0 1
b
b
b
. For this reason it is common practice to rewrite the matrix without
the rows which contain all zeros, so that the matrix of the given system is 1 1 3 1
0 1 1 0
and the
reader can show that this matrix can be reduced to 1 0 2 1
0 1 1 0
which gives us the system
x1 - 2x3 = -1
x2 - x3 = 0
5
The second equation tells us that x3 = x2
Substitute this value of x2 in first equation to obtain
x1 = 2x2 - 1
So all solutions of the given system are ordered triples of the form (2x2 – 1, x2, x2)
where x2 can be any real number. If we chose x2 = 0 then we find that (-1, 0, 0) is one example
of a solution to this system of equations and if we replace x3 by 2 we obtain (3, 2, 2) as another
solution.
Keep in mind that there are often many different ways to describe the solutions of the same system. For example, I claim that the solutions of the system given in this example 9 can also be described as any ordered triple of the form (x1, ½(x1+ 1), ½(x1+ 1) where x1 is any real
number. So if x1= 1 we obtain the triple (1, 1, 1) as a solution, which certainly satisfies the
original system of equations. To obtain my form of the solution set just take the equation x1 =
2x2 – 1 and solve for x2 and then for x3.
Exercises
1. Determine the solutions of the system of equations whose matrix is row equivalent to
. Give three examples of the solutions. Verify that your solutions satisfy the original system of equations.
2. Determine the solutions of the system of equations whose matrix is row equivalent to
. Give three examples of the solutions. Verify that your solutions satisfy the original system of equations.
3. Determine the solutions of the system of equations whose matrix is row equivalent to
. Give three examples of the solutions. Verify that your solutions satisfy the original system of equations.
1 0 0 1
0 1 2 0
0 0 0 0
1 0 0 1
0 1 2 3
0 0 0 0
1 0 1 1 0 1 3 1
0 0 0 0
6
4. Determine the solutions of the system of equations whose matrix is row equivalent to
. Give three examples of the solutions. Verify that your solutions satisfy the original system of equations.
5. Determine the solutions of the system of equations whose matrix is row
equivalent to . Give three examples of the solutions.
Verify that your solutions satisfy the original system of equations.
1 1 1 1 0 2 3 2 0 0 0 0
1 1 1 1 1 2 2 2 0 0 0 0
7