Section 1.4 Pivots and their influence on solution spaces
By now, we have seen several examples illustrating how the reduced row echelon matrix leads to a convenient description of the solution space to a linear system. In this section, we will use this understanding to make some general observations about how certain features of the reduced row echelon matrix reflect the nature of the solution space.
Remember that a leading entry in a reduced row echelon matrix is the leftmost nonzero entry in a row of the matrix. As we'll see, the positions of these leading entries encode a lot of information about the solution space of the corresponding linear system. For this reason, we make the following definition.
Definition 1.4.1.
A pivot position in a matrix \(A\) is the position of a leading entry in the reduced row echelon matrix of \(A\text{.}\)
For instance, in this reduced row echelon matrix, the pivot positions are indicated in bold:
We can refer to pivot positions by their row and column number saying, for instance, that there is a pivot position in the second row and fourth column.
Preview Activity 1.4.1. Some basic observations about pivots.

Shown below is a matrix and its reduced row echelon form. Indicate the pivot positions.
\begin{equation*} \left[ \begin{array}{rrrr} 2 \amp 4 \amp 6 \amp 1 \\ 3 \amp 1 \amp 5 \amp 0 \\ 1 \amp 3 \amp 5 \amp 1 \\ \end{array} \right] \sim \left[ \begin{array}{rrrr} 1 \amp 0 \amp 1 \amp 0 \\ 0 \amp 1 \amp 2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ \end{array} \right]\text{.} \end{equation*} How many pivot positions can there be in one row? In a \(3\times5\) matrix, what is the largest possible number of pivot positions? Give an example of a \(3\times5\) matrix that has the largest possible number of pivot positions.
How many pivots can there be in one column? In a \(5\times3\) matrix, what is the largest possible number of pivot positions? Give an example of a \(5\times3\) matrix that has the largest possible number of pivot positions.
Give an example of a matrix with a pivot position in every row and every column. What is special about such a matrix?
When we have looked at solution spaces of linear systems, we have frequently asked whether there are infinitely many solutions, exactly one solution, or no solutions. We will now break this question into two separate questions.
Question 1.4.2. Two Fundamental Questions.
When we encounter a linear system, we often ask
 Existence
Is there a solution to the linear system? If so, we say that the system is consistent; if not, we say it is inconsistent.
 Uniqueness
If the linear system is consistent, is the solution unique or are there infinitely many solutions?
These two questions represent two sides of a coin that appear in many variations throughout our explorations. In this section, we will study how the location of the pivots influence the answers to these two questions. We begin by considering the first question concerning the existence of solutions.
Subsection 1.4.1 The existence of solutions
Activity 1.4.2.

Shown below are three augmented matrices in reduced row echelon form.
\begin{equation*} \left[ \begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 3 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}\begin{equation*} \left[ \begin{array}{rrrr} 1 \amp 0 \amp 2 \amp 3 \\ 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}\begin{equation*} \left[ \begin{array}{rrrr} 1 \amp 0 \amp 2 \amp 0 \\ 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}For each matrix, identify the pivot positions and determine if the corresponding linear system is consistent. Explain how the location of the pivots determines whether the system is consistent or inconsistent.
Each of the augmented matrices above has a row in which each entry is zero. What, if anything, does the presence of such a row tell us about the consistency of the corresponding linear system?
Give an example of a \(3\times5\) augmented matrix in reduced row echelon form that represents a consistent system. Indicate the pivot positions in your matrix and explain why these pivot positions guarantee a consistent system.
Give an example of a \(3\times5\) augmented matrix in reduced row echelon form that represents an inconsistent system. Indicate the pivot positions in your matrix and explain why these pivot positions guarantee an inconsistent system.
Write the reduced row echelon form of the coefficient matrix of the corresponding linear system in Item d? (Remember that the Augmentation Principle says that the reduced row echelon form of the coefficient matrix simply consists of the first four columns of the augmented matrix.) What do you notice about the pivot positions in this coefficient matrix?

Suppose we have a linear system for which the coefficient matrix has the following reduced row echelon form.
\begin{equation*} \left[ \begin{array}{rrrrr} 1 \amp 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp 3 \\ \end{array} \right] \end{equation*}What can you say about the consistency of the linear system?
Let's summarize the results of this activity by considering the following reduced row echelon matrix:
In terms of variables \(x\text{,}\) \(y\text{,}\) and \(z\text{,}\) the final equation says
If we evaluate the lefthand side with any values of \(x\text{,}\) \(y\text{,}\) and \(z\text{,}\) we get 0, which means that the equation always holds. Therefore, its presence has no effect on the solution space defined by the other three equations.
The third equation, however, says that
Again, if we evaluate the lefthand side with any values of \(x\text{,}\) \(y\text{,}\) and \(z\text{,}\) we get 0 so this equation cannot be satisfied for any \((x,y,z)\text{.}\) This means that the entire linear system has no solution and is therefore inconsistent.
An equation like this appears in the reduced row echelon matrix as
The pivot positions make this condition clear: the system is inconsistent if there is a pivot position in the rightmost column of the corresponding augmented matrix.
In fact, we will soon see that the system is consistent if there is not a pivot in the rightmost column of the corresponding augmented matrix. This leaves us with the following
Proposition 1.4.3.
A linear system is inconsistent if and only if there is a pivot position in the rightmost column of the corresponding augmented matrix.
This also says something about the pivot positions of the coefficient matrix. Consider an example of an inconsistent system corresponding to the reduced row echelon form of the following augmented matrix
The Augmentation Principle says that that the reduced row echelon form of the coefficient matrix is
which shows that the coefficient matrix has a row without a pivot position. To turn this around, we see that if every row of the coefficient matrix has a pivot position, then the system must be consistent. For instance, if our linear system has a coefficient matrix whose reduced row echelon form is
then we can guarantee that the linear system is consistent because there is no way to obtain a pivot in the rightmost column of the augmented matrix.
Proposition 1.4.4.
If every row of the coefficient matrix has a pivot position, then the corresponding system of linear equations is consistent.Subsection 1.4.2 The uniqueness of solutions
Now that we have studied the role that pivot positions play in the existence of solutions, let's turn to the question of uniqueness.
Activity 1.4.3.

Here are the three augmented matrices in reduced row echelon form that we considered in the previous section.
\begin{equation*} \left[ \begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 3 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}\begin{equation*} \left[ \begin{array}{rrrr} 1 \amp 0 \amp 2 \amp 3 \\ 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}\begin{equation*} \left[ \begin{array}{rrrr} 1 \amp 0 \amp 2 \amp 0 \\ 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}For each matrix, identify the pivot positions and state whether the corresponding linear system is consistent. If the system is consistent, explain whether the solution is unique or whether there are infinitely many solutions.
If possible, give an example of a \(3\times5\) augmented matrix that corresponds to a linear system having a unique solution. If it is not possible, explain why.
If possible, give an example of a \(5\times3\) augmented matrix that corresponds to a linear system having a unique solution. If it is not possible, explain why.
What condition on the pivot positions guarantees that a linear system has a unique solution?
If a linear system has a unique solution, what can we say about the relationship between the number of equations and the number of variables?
Let's consider what we've learned in this activity. Since we are interested in the question of whether a consistent linear system has a unique solution or infinitely many, we will only consider consistent systems. By the results of the previous section, this means that there is not a pivot in the rightmost column of the augmented matrix. Here are two possible examples:
In the first example, we have the equations
demonstrating the fact that there is a unique solution \((x_1,x_2,x_3) = (4,1,2)\text{.}\)
In the second example, we have the equations
that we may rewrite in parametric form as
Here we see that \(x_1\) and \(x_2\) are basic variables that may be expressed in terms of the free variable \(x_3\text{.}\) In this case, the presence of the free variable leads to infinitely many solutions.
Remember that every column of the coefficient matrix corresponds to a variable in our linear system. In the first example, we see that every column of the coefficient contains a pivot position, which means that every variable is uniquely determined. In the second example, the column of the coefficient matrix corresponding to \(x_3\) does not contain a pivot position, which results in \(x_3\) appearing as a free variable. This illustrates the following principle.
Principle 1.4.5.
Suppose that we consider a consistent linear system.
If every column of the coefficient matrix contains a pivot position, then the system has a unique solution.
If there is a column in the coefficient matrix that contains no pivot position, then the system has infinitely many solutions.
Columns that contain a pivot position correspond to basic variables while columns that do not correspond to free variables.
When a linear system has a unique solution, every column of the coefficient matrix has a pivot position. Since every row contains at most one pivot position, there must be at least as many rows as columns in the coefficient matrix. Therefore, the linear system has at least as many equations as variables, which is something we intuitively suspected in Section 1.1.
It is reasonable to ask how we choose the free variables. For instance, if we have a single equation
then we may write
or, equivalently,
Clearly, either variable may be considered as a free variable in this case.
As we'll see in the future, we are more interested in the number of free variables rather than in their choice. For convenience, we will adopt the convention that free variables correspond to columns without a pivot position, which allows us to quickly identify them. For example, the variables \(x_2\) and \(x_4\) appear as free variables in the following linear system:
Subsection 1.4.3 Summary
We have seen how the locations of pivot positions, in both the augmented and coefficient matrices, give vital information about the existence and uniqueness of solutions to linear systems. More specifically,
A linear system is inconsistent exactly when a pivot position appears in the rightmost column of the augmented matrix.
If a linear system is consistent, the solution is unique when every column of the coefficient matrix contains a pivot position. There are infinitely solutions when there is a column of the coefficient matrix without a pivot position.
If a linear system is consistent, the columns of the coefficient matrix containing pivot positions correspond to basic variables and the columns without pivot positions correspond to free variables.
Exercises 1.4.4 Exercises
1.
For each of the augmented matrices in reduced row echelon form given below, determine whether the corresponding linear system is consistent and, if so, determine whether the solution is unique. If the system is consistent, identify the free variables and the basic variables and give a description of the solution space in parametric form.
 \begin{equation*} \left[ \begin{array}{rrrrr} 0 \amp 1 \amp 0 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 0 \amp 3 \\ 0 \amp 0 \amp 0 \amp 1 \amp 2 \\ \end{array} \right]\text{.} \end{equation*}
 \begin{equation*} \left[ \begin{array}{rrrrr} 1 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right]\text{.} \end{equation*}
 \begin{equation*} \left[ \begin{array}{rrrrr} 1 \amp 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \\ \end{array} \right]\text{.} \end{equation*}
 \begin{equation*} \left[ \begin{array}{rrrr} 1 \amp 0 \amp 0 \amp 3 \\ 0 \amp 1 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \amp 2 \\ \end{array} \right]\text{.} \end{equation*}
2.
For each of the following linear systems, determine whether the system is consistent, and, if so, determine whether there are infinitely many solutions.
 \begin{equation*} \begin{alignedat}{5} 2x_1\amp{}{}\amp x_2\amp {}+{}\amp 3x_3 \amp {}{} \amp \amp {}={} \amp 10 \\ x_1\amp{}+{}\amp x_2\amp \amp \amp {}+{} \amp 3x_4 \amp {}={} \amp 8 \\ \amp{}{}\amp 2x_2\amp {}+{}\amp 2x_3 \amp {}{} \amp x_4 \amp {}={} \amp 4 \\ 3x_1\amp{}+{}\amp 2x_2\amp {}{}\amp x_3 \amp {}+{} \amp x_4 \amp {}={} \amp 10 \\ \end{alignedat} \end{equation*}
 \begin{equation*} \begin{alignedat}{5} 2x_1\amp{}{}\amp x_2\amp {}+{}\amp 3x_3 \amp {}+{} \amp 3x_4 \amp {}={} \amp 8 \\ x_1\amp{}+{}\amp x_2\amp \amp \amp {}+{} \amp 2x_4 \amp {}={} \amp 1 \\ \amp{}{}\amp 2x_2\amp {}+{}\amp 2x_3 \amp {}+{} \amp 6x_4 \amp {}={} \amp 4 \\ 3x_1\amp{}+{}\amp 2x_2\amp {}{}\amp x_3 \amp {}{} \amp 3x_4 \amp {}={} \amp 1 \\ \end{alignedat} \end{equation*}
 \begin{equation*} \begin{alignedat}{5} 2x_1\amp{}{}\amp x_2\amp {}+{}\amp 3x_3 \amp {}+{} \amp 3x_4 \amp {}={} \amp 8 \\ x_1\amp{}+{}\amp x_2\amp \amp \amp {}+{} \amp 2x_4 \amp {}={} \amp 1 \\ \amp{}{}\amp 2x_2\amp {}+{}\amp 2x_3 \amp {}+{} \amp 6x_4 \amp {}={} \amp 4 \\ 3x_1\amp{}+{}\amp 2x_2\amp {}{}\amp x_3 \amp {}{} \amp 3x_4 \amp {}={} \amp 0 \\ \end{alignedat} \end{equation*}
3.
Include an example of an appropriate matrix as you justify your responses to the following questions.
Suppose a linear system having six equations and three variables is consistent. Can you guarantee that the solution is unique? Can you guarantee that there are infinitely solutions?
Suppose that a linear system having three equations and six variables is consistent. Can you guarantee that the solution is unique? Can you guarantee that there are infinitely solutions?
Suppose that a linear system is consistent and has a unique solution. What can you guarantee about the pivot positions in the augmented matrix?
4.
Determine whether the following statements are true or false and provide a justification for your response.
If the coefficient matrix of a linear system has a pivot in the rightmost column, then the system is inconsistent.
If a linear system has two equations and four variables, then it must be consistent.
If a linear system having four equations and three variables is consistent, then the solution is unique.
Suppose that a linear system has four equations and four variables and that the coefficient matrix has four pivots. Then the linear system is consistent and has a unique solution.
Suppose that a linear system has five equations and three variables and that the coefficient matrix has a pivot position in every column. Then the linear system is consistent and has a unique solution.
5.
We began our explorations in Section 1.1 by noticing that the solution spaces of linear systems with more equations seem to be smaller. Let's reexamine this idea using what we know about pivot positions.
Remember that the solution space of a single linear equation in three variables is a plane. Can two planes ever intersect in a single point? What are the possible ways in which two planes can intersect? How can our understanding of pivot positions help answer these questions?
Suppose that a consistent linear system has more variables than equations. By considering the possible pivot positions, what can you say with certainty about the solution space?
If a linear system has many more equations than variables, why is it reasonable to expect the system to be inconsistent?
6.
The following linear systems contain either one or two parameters.

For what values of the parameter \(k\) is the following system consistent? For which of those values is the solution unique?
\begin{equation*} \begin{alignedat}{3} x_1 \amp {}+{} \amp 2x_2 \amp {}={} \amp 3 \\ 2x_1 \amp {}{} \amp 4x_2 \amp {}={} \amp k \\ \end{alignedat}\text{.} \end{equation*} 
For what values of the parameters \(k\) and l is the following system consistent? For which of those values is the solution unique?
\begin{equation*} \begin{alignedat}{3} 2x_1 \amp {}+{} \amp 4x_2 \amp {}={} \amp 3 \\ x_1 \amp {}+{} \amp kx_2 \amp {}={} \amp l \\ \end{alignedat}\text{.} \end{equation*}
7.
Consider the linear system described by the following augmented matrix.
Find a choice for the parameters \(a\text{,}\) \(b\text{,}\) and \(c\) that causes the linear system to be inconsistent. Explain why your choice has this property.
Find a choice for the parameters \(a\text{,}\) \(b\text{,}\) and \(c\) that causes the linear system to have a unique solution. Explain why your choice has this property.
Find a choice for the parameters \(a\text{,}\) \(b\text{,}\) and \(c\) that causes the linear system to have infinitely many solutions. Explain why your choice has this property.
8.
A linear system where the right hand side of every equation is 0 is called homogeneous. The augmented matrix of a homogeneous system, for instance, has the following form:
Using the concepts we've seen in this section, explain why a homogeneous linear system must be consistent.
What values for the variables are guaranteed to give a solution? Use this to offer another explanation for why a homogeneous linear system is consistent.

Suppose that a homogeneous linear system has a unique solution.
Give an example of such a system by writing its augmented matrix in reduced row echelon form.
Write just the coefficient matrix for the example you gave in the previous part. What can you say about the pivot positions in the coefficient matrix? Explain why your observation must hold for any homogeneous system having a unique solution.
If a homogeneous system of equations has a unique solution, what can you say about the number of equations compared to the number of variables?
9.
In a previous math class, you have probably seen the fact that, if we are given two points in the plane, then there is a unique line passing through both of them. In this problem, we will begin with the four points on the left below and ask to find a polynomial that passes through these four points as shown on the right.
A degree three polynomial can be written as
where \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are coefficients that we would like to determine. Since we want the polynomial to pass through the point \((3,1)\text{,}\) we should require that
In this way, we obtain a linear equation for the coefficients \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\text{.}\)
Write the four linear equations for the coefficients obtained by requiring that the graph of the polynomial \(p(x)\) passes through the four points above.
Write the augmented matrix corresponding to this system of equations and use the Sage cell below to solve for the coefficients.
Write the polynomial \(p(x)\) that you found and check your work by graphing it in the Sage cell below and verifying that it passes through the four points. To plot a function over a range, you may use a command like
plot(1 + x 2*x^2, xmin = 1, xmax = 4)
.
Rather than looking for a degree three polynomial, suppose we wanted to find a polynomial that passes through the four points and that has degree two, such as
\begin{equation*} p(x) = a + bx + cx^2\text{.} \end{equation*}Solve the linear system for the coefficients. What can you say about the existence and uniqueness of a degree two polynomial passing through these four points?

Rather than looking for a degree three polynomial, suppose we wanted to find a polynomial that passes through the four points and that has degree four, such as
\begin{equation*} p(x) = a + bx + cx^2 + dx^3 + ex^4\text{.} \end{equation*}Solve the linear system for the coefficients. What can you say about the existence and uniqueness of a degree four polynomial passing through these four points?
Suppose you had 10 points and you wanted to find a polynomial passing through each of them. What should the degree of the polynomial be to guarantee that there is exactly one such polynomial? Explain your response.