An Inquiry-Based Introduction to Linear Algebra and Applications

Subsection Application: Balancing Chemical Reactions

Linear systems have applications in chemistry when balancing chemical equations. When a chemical reaction occurs, molecules of different substances combine to create molecules of other substances. Chemists represent such reactions with chemical equations. To balance a chemical equation means to find the number of atoms of each element involved that will preserve the number of atoms in the reaction. As an example, consider the chemical equation

This equation asks about what will happen when the chemicals ethane (\(\text{C} _2\text{H} _6\)) and oxygen (\(\text{O} _2\)), called the reactants of the reaction, combine to produce carbon dioxide (\(\text{ CO } _2\)) and water (\(\text{H} _2\text{O}\)), called the products of the reaction (note that oxygen gas is diatomic, so that oxygen atoms are paired). The arrow indicates that it is the reactants that combine to form the products. Any chemical reaction has to obey the Law of Conservation of Mass that says that mass can neither be created nor destroyed in a chemical reaction. Consequently, a chemical reaction requires the same number of atoms on both sides of the reaction. In other words, the total mass of the reactants must equal the total mass of the products. In reaction (3.1) the chemicals involved are made up of carbon (C), hydrogen (H), and oxygen (O) atoms. To balance the equation, we need to know how many molecules of each chemical are combined to preserve the number of atoms of C, H, and O. This can be done by setting up a linear system of equations of the form

where \(x_1\text{,}\) \(x_2\text{,}\) \(x_3\text{,}\) and \(x_4\) represent the number of molecules of \(\text{C} _2\text{H} _6\text{,}\) \(\text{O} _2\text{,}\) \(\text{ CO } _2\text{,}\) and \(\text{H} _2\text{O}\text{,}\) respectively, in the reaction and then solving the system. Specific details can be found at the end of this section.

Subsection Introduction

In the previous sections, we identified operations on a given linear system with corresponding equivalent operations on the matrix representations which simplify the system and its matrix representation without changing the solutions of the system. Our end goal was to obtain a system which could be solved using back substitution, such as

The augmented matrix for this system is

The matrices of linear systems which can be solved via back substitution are said to be in row echelon form (or simply echelon form). We will define the properties of matrices in this form precisely in this section. Our goal will be to prescribe a precise procedure for converting any matrix to an equivalent one in row echelon form without having to convert back to the system representation.

Subsection The Echelon Forms of a Matrix

In the previous sections we saw how to simplify a linear system and its matrix representation via the elimination method without changing the solution set. This process is more efficient when performed on the matrix representation rather than on the system itself. Furthermore, the process of applying row operations to any augmented matrix is one that can be automated. In order to write an algorithm that can be used with any size augmented matrix to the extent that it can be applied even by a computer program, it is necessary to have a consistent procedure and a stopping point for the simplification process. The two main properties that we want the simplified augmented matrix to satisfy are that it should be easy to see if the system has solutions from the simplified matrix, and in cases when there are solutions, the general form of the solutions can be easily found. Hence the topic of this section is to define the process of elimination completely and generally.

We begin by discussing the row echelon or, simply, echelon form of a matrix. We know that the forward phase of the elimination on a linear system produces a system which can be solved by back substitution. The matrix representation of such a simplified system is said to be in row echelon or simply echelon form. Note that matrices in this form have the first nonzero entry in each row to the right of and below the first nonzero entry in the preceding row. Our next step is to formally describe this form — one that you tried to explain in problem 3 of Preview Activity 3.1.

A pivot is also called a leading entry of a row. Note that properties (1) and (2) above imply that all entries in a column below a pivot are zeros. It can be shown that the positions of these pivots, called pivot positions, are unique and tell us quite a bit about a matrix and the solutions of the linear system it corresponds to. The columns that the pivots are in, called pivot columns, will also have useful properties as we will see soon.

Once an augmented matrix is in row echelon form, we can use back substitution to solve the corresponding system. However, we can make solving much easier with just a little more elimination work.

Row operations are easy to apply, so if we are inclined, there is no reason to stop at the row echelon form. For example, starting with the following matrix

in row echelon form, we could take the row operations even farther and avoid the process of back substitution altogether. First, we multiply the last row by \(1/2\) to simplify that row:

Then we use the third row to eliminate entries above the third pivot:

We can continue in this manner (we call this process backward elimination) to make 0 all of the entries above the pivots (one in the second column, and one in the fourth) with the pivots being 1, to ultimately obtain the equivalent augmented matrix

The system corresponding to this augmented matrix is

so we can just directly read off the solution to the system: \(x_3\) free and \(x_1=2- \frac{5}{2}x_3\text{,}\) \(x_2=1-3x_3\text{,}\) \(x_4=2\text{.}\) This final row reduced form makes solving the system very easy, and this form is called the reduced row echelon form.

In short, the reduced row echelon form of a matrix is a row echelon form in which all the pivots are 1 and any entries below and above the pivots are 0.

If we use either of these two row echelon forms, solving the original system becomes straightforward and, as a result, these matrix forms are stopping points for the row operation algorithm to solve a system. It is also very easy to write a computer program to perform row operations to obtain and row echelon or reduced row echelon form of the matrix, making hand computations unnecessary. We will discuss this shortly.

Subsection Determining the Number of Solutions of a Linear System

Consider the system

The augmented matrix for this system is

Note that this matrix is already in row echelon form. The reduced row echelon form of this augmented matrix is

Since there are leading 1s in the first three columns, we can use those entries to write \(x_1\text{,}\) \(x_2\text{,}\) and \(x_3\) in terms of \(x_4\text{.}\) We then choose \(x_4\) to be arbitrary and write the remaining variables in terms of \(x_4\text{.}\) Let \(x_4 = t\text{.}\) Solving the third equation for \(x_3\) gives us \(x_3 = 4+2t\text{.}\) The second equation shows that \(x_2 = 2+t\text{,}\) and the first that \(x_1 = 0\text{.}\) Each value of \(t\) provides a solution to the system, so our system has infinitely many solutions. These solutions are

where \(t\) can have any value.

When solving a linear system of equations, the free variables can be chosen arbitrarily and we can write the basic variables in terms of the free variables. Therefore, the existence of a free variable leads to infinitely many solutions for consistent systems. However, it is possible to have a system with free variables which is inconsistent. (Can you think of an example?)

Subsection Producing the Echelon Forms

In this part, we consider the formal process of creating the row and reduced row echelon forms of matrices. The process of creating the row echelon form is the equivalent of the elimination method on systems of linear equations.

The complete process of applying row operations to reduce an augmented matrix to a row or reduced row echelon form can be expressed as a recursive process in an algorithmic fashion, making it possible to program computers to solve linear systems. Here are the steps to do so:

Step 1

Begin with the leftmost nonzero column (if there is one). This will be a pivot column.

Step 2

Select a nonzero entry in this pivot column as a pivot. If necessary, interchange rows to move this entry to the first row (this entry will be a pivot).

Step 3

Use row operations to create zeros in all positions below the pivot.

Step 4

Cover (or ignore) the row containing the pivot position and cover all rows, if any, above it. Apply steps 1-3 to the submatrix that remains. Repeat the process until there are no more nonzero rows to modify.

To obtain the reduced row echelon form we need one more step.

Step 5

Beginning with the rightmost pivot and working upward and to the left, create zeros above each pivot. If a pivot is not 1, make it 1 by an appropriate row multiplication.

The algorithm described in steps 1-4 will produce the row echelon form of the matrix. This algorithm is called Gaussian elimination. When we add step 5 to produce the reduced row echelon form, the algorithm is called Gauss-Jordan elimination.

If we compare row echelon forms from Activity 3.5, it is likely that different groups or individuals produced different row echelon forms. That is because the row echelon form of a matrix is not unique. (Is the row echelon form ever unique?)

However, if row operations are applied correctly, then we will all arrive at the same reduced row echelon form in Activity 3.5:

It turns out that the reduced row echelon form of a matrix is unique.

Two matrices who are connected by row operations are said to be row equivalent.

Since every elementary row operation is reversible, if \(B\) is row equivalent to \(A\text{,}\) then \(A\) is also row equivalent to \(B\text{.}\) Thus, we just say that \(A\) and \(B\) are row equivalent. While the row echelon form of a matrix is not unique, it is the case that the reduced row echelon form of a matrix is unique.

The reduced row echelon form of a matrix that corresponds to a system of linear equations provides us with an equivalent system whose solutions are easy to find. As an example, consider the system

with augmented matrix

Notice that the coefficient matrix (the left hand side portion of the augmented matrix) of this system is same as the matrix we considered in Activity 3.5. Since we are augmenting with a column of zeros, no row operations will change those zeros in the augmented column. So the row operations applied in Activity 3.5 will give us the reduced row echelon form of this augmented matrix as

Note that the third column is not a pivot column. That means that the variable \(x_3\) is a free variable. There are pivots in the other three columns of the coefficient matrix, so we can solve for \(x_1\text{,}\) \(x_2\text{,}\) and \(x_4\) in terms of \(x_3\text{.}\) These variables are the basic variables. The third row of the augmented matrix tells us that \(x_4=0\text{.}\) The second row corresponds to the equation \(x_2+2x_3 = 0\text{,}\) and solving for \(x_2\) shows that \(x_2 = -2x_3\text{.}\) Finally, the first row tells us that \(x_1+6x_3 = 0\text{,}\) so \(x_1 = -6x_3\text{.}\) Therefore, the general solution to this system of equations is

The fact that \(x_3\) is free means that we can choose any value for \(x_3\) that we like and obtain a specific solution to the system. For example, if \(x_3=-1\text{,}\) then we have the solution \(x_1 = 6\text{,}\) \(x_2 = 2\text{,}\) \(x_3 = -1\text{,}\) and \(x_4 = 0\text{.}\) Check this to be sure.

Recall that in the previous section, we determined the criteria for when a system has a unique solution, or infinitely many solutions, or no solution. With the use of the row echelon form of the augmented matrix, we can rewrite these criteria as follows:

Subsection Examples

Subsection Summary

In this section we learned about the row echelon and reduced row echelon forms of a matrix and some of the things these forms tell us about solutions to systems of linear equations.

• A matrix is in row echelon form if

  1. All nonzero rows are above any rows of all zeros.

  2. Each pivot (the first nonzero entry) of a row is in a column to the right of the pivot of the row above it.

• Once an augmented matrix is in row echelon form, we can use back substitution to solve the corresponding linear system.

• To reduce a matrix to row echelon form we do the following:

  • Begin with the leftmost nonzero column (if there is one). This will be a pivot column.

  • Select a nonzero entry in this pivot column as a pivot. If necessary, interchange rows to move this entry to the first row (this entry will be a pivot).

  • Use row operations to create zeros in all positions below the pivot.

  • Cover (or ignore) the row containing the pivot position and cover all rows, if any, above it. Apply the preceding steps to the submatrix that remains. Repeat the process until there are no more nonzero rows to modify.

• A matrix is in reduced row echelon form if it is in row echelon form and

  1. The pivot in each nonzero row is 1.

  2. Each pivot is the only nonzero entry in its column.

• To obtain the reduced row echelon form from the row echelon form, beginning with the rightmost pivot and working upward and to the left, create zeros above each pivot. If a pivot is not 1, make it 1 by an appropriate row multiplication.

• Both row echelon forms of an augmented matrix tell us about the number of solutions to the corresponding linear system.

  • A linear system is inconsistent if and only if a row echelon form of the augmented matrix of the system contains a row of the form

    \begin{equation*} [0 \ 0 \ 0 \ \cdots \ 0 \ *]\text{,} \end{equation*}

    where \(*\) is not zero. Another way to say this is that a linear system is inconsistent if and only if the last column of the augmented matrix of the system is a pivot column.

  • A consistent linear system will have a unique solution if and only if each column but the last in the augmented matrix of the system is a pivot column. This is equivalent to saying that a consistent linear system will have a unique solution if and only if the consistent system has no free variables.

  • A consistent linear system will have infinitely many solutions if and only if the coefficient matrix of the system contains a non-pivot column. In that case, the free variables corresponding to the non-pivot columns can be chosen arbitrarily and the basic variables corresponding to pivot columns can be written in terms of the free variables.

  • A linear system can have no solutions, exactly one solution, or infinitely many solutions.

Subsection Project: Modeling a Chemical Reaction

Recall the chemical equation

from the beginning of this section. This equation illustrates the reaction between ethane (\(\text{C} _2\text{H} _6\)) and oxygen (\(\text{O} _2\)),called the reactants, to produce carbon dioxide (\(\text{ CO } _2\)) and water (\(\text{H} _2\text{O}\)), called the products of the reaction. In any chemical reaction, the total mass of the reactants must equal the total mass of the products. In our reaction the chemicals involved are made up of carbon (C), hydrogen (H), and oxygen (O) atoms. To balance the equation, we need to know how many molecules of each chemical are combined to preserve the number of atoms of C, H, and O.

Let \(x_1\) be the number of molecules of \(\text{C} _2\text{H} _6\text{,}\) \(x_2\) the number of molecules of \(\text{O} _2\text{,}\) \(x_3\) the number of molecules of \(\text{ CO } _2\text{,}\) and \(x_4\) the number of molecules of \(\text{H} _2\text{O}\) in the reaction. We can then represent this reaction as

In each molecule (e.g., ethane \(\text{C} _2\text{H} _6\)), the subscripts indicate the number of atoms of each element in the molecule. So 1 molecule of ethane contains 2 atoms of carbon and 6 atoms of hydrogen. Thus, there are 2 atoms of carbon in \(\text{C} _2\text{H} _6\) and 0 atoms of carbon in \(\text{O} _2\text{,}\) giving us \(2x_1\) carbon atoms in \(x_1\) molecules of \(\text{C} _2\text{H} _6\) and 0 carbon atoms in \(x_2\) molecules of \(\text{O} _2\text{.}\) On the product side of the reaction there is 1 carbon atom in \(\text{ CO } _2\) and 0 carbon atoms in \(\text{H} _2\text{O}\text{.}\) To balance the reaction, we know that the number of carbon atoms in the products must equal the number of carbon atoms in the reactants.