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Inside Collection: Applied Finite Mathematics
Summary: This chapter covers principles of matrices. After completing this chapter students should be able to: complete matrix operations; solve linear systems using Gauss-Jordan method; Solve linear systems using the matrix inverse method and complete application problems.
In this chapter, you will learn to:
In this section you will learn to:
A matrix is a rectangular array of numbers. Matrices are useful in organizing and manipulating large amounts of data. In order to get some idea of what matrices are all about, we will look at the following example.
Fine Furniture Company makes chairs and tables at its San Jose, Hayward, and Oakland factories. The total production, in hundreds, from the three factories for the years 1994 and 1995 is listed in the table below.
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| San Jose | 30 | 18 | 36 | 20 | ||||||||
| Hayward | 20 | 12 | 24 | 18 | ||||||||
| Oakland | 16 | 10 | 20 | 12 | ||||||||
Before we go any further, we need to familiarize ourselves with some terms that are associated with matrices. The numbers in a matrix are called the entries or the elements of a matrix. Whenever we talk about a matrix, we need to know the size or the dimension of the matrix. The dimension of a matrix is the number of rows and columns it has. When we say a matrix is a 3 by 4 matrix, we are saying that it has 3 rows and 4 columns. The rows are always mentioned first and the columns second. This means that a
Matrix Addition and Subtraction
If two matrices have the same size, they can be added or subtracted. The operations are performed on corresponding entries.
Given the matrices
Find, if possible.
Multiplying a Matrix by a Scalar
If a matrix is multiplied by a scalar (a constant number), each entry is multiplied by that scalar.
Given the matrix
Multiplication of Two Matrices
To multiply a matrix by another is not as easy as the addition, subtraction, or scalar multiplication of matrices. Because of its wide use in application problems, it is important that we learn it well. Therefore, we will try to learn the process in a step by step manner. We first begin by finding a product of a row matrix and a column matrix.
Given
Given
Given
Given
Given the matrices
Find, if possible.
We summarize matrix multiplication as follows:
In order for product
Given the matrices
Find
In this chapter, we will be using matrices to solve linear systems. In Section 7, we will be asked to express linear systems as the matrix equation
Verify that the system of two linear equations with two unknowns:
can be written as
Express the following system as
In this section, we learn to solve systems of linear equations using a process called the Gauss-Jordan method. The process begins by first expressing the system as a matrix, and then reducing it to an equivalent system by simple row operations. The process is continued until the solution is obvious from the matrix. The matrix that represents the system is called the augmented matrix, and the arithmetic manipulation that is used to move from a system to a reduced equivalent system is called a row operation.
Write the following system as an augmented matrix.
In the Section 2, we expressed the system of equations as
For the following augmented matrix, write the system of equations it represents.
Once a system is expressed as an augmented matrix, the Gauss-Jordan method reduces the system into a series of equivalent systems by employing the row operations. This row reduction continues until the system is expressed in what is called the reduced row echelon form. The reduced row echelon form of the coefficient matrix has 1's along the main diagonal and zeros elsewhere. The solution is readily obtained from this form.
The method is not much different form the algebraic operations we employed in the elimination method in the first chapter. The basic difference is that it is algorithmic in nature, and, therefore, can easily be programmed on a computer.
We will next solve a system of two equations with two unknowns, using the elimination method, and then show that the method is analogous to the Gauss-Jordan method.
Solve the following system by the elimination method.
Solve the following system from Example 14 by the Gauss-Jordan method, and show the similarities in both methods by writing the equations next to the matrices.
Now we list the three row operations the Gauss-Jordan method employs.
One can easily see that these three row operation may make the system look different, but they do not change the solution of the system.
The first row operation states that if any two rows of a system are interchanged, the new system obtained has the same solution as the old one. Let us look at an example in two equations with two unknowns. Consider the system
We interchange the rows, and we get,
Clearly, this system has the same solution as the one above.
The second operation states that if a row is multiplied by any non-zero constant, the new system obtained has the same solution as the old one. Consider the above system again,
We multiply the first row by –3, we get,
Again, it is obvious that this new system has the same solution as the original.
The third row operation states that any constant multiple of one row added to another preserves the solution. Consider our system,
If we multiply the first row by – 3, and add it to the second row, we get,
And once again, the same solution is maintained.
Now that we understand how the three row operations work, it is time to introduce the Gauss-Jordan method to solve systems of linear equations.
As mentioned earlier, the Gauss-Jordan method starts out with an augmented matrix, and by a series of row operations ends up with a matrix that is in the reduced row echelon form. A matrix is in the reduced row echelon form if the first nonzero entry in each row is a 1, and the columns containing these 1's have all other entries as zeros. The reduced row echelon form also requires that the leading entry in each row be to the right of the leading entry in the row above it, and the rows containing all zeros be moved down to the bottom.
We state the Gauss-Jordan method as follows.
Solve the following system by the Gauss-Jordan method.
Before we leave this section, we mention some terms we may need in the fourth chapter. The process of obtaining a 1 in a location, and then making all other entries zeros in that column, is called pivoting. The number that is made a 1 is called the pivot element, and the row that contains the pivot element is called the pivot row. We often multiply the pivot row by a number and add it to another row to obtain a zero in the latter. The row to which a multiple of pivot row is added is called the target row.
In this section you will learn to:
If we consider the intersection of two lines in a plane, three things can happen.
The figures below shows all three cases.
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Every system of equations has either one solution, no solution, or infinitely many solutions.
In the Section 4, we used the Gauss-Jordan method to solve systems that had exactly one solution. In this section, we will determine the systems that have no solution, and solve the systems that have infinitely many solutions.
Solve the following system of equations.
At this stage, we are going to start using a calculator to row reduce the augmented matrix.
Solve the following system of equations.
Solve the following system of equations.
Solve the following system of equations.
Solve the following system of equations.
We summarize our discussion in the following table.
In this section you will learn to:
In this section, we will learn to find the inverse of a matrix, if it exists. Later, we will use matrix inverses to solve linear systems.
Definition of an Inverse: An
Given matrices
Find the inverse of the following matrix.
What you just witnessed is no coincidence. This is the method that is often employed in finding the inverse of a matrix.
We list the steps, as follows:
Given the matrix
Now that we know how to find the inverse of a matrix, we will use inverses to solve systems of equations. The method is analogous to solving a simple equation like the one below.
Solve the following equation .
To solve a linear system, we first write the system in the matrix equation
Consider the following example.
Solve the following system
Solve the following system
Once again, we remind the reader that not every system of equations can be solved by the matrix inverse method. Although the Gauss-Jordan method works for every situation, the matrix inverse method works only in cases where the inverse of the square matrix exists. In such cases the system has a unique solution.
We summarize our discussion in the following table.
In this section, we see a use of matrices in encoding and decoding secret messages. There are many techniques used, but we will use a method that first converts the secret message into a string of numbers by arbitrarily assigning a number to each letter of the message. Next we convert this string of numbers into a new set of numbers by multiplying the string by a square matrix of our choice that has an inverse. This new set of numbers represents the coded message. To decode the message, we take the string of coded numbers and multiply it by the inverse of the matrix to get the original string of numbers. Finally, by associating the numbers with their corresponding letters, we obtain the original message.
In this section, we will use the correspondence where the letters A to Z correspond to the numbers 1 to 26, as shown below, and a space is represented by the number 27, and all punctuation is ignored.
| A | B | C | D | E | F | G | H | I | J | K | L | M |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
| N | O | P | Q | R | S | T | U | V | W | X | Y | Z |
| 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 |
Use the matrix
Decode the following message that was encoded using matrix
Now suppose we wanted to use a
Using the matrix
Decode the following message that was encoded using matrix
We summarize.
In the 1930's, Wassily Leontief used matrices to model economic systems. His models, often referred to as the input-output models, divide the economy into sectors where each sector produces goods and services not only for itself but also for other sectors. These sectors are dependent on each other and the total input always equals the total output. In 1973, he won the Nobel Prize in Economics for his work in this field. In this section we look at both the closed and the open models that he developed.
As an example of the closed model, we look at a very simple economy, where there are only three sectors: food, shelter, and clothing.
We assume that in a village there is a farmer, carpenter, and a tailor, who provide the three essential goods: food, shelter, and clothing. Suppose the farmer himself consumes 40% of the food he produces, and gives 40% to the carpenter, and 20% to the tailor. Thirty percent of the carpenter's production is consumed by himself, 40% by the farmer, and 30% by the carpenter. Fifty percent of the tailor's production is used by himself, 30% by the farmer, and 20% by the tailor. Write the matrix that describes this closed model.
In Example 36 above, how much should each person get for his efforts?
The Open Model
The open model is more realistic, as it deals with the economy where sectors of the economy not only satisfy each others needs, but they also satisfy some outside demands. In this case, the outside demands are put on by the consumer. But the basic assumption is still the same; that is, whatever is produced is consumed.
Let us again look at a very simple scenario. Suppose the economy consists of three people, the farmer
Let us assume that whatever the farmer produces, 20% is used by him, 15% by the carpenter, 10% by the tailor, and the consumer uses the other 40 billion dollars worth of the food. Ten percent of the carpenter's production is used by him, 25% by the farmer, 5% by the tailor, and 50 billion dollars worth by the consumer. Fifteen percent of the clothing is used by the tailor, 10% by the farmer, 5% by the carpenter, and the remaining 60 billion dollars worth by the consumer. We write the internal consumption in the following table, and express the demand as the matrix D.
| .20 | .25 | .10 | |
| .15 | .10 | .05 | |
| .10 | .05 | .15 |
The consumer demand for each industry in billions of dollars is given below.
In Example 37, what should be, in billions of dollars, the required output by each industry to meet the demand given by the matrix D?
Suppose an economy consists of three industries
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Demand | Total | |
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40 | 50 | 60 | 100 | 250 |
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30 | 40 | 40 | 110 | 220 |
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20 | 30 | 30 | 120 | 200 |
The first row says that of the $250 dollars worth of production by the industry
Find the proportion of the amounts consumed by each of the industries. In other words, find the matrix A.
Once again, the total input equals the total output.
We summarize as follows:
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Last edited by Collins Williamson on Jul 14, 2011 5:32 pm -0500.
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