The difference between two matrices, A – B, can be thought of as the sum A + (-1B). If
then
If we combine these two matrices together, we get
Looking closely, we see that each entry in the matrix is the difference between the corresponding entries in A and B. We can simplify the process of subtracting matrices by simply subtracting the corresponding entries in the matrices being subtracted. Keep in mind that we can only do this if the matrices being subtracted have the same size.
The difference of two m x n matricesis
The circulation matrices
indicate the quarterly circulation of Ed Magazine for renewing and non-renewing subscribers. The difference of these matrices,
is found by subtracting the corresponding entries in each matrix. This difference indicates how the circulation for renewing subscribers differs from the circulation for non-renewing subscribers. For instance, the entry in the first row tells us that 4000 more issues were produced and sent to renewing subscribers than to non-renewing subscribers in the first quarter.
Example 8 Subtract Two Matrices
The total cash disbursement matrix D was calculated to be
In the next section, we’ll show that the total cash receipts for Ed Magazine are given by the matrix
The business’s cash flow (also called the total net cash) is defined as the difference between the total cash receipts and the total cash disbursements. Compute the cash flow matrix, CF = R – D, and explain its meaning.
Solution The cash flow matrix is computed by subtracting the corresponding entries in R and D:
Notice that some of the entries in the cash flow matrix are negative. Negative entries in the cash flow matrix occur when the cash disbursements for expenses are greater than the cash receipts from subscribers. In the second and fourth quarters, more money is being spent than is coming in. This is not a serious issue as long as the magazine has adequate cash reserves to cover the deficit.
If we assume that the company begins the year with no cash reserves, then after the end of the first quarter of the year they will have cash on hand of $113,560 due to the fact that cash receipts from subscribers exceed cash disbursements. As long as they reserve at least $13,160 for the next quarter, they will be able to cover the negative cash flow in the second quarter. The rest of the cash receipts can be used for other purposes such as investing or purchasing new equipment.
Using matrices we can model the cash flow in each quarter to insure that there is always enough cash on hand to handle any times when the cash disbursements exceed the cash receipts.
The sum of two matrices is accomplished by adding the corresponding entries in the matrix. Because of this, the matrices being added must have the same size.
The sum of two m x n matricesandis
The circulation matrices
indicate the quarterly circulation of Ed Magazine for renewing and non-renewing subscribers. The sum of these matrices,
is found by adding the corresponding entries in each matrix. This sum shows the total circulation for these two subscriber groups together. For instance, the entry in the second row tells us that 34,400 issues were produced and sent to renewing and non-renewing subscribers in the second quarter.
Example 5 Add Two Matrices
The cash receipts from subscribers for Ed Magazine are
Find and interpret the sum R = R1 + R2.
Solution Each matrix is a 4 x 1 matrix so it is possible to calculate the sum. The sum is found by adding the corresponding entries,
To discover what the entries in the sum represent, let’s look at the second row. It is calculate from the entries in the second row of R1 and R2,
Since the magazine receives money from new subscribers or renewing subscribers, the sum is the total cash receipts from all subscribers. This particular sum comes from the row corresponding to the second quarter. Other rows give the total cash receipts from subscribers for other quarters.
The sum of several matrices is carried out in a similar fashion. As long as each matrix in the sum has the same size, the entries in the sum are calculated by adding the corresponding entries of each matrix.
Example 6 Add Several Matrices
The circulation for Ed Magazine is broken into three matrices: the circulation for new subscribers C1, the circulation for subscribers who renew their subscription C2, and the circulation for subscribers who do not renew their subscription C3. When organized into matrices, this information becomes
Find and interpret C = C1 + C2 + C3.
Solution We can find the sum of three matrices as long as each matrix has the same size. Since each matrix in this sum is 4 x 1, the sum of these three matrices is found by adding the corresponding entries of each matrix:
Each row of the sum gives the total circulation in the corresponding quarter. For instance, in the spring quarter Ed Magazine produced 42400 issues of the magazine.
Example 7 Matrix Arithmetic
Ed Magazine maintains production, editorial and advertising staff that help publish the magazine each quarter. These disbursements do not change even if the number of issues produced increases. Because these disbursements are fixed, they are called fixed disbursements. The table below gives these disbursements for each quarter.
Recall that the magazine pays $0.60 to print and mail each issue. In Example 6, we found that the total circulation by quarter is given by the matrix
Write the fixed disbursements in a matrix F and use it to find a matrix for the total cash disbursements
Solution The circulation matrix gives the number of issues printed and mailed each quarter. If we multiply this matrix C by the expense to print and mail each issue, the product tells us how much is being spent to print and mail the magazine each quarter. The total cash disbursements D are calculated by adding the fixed disbursements in each quarter, F, to the expense of printing and mailing the magazine in each quarter.
We summarize this relationship by writing
The second term gets larger as more magazines are printed and mailed so it is referred to as the variable expenses. Since the circulation matrix C is a 4 x 1 matrix with the rows corresponding to quarters, the fixed disbursements F must also be a 4 x 1 matrix. This insures that we’ll be able to carry out the sum of F and 0.60 C.
The fixed disbursements in the table yield the matrix
With these matrices, the total cash disbursements are
Each entry in D gives the total cash spent by Ed Magazine on fixed expenses and the variables expenses in a particular quarter. These disbursements vary by quarter since different issues may require more or less production and editing and the circulation changes throughout the year.
In mathematics, real numbers are often called scalars. The process of multiplying a matrix by a scalar is simply a matter of multiplying each entry in the matrix by the scalar.
The product of a scalar c and a matrixis
For instance, the circulation of issues to new subscribers at Ed Magazine is given by the column matrix
In this matrix, the rows correspond to the different quarters. This means the entry in the second row, first column tells us that in the second quarter 8200 issues of Ed Magazine was distributed to new subscribers. If each issues costs $0.60 to print and mail, the product
gives the expense of printing and mailing the magazine to new subscribers in each quarter. We can calculate the expense of printing and mailing the magazine to other subscribers similarly.
Example 4 Multiply a Matrix by a Scalar
It costs Ed Magazine $0.60 to print and mail one issue of the magazine. The number of issues distributed to renewing subscribers is given by the matrix C2 and the number of issues distributed to non-renewing subscribers is given by the matrix C3 where
The rows of each matrix refer to the first, second, third and fourth quarters of the year.
a. Find the product 0.60 C2.
Solution To multiply the circulation matrix by the scalar 0.60, multiply each entry in the matrix by 0.60:
b. Find the product 0.60 C3.
Solution Multiply each entry in the circulation by the cost per issue 0.60 to give
c. Interpret what the products in parts a and b mean.
Solution Each entry in the product comes from multiplying the expense of printing and mailing a single issue in dollars per issue by the number of issues printed and sent to renewing subscribers. In the product 0.60 C2, each entry is
This is the expense of printing and mailing all issues to the renewing subscribers in the quarter corresponding to the numbers. Since these subscribers are subscribed for the entire year, the expense is the same through all quarters. For the product
each entry in the product yields the expense to print and mail issue of the magazine to non-renewing subscribers in each quarter. Note that this expense drops throughout the year as the number of non-renewing subscribers drops off.
When we use a matrix to organize data, all of the definitions we developed in Chapter 2 still apply. You’ll recall that a matrix can have any number of rows and columns and is typically named with a capital letter. A matrix with m rows and ncolumns named A would look like
Notice that this matrix does not contain a dashed vertical line in front of the last column. The dashed line is unique to augmented matrices and is used to separate the coefficients from the constants. The dots in the matrix indicated a pattern in the matrix. In this case, the dots indicate the arbitrary number of rows m and columns n in the matrix.
The individual entries (also called elements) of the matrix are symbolized with lowercase letters, like amn, and these symbols represent numbers. The subscript on the lowercase letter indicates the location of the entry in the matrix. The symbol a23 represents the number in the second row, third column of the matrix.
The size of a matrix (also called the dimensions of the matrix) is the number of rows and columns in a matrix. For the matrix A with m rows and n columns, we would say the size of the matrix is m x n (read m by n).
Several sizes of matrices are given special names. A matrix with the same number of rows and column is called a square matrix. An example of a square matrix is the 2 x 2 matrix
The exact number of rows and columns in a square matrix is not important, only the fact that the number of rows and columns is the same.
Matrices with a single row or a single column are also given special names. Row matrices like or are matrices with only a single row, but any number of columns. Column matrices like or are matrices with any number of rows, but a single column.
The size of a matrix is an important prerequisite in determining if two matrices are equal.
Two matrices are equal if they have the same size and each entry in one matrix is equal to the corresponding entry in the other matrix.
Example 1 Matrix Terminology
The matrices
are 3 x 3 square matrices.
a. What is the value of the entry b32?
Solution The subscript on b32 refers to the entry in the third row, second column of the matrix B. Therefore, b32= -4.
b. Is a23 = b23?
Solution The subscripts on a23 and b23 refer to the corresponding entries in the second row, third column of A and B. Since the entry in that location is 4 in both matrices, a23 = b23 and is equal to 4.
c. Is A = B?
Solution For the matrices to be equal, they must have the same size and each entry in A must be equal to the corresponding entry in B. Both matrices are 3 x 3. In part b, we determined that the entries in the second row, third column of each matrix were equal. However, a13 in
and b13 in
are not the same so A ≠ B. This is in spite of the fact that every other set of corresponding entries are equal.
Ed Magazine is a fictional magazine that publishes four issues each year. It has a loyal base of subscribers and twice a year it conducts subscription drives for new subscribers. At the same time they are acquiring new subscribers, the current subscriber’s subscriptions are expiring. Some of these expiring subscriptions belong to first time subscribers and others are long time subscribers who have renewed their subscriptions in the past. The table below shows the numbers of new and expiring subscribers by quarter.
Although this information could be placed in a matrix in several different ways, two approaches stand out. Since the rows in the table correspond to the four different quarters during the year and the columns correspond to numbers of subscribers, we could use a matrix with four rows and three columns:
Normally we don’t include the red labels on a matrix. However, they are often included to help clarify how the information in the matrix is organized. To name this matrix of subscribers, we could use the letter S and write
This organization capitalizes on the fact that all of the numbers in the table indicate the number of subscribers in a certain category.
Let’s look at the table differently.
The rows in the table still refer to quarters, but now the shading in the table emphasizes a difference in the numbers. The numbers in the blue region corresponds to the number of new subscribers by quarter and the red region corresponds to subscribers whose subscriptions are expiring.
With this difference in mind, we could define two matrices for this table,
The matrix Nis a 4 x 1 column matrix representing the number of new subscribers of Ed Magazine. The matrix E is a 4 x 2 matrix representing the number of expiring subscribers in two categories by quarter. Depending on the application, these matrices may be more useful than the 4 x 3 matrix S.
Example 2 Organize Information in a Matrix
A magazine’s circulation is the number of issues it distributes. Ed Magazine is distributed to three categories of subscribers each quarter.
Use this information to define three matrices named C1, C2, and C3, where C1 describes the number of issues distributed to new subscribers, C2 describes the number of issues distributed to subscribers who have renewed their subscription, and C3 describes the number of issues distributed to subscribers who have not renewed their subscriptions.
Solution The first column in the table corresponds to issues distributed to new subscribers
If we let the rows in the matrix C1 correspond to the quarters, then can organize the information in the table in a 4 x 1 matrix as
Alternatively, we could also let the quarters correspond to the columns in a 1 x 4 matrix and define
Either matrix organizes the information appropriately. Since the original table matches each row with a quarter, we’ll follow the same principal and let the rows of the matrices correspond to the quarters.
Letting the rows match the quarters, the other columns in the table give the entries in C2 and C3,
In Example 2, we mentioned the fact that the data in the first column of the table could be written as a row matrix or a column matrix. These matrices are examples of transposes. In other words, the matrix
is the transpose of the matrix
The transpose of the matrix A, written AT, is obtained by writing the columns of the matrix A as rows in the matrix AT. Alternatively, we could also write the rows of the matrix A as columns in the matrix AT.
Example 3 Find the Transpose of a Matrix
Find and label the transpose of the expiring subscriber matrix
Solution To get the transpose of the expiring subscriber matrix, we interchange the rows and columns. In other words, the columns of E become the rows of the transpose ET or the rows of E become the columns of the transpose ET to yield
In the original matrix E, the rows of the subscriber matrix correspond to the quarters and the columns tell us the subscriber category.
In the transpose, these roles are reversed.
The information in each matrix is the same, but organized differently.
How do you mix different grades of ethanol to create a new grade of ethanol?
In the next example we examine how to use matrices to solve the ethanol mixing problem from Section 2.2 and extend the example to more types of ethanol.
Example 9 Mixing Ethanol Blends
In Example 12 of section 2.2 we created a system of equations to describe a mix of E10 and E85 ethanol,
where E10 is the amount of 10% ethanol pumped in gallons and E85 is the amount of 85% ethanol pumped in gallons. The first equation describes the total amount in the mixture, 10 gallons. The second equation describes the total amount of ethanol in the mixture, 20% of 10 gallons or 2 gallons.
Solve this system of equations by finding the reduced row echelon form for the augmented matrix.
Solution The augmented matrix for this system is
where the first column corresponds to E10 and the second column corresponds to E85. To convert this matrix to reduced row echelon form, we must create pivots in the first and second columns and put zeros above and below the pivots. There is already a 1 in the first row and column, so we need to use row operations to put a 0 below it:
We can place a pivot in the second row and column by multiplying the row by the reciprocal of 0.75,
A 0 is placed above the pivot in the second column by multiplying the second row by -1, adding it to the first row, and putting the result in the first row:
Based on this reduced row echelon form, E10 = 26/3 and E85 = 4/3. This means that mixing 26/3 gallons of 10% ethanol with 4/3 gallons of 85% ethanol will yield 10 gallons of 20% ethanol.
Example 10 Mixing Three Types of Ethanol
In some Midwestern states, three different blends of ethanol are available. E10 contains 10% ethanol, E30 contains 30% ethanol, and E85 contains 85% ethanol. How much of each of the blends must be mixed to make 10 gallons of 20% ethanol? If the price per gallon for each type is given in the table below, what is the least that the 10 gallon blend would cost?
Solution This problem adds another type of ethanol that should contribute to the total number of gallons in the mixture and the total amount of ethanol in the mixture. Each of the two equations in Example 9 needs another term corresponding to E30:
where E10 is the number of gallons of 10% ethanol, E30 is the number of gallons of 30% ethanol, and E85 is the number of gallons of 85% ethanol.
The augmented matrix for this system is
As in the previous example, the pivot in the first column is already in place so we simply need to use row operations to put a 0 below it:
To put a pivot in the second column, multiply the second row by the reciprocal of 0.20:
To put a 0 above the pivot,
Writing this matrix as an a system of equations leads to
E85 appears in both equations so we’ll solve for the other variables in terms of E85,
This system has many solutions, but we must be careful. These variables represent gallons of different ethanol blends. Because of this, the variables cannot be negative. As long we pick reasonable values for E85, we can insure that all the variables are nonnegative. For instance, if we set E85 = 0 we get
This solution makes sense since mixing equal amounts of E10 and E30 should yield a mixture with a percent ethanol midway between 10% and 30%.
Notice that as we increase the amount of E85, the amount of E10 increases (the E85 term is added in the E10 equation) and the amount of E30 decreases (the E85 term is subtracted in the E30 equation). Eventually the amount of E30 will equal 0 as E85 is increased. This occurs when
Reasonable values for E85 are from 0 to 4/3 including 0 and 4/3 since this leads to non-negative values for E10 and E30 as well.
We can calculate the total cost of any combination using cost per gallon of each type of ethanol. For instance, 5 gallons of E10 and 5 gallons of E30 would cost
Here is a table of some of the possible solutions based on
and their corresponding total costs:
As the amount of E85 is increased, the total cost drops. Each increase of 0.25 gallons of E85 leads to a drop in the total cost of about 0.02 dollars. The lowest cost appears to come from mixing 4/3 gallons of E85 and 26/3 gallons of E10. Using any more E85 would require us to use a negative amount of E30. Even though this would yield a lower cost, it is not a reasonable value for this problem.
