2.4 Solving a System of Linear Equations with Matrices

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1 .4 Solving a System of Linear Equations with Matrices Question : What is a matrix? Question : How do you form an augmented matrix from a system of linear equations? Question : How do you use row operations to determine the reduced row echelon form of a matrix? Question 4: Do all systems of linear equations have unique solutions? Question : How do you mix different grades of ethanol to create a new grade of ethanol? In section., you learned how to solve systems of equations in two variables using the Substitution Method or the Elimination Method. These methods worked well for problems involving two variables, but are cumbersome for problems involving more than two variables. Of course, problems involving more than two variables are what you are most likely to encounter in business and finance so we ll need more efficient techniques involving matrices to solve these systems.

2 Question : What is a matrix? A matrix is simply a table of numbers enclosed by a set of square brackets. These numbers may correspond to inventory levels, production quotas, or almost anything. We specify the size of a matrix by giving the number of rows and columns in the matrix. An m x n matrix (read m by n) is a table of numbers with m rows and n columns enclosed by a set of square brackets. The plural of matrix is matrices. The size of a matrix is always listed as row by column. Shown below are several matrices of various sizes Any matrix with only one row is also called a row matrix. The matrix in the center is an example of a row matrix. A matrix with only one column is called a column matrix. The matrix on the far right is an example of a column matrix. Capital letters are used to name matrices. For instance, we might name the x matrix given above with the letter A, 0 A 4 7 The individual entries in the matrix are denoted by the corresponding lower case letter with a subscript. The number 4 in the third row and first column is called a and the number - in the second row and second column is called a. In fact, we can match any entry to its name using the lowercase letter matching its name with a subscript. In general, a mn is the entry in the m th row and n th column.

3 Example Find the Matrix Entry For the matrices B C find the entries indicated in each part. a. b Solution Let s examine the matrix entry in detail. The entries are in matrix B b First row Second column The entry in the first row and second column of the matrix B is 0.7. b. c Solution The entry in the second row and first column of C is. c. b Solution This entry matches with the number in the third row and second column of B. Since B only has two rows, b does not exist for the given matrix B.

4 Question : How do you form an augmented matrix from a system of linear equations? The entries in a matrix are not just random values without meaning. In general, the rows and columns in a matrix have labels that help you to match numbers with an application. For an augmented matrix, the rows in a matrix correspond to the equations in a system of linear equations. The columns match with the variables and constants in each equation. The augmented matrix that corresponds to the system of linear equations x y 8 x y is 8 For a system to be written as an augmented matrix, all of the variables in the system must be on one side of the equal sign and in the same order for each equation. The constants must be on the other side of the equal sign. If we rewrite the system as x y 8 x ( ) y we see that the coefficients and constants in the system match with the entries in the augmented matrix. The dashed line between the second and third columns separates the variables from the constants in the system of equations. On the left side of the dashed line are the coefficients of the variables and on the right sides are the constants from each equation, Coefficients on variables in the first equation Constant in the first equation 8 Coefficients on variables in the second equation Constant in the second equation 4

5 Following this template, we can write any system of linear equations as an augmented matrix or any augmented matrix as a system of linear equations. Example Convert a System to an Augmented Matrix Write each of the systems of linear equations as an augmented matrix. a. xy xy 0 Solution Since this system has two equations, the augmented matrix must have two rows. There are two variables and constants in the system so we need three columns in the augmented matrix. We ll match the first column to the coefficients of x, the second column to the coefficients of y and the third column to the constants. This leaves us with the augmented matrix 0 x y z b. x y yz 9 Solution Since there are three equations in the system the augmented matrix needs to have three rows. Although each equation does not have three variables, there are a total of three variables in the system. This means that there must be four columns in the augmented matrix. The last column will correspond to the constants in the system. The augmented matrix is 0 0 9

6 If a variable is missing from an equation, the augmented matrix contains a 0 in the row and column that matches the equation and missing variable. c. x x x 0 0.x 0.x 0.0x 0. x x x Solution The first equation has the appropriate form to convert into an augmented matrix. The variables are on the left side and the constants are on the right side of the equals sign. However, the second equation has variables on both sides of the equal sign. To convert the second equation to the proper format, remove the parentheses 0.x 0.x 0.0x 0.x 0.x 0.x and move all of the variables to the left side, 0.0x 0.08x 0.07x 0 eplace the second equation with this equivalent equation to give x x x 0 0.0x 0.08x 0.07x 0 If we let the first column correspond to x, the second column correspond to x, and the third column correspond to x, the augmented matrix is

7 Example Convert an Augmented Matrix to a System of Equations For the augmented matrix, write the corresponding system of linear equations with the variables x, y, z. Solution The three columns to the left of the dashed lines must correspond with the three given variables, x, y, and z. The numbers in each row to the left of the dashed line are the coefficients in the equation. The numbers to the right of the dashed line are the constants in each equation. The first row of the augmented matrix corresponds to the equation xz 6. Since there is a 0 in the second column, no y appears in this equation. The second row of the augmented matrix indicates that 0.x yz 4. The last row gives the equation xyz 0. The complete system of equations is 0 6 x z x y z 4 0 x y z 0 augmented matrix system of equations Once a system of linear equations is converted to an augmented matrix, we can modify the augmented matrix to determine each variable. Next we ll look at a strategy for solving the system of equations using the augmented matrix. 7

8 Question : How do you use row operations to determine the reduced row echelon form of a matrix? To solve a system of linear equations using matrices, we must convert the augmented matrix to another matrix that is in reduced row echelon form. An augmented matrix corresponding to a system of m equations in n variables is in reduced row echelon form if the leading nonzero entry in each row is a one and all entries above and below the leading ones are zeros. Each of the augmented matrices below is in reduced row echelon form: When an augmented matrix is in reduced row echelon form, it is easy to find the solution to the system of linear equations. The system of equations corresponding to an augmented matrix and the system of equations corresponding to its reduced row echelon form are equivalent. This means that the systems of equations have the same solution set. For instance, suppose we have the system of linear equations 4x yz x y z. xyz 4 This system yields the augmented matrix 8

9 4. 4 Later in this section we see how to get the reduced row echelon form for this matrix, If we rewrite this matrix as a system of linear equations, we get x 0y0z 0xy0z. 0x0yz Dropping all of the terms with a 0 coefficient leads to the solution, x, y and z. We can check to see that these values satisfy the original system to insure it is equivalent to the system of equations corresponding to the reduced row echelon form: 4? TUE 4xyz x y z xyz 4 x y z? TUE 4? 4 4 TUE 9

10 To solve a system of m linear equations in n variables, we will follow a simple strategy. Strategy for Solving a System of Linear Equations with Matrices. Determine the augmented matrix that corresponds to the given system of equations.. Find the reduced row echelon form of the augmented matrix.. Use the reduced row echelon form to obtain the solution of the original system of equations. An augmented matrix is transformed to its reduced row echelon form by using three different operations on the rows. These row operations change the entries in the rows without changing the solutions to the corresponding system of linear equations. If these operations seem familiar, it is no mistake. Each operation corresponds to an equation transformation discussed in Section.. ow Operations on Matrices Each of the row operations below changes a matrix to an equivalent matrix:. Interchange any two rows of a matrix.. Multiply a row by a nonzero constant.. eplace a row with the sum of a nonzero multiple of one row to a nonzero multiple of another row. 0

11 Our goal is to use these row operations to transform the original augmented matrix to its reduced row echelon form. It is helpful to symbolize the row operations to help identify what is being done. If we want to interchange two rows like the first and second, we will write Interchange and To symbolize that a row is being multiplied by a constant and replacing another row we use a similar notation. For instance, if we want to indicate that row is being replaced by 0 of row we would write 0 If we need to indicate that multiples of rows are being added and replacing another row, we would write something like This particular notation indicates that you are multiplying row by - and adding it to row. The sum is placed in the third row. Notes like these help us to document our row operations. You are encouraged to adopt the same notes so that you can follow your own work easily.

12 Example 4 Solve a System of Linear Equations Using Matrices Solve the system of equations 4x yz x y z xyz 4 by transforming its augmented matrix to reduced row echelon form. Solution The augmented matrix for this system is 4. 4 The strategy for finding its reduced row echelon form is similar to the strategy for the Elimination Method. To put this matrix into reduced row echelon form, we ll work with each column, one at a time. Instead of using equation transformations to change the leading coefficient to a, we ll use row operations to create a in the entry in the first row and first column of the matrix. After placing the, we ll use row operations to put zeros in the rest of the column. This step corresponds to eliminating the first variable in all of the equations except the first equation. We ll continue through the rest of the columns of the matrix placing ones and zeros in a manner similar to the Elimination Method. First we ll use row operations to transform the original augmented matrix to??? 0??? 0???

13 In each column we ll use row operations to put the in the column and then continue to use row operations so that the zeros appear below the. The is called the pivot in the column. For this augmented matrix, we could multiply the first row by or 4 interchange the first and second rows. If we multiply the first row by 4 and replace the first row with the result we get The symbols on the left indicate that each entry in the first row is multiplied by 4. If we interchange the first and second rows we get 4. 4 Either row operation puts a at the top of the first column. However, interchanging the rows does not introduce any fractions into the problem, so that is the best row operation to use. Now that the pivot has been established, we ll use it to put zeros everywhere else in the column. To put a 0 below the one, multiply the first row by -4 and add the result to the second row. Place the sum in the second row: 4 : :

14 The symbols on the left show -4 times the first row of the matrix and the second row below it. Adding the columns vertically gives the sum that is placed underneath the horizontal bar. The blue arrow indicates where that sum is placed in the matrix. To put a 0 in the bottom of the first column, multiply the first row in this new matrix by and add it to the third row. Place the sum in the third row: : 4 : The first column is in the proper format so now we continue onto the second column. We need to use more row operations to change this new matrix into 0?? 0??. 0 0?? Start by putting a in the middle of the second column. This can be done in many ways, but we multiply the second row by 6 to place the pivot. This gives us With the pivot in place, we now put zeros in the rest of the second column: 4

15 : 0 : 0 : : Next we place the pivot in the third column. To put a in the bottom of the third column, multiply the third row by and place the result in the third column: Now that the pivot has been established in the third column, we can place zeros in the rest of the column. : 0 0 : 0 : : This matrix is in reduced row echelon form and indicates that the solution is x, y, and z.

16 Example Solve a System of Linear Equations Using Matrices Solve the system of equations x y z x y4z 8 x yz 8 by transforming its augmented matrix to reduced row echelon form. Solution The augmented matrix for this system is To place a at the top of the first column, interchange the first and second rows. This transforms the original augmented matrix to A zero is created below the pivot by multiplying the first row by -, adding this to the second row, and placing the result in the second row. A zero is also created at the bottom of the first column by multiplying the first row by -, adding this to the third row, and placing the result in the third row. These two row operations yield 6

17 : : 0 9 : 0 40 : In the second column, we need to place a pivot (a ) in the second row. Multiply the second row by and place the result in the second row: Once the pivot is established, we can use it to put zeros in the rest of the column: : : : : With these two row operations, the second column is in the proper format for reduced row echelon form. In the third column, we need to place a pivot in the third row. To do this, multiply the third row by 4 : 7

18 Create zeros above the pivot to put the matrix in reduced row echelon form: 7 9 : : 0 7 : : Since the first column corresponds to x, the second column to y and the third column to z, the solution to the original system is x y z 0 In each of the last two examples, the same strategy is used to find the reduced row echelon form. Starting with the first column, we place the pivot in the proper position by interchanging rows or multiplying a row by a constant. Once the pivot is in place, we can multiply the pivot row by a constant and add it to the other rows to put zeros in the rest of the column. By doing this in each column, we can efficiently get the reduced row echelon form and the solution to the system of linear equations. 8

19 Question 4: Do all systems of linear equations have unique solutions? Earlier we examined two systems where the numbers of variables was equal to the number of equations. Often there are fewer equations than variables or more equations than variables. Even in systems where the number of variables initially equals the number of equations, one of the rows may become all zeros, meaning that the equation was unnecessary. In all of these cases, we can still use the strategy from Question to solve the problem. Example 6 Solve a System with More Variables Than Equations Solve the system of equations x x x 70 x x.x 0 by transforming its augmented matrix to reduced row echelon form. Solution This system does not use x, y, and z like earlier examples. However, if we let the first column in the augmented matrix correspond to x, the second column to x, and the third column to x, we can write the augmented matrix for this system as To put this matrix into reduced row echelon form, we must transform it using row operations to 0?? 0?? Since there is no third row, we have no hope of solving for a solution where x, x, and x are each a unique value. Instead, we ll be able to solve for x and x in terms of x. 9

20 The original augmented matrix already has a in the first column and row, so there is no need to put a pivot there. To put a 0 below the pivot, multiply the first row by -, add it to the second row, and put the result in place of the second row: : 40 : The previous row operation not only put a 0 in the first column, but also placed the pivot in the second row and column (very lucky!). To put this matrix into reduced row echelon form, we need to put a zero above the pivot in the second column. Multiply the second row by -, add it to the first row, and place the result in the first row: : 0. 0 : This matrix is in reduced row echelon form, but we can t read the solution off as we have in earlier examples. If we convert this back to a system of equations, we get x 0.x 00 x.x 0 Notice that each of these equations contains x and it is easy to solve for x in the first equation and x x and x we get in the second equation. If we solve for x 0.x 00 x.x 0 0

21 Although we don t have specific numbers for each variable, we do have a recipe for finding values. If we choose any value for x, we can find the corresponding values for x and x. For instance, if we choose x 00, then x x If x 0, then x x Since x can be any number, there are an infinite number of solutions to the system. However, not just any combination of numbers works. Once a value for x is chosen, the equations above must be used to calculate corresponding values for x and x. This can be summarized by writing x can be any real number x 0.x 00 x.x 0 In this case, we have more variables than equations so we would expect to be able to solve for only some of the variables explicitly. In each row we can solve for a variable explicitly as long as the row is not entirely zeros. Any variables beyond the number of nonzero rows, in this case one, will be values that we can pick. These variables are called parameters. Parameters are variables whose values are arbitrary and can be picked to be anything that is reasonable for the system of equations.

22 Example 7 Solve a System with More Equations Than Variables Solve the system of equations x x x 8x 6 x x by transforming its augmented matrix to reduced row echelon form. Solution The augmented matrix for this system is 8 6 To put this matrix into reduced row echelon form, we need to place pivots in the first and second columns and zeros in the rest of these columns. The entry in the first column and row is a so the pivot is already in place. To put zeros below the pivot, : 4 4 : : : In the second column and row, we need to transform the to a by multiplying the row by :

23 Now place zeros in the rest of the column, : 0 : 0 4 : 0 : If we convert this matrix back to a system of equations, we find that x 4 and x. Notice that the last row of the matrix is all zeros and does not contribute to the solution. This means that there are two variables and two nonzero rows so no parameters are needed in the solution. We can check the solution by substituting 4, into each equation: First equation: 4? TUE Second equation: 4 8 6? 6 6 TUE Third equation: 4? TUE

24 Since, 4, x x satisfies each equation in the system, it is the solution to the original system of equations. Example 8 Solve a System with More Equations Than Variables Solve 0x x 44 x x x x 8 by transforming its augmented matrix to reduced row echelon form. Solution Like the last example, we ll let the first column in the augmented matrix correspond to x and the second column to x. The augmented matrix for this system is To create a pivot in the first row and column, we have two possibilities. We could multiply the first row by 0 or interchange the first and third row. Since interchanging rows does not introduce and fractions into the matrix, we ll do that to yield Now we ll use row operations to fill the rest of the first column with zeros: 4

25 : : : : To create a pivot in the second row and column, multiply the second row by : Now use row operations to place zeros above and below the pivot in the second column: : 0 6 : 8 0 : 0 6 : The first two rows in the reduced row echelon form suggest a solution, but the third is problematic. If we write this row as an equation, we get 0x 0x 7

26 The left side is simply 0. Since 0 cannot equal 7, this implies that there are no solutions to this system. This system is an inconsistent system. These examples illustrate what may happen when there are more equations than variables or when there are more variables than equations. It is possible for the system to have a unique solution as in Example. The system may be a system with infinitely many solutions like Example 6 or have no solutions like the inconsistent system in Example 8. 6

27 Question : 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. and extend the example to more types of ethanol. Example 9 Mixing Ethanol Blends In Example of section. we created a system of equations to describe a mix of E0 and E8 ethanol, E0 E E0 0.8 E8 where E0 is the amount of 0% ethanol pumped in gallons and E8 is the amount of 8% ethanol pumped in gallons. The first equation describes the total amount in the mixture, 0 gallons. The second equation describes the total amount of ethanol in the mixture, 0% of 0 gallons or 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 E0 and the second column corresponds to E8. 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 in the first row and column, so we need to use row operations to put a 0 below it: 7

28 0.0 : : We can place a pivot in the second row and column by multiplying the row by the reciprocal of 0.7, A 0 is placed above the pivot in the second column by multiplying the second row by -, adding it to the first row, and putting the result in the first row: : 0 : Based on this reduced row echelon form, E0 and E8 means that mixing 6 gallons of 0% ethanol with 4 gallons of 8% ethanol will yield 0 gallons of 0% ethanol.. This Example 0 Mixing Three Types of Ethanol In some Midwestern states, three different blends of ethanol are available. E0 contains 0% ethanol, E0 contains 0% ethanol, and E8 contains 8% ethanol. How much of each of the blends must be mixed to make 0 gallons of 0% ethanol? If the price per gallon for each type 8

29 is given in the table below, what is the least that the 0 gallon blend would cost? Blend Cost per Gallon E0 $.7 E0 $.6 E8 $.4 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 E0: Total amount of mixture: E0 E0 E8 0 Total amount of ethanol: 0.0E0 0.0E0 0.8E8.0 0 where E0 is the number of gallons of 0% ethanol, E0 is the number of gallons of 0% ethanol, and E8 is the number of gallons of 8% 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: 0.0 : :

30 To put a pivot in the second column, multiply the second row by the reciprocal of 0.0: To put a 0 above the pivot, : 0.7 : Writing this matrix as an a system of equations leads to E0.7E8 E0.7E8 E8 appears in both equations so we ll solve for the other variables in terms of E8, E0.7E8 E0.7E8 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 E8, we can insure that all the variables are nonnegative. For instance, if we set E8 0 we get E0.7 0 E

31 This solution makes sense since mixing equal amounts of E0 and E0 should yield a mixture with a percent ethanol midway between 0% and 0%. Notice that as we increase the amount of E8, the amount of E0 increases (the E8 term is added in the E0 equation) and the amount of E0 decreases (the E8 term is subtracted in the E0 equation). Eventually the amount of E0 will equal 0 as E8 is increased. This occurs when 0.7E8.7E8 E8 or 4.7 easonable values for E8 are from 0 to 4 including 0 and 4 since this leads to non-negative values for E0 and E0 as well. We can calculate the total cost of any combination using cost per gallon of each type of ethanol. For instance, gallons of E0 and gallons of E0 would cost total cost Here is a table of some of the possible solutions based on E0.7E8 E0.7E8 and their corresponding total costs:

32 E8 (gallons) E0 (gallons) E0 (gallons) Approximate Total Cost (dollars) As the amount of E8 is increased, the total cost drops. Each increase of 0. gallons of E8 leads to a drop in the total cost of about 0.0 dollars. The lowest cost appears to come from mixing 4 gallons of E8 and 6 gallons of E0. Using any more E8 would require us to use a negative amount of E0. Even though this would yield a lower cost, it is not a reasonable value for this problem.