Category Archives: Chapter 2

How Do You Write Out The Solutions To A Dependent System of Linear Equations?

Dependent systems of equations are systems of equations that have many solutions. Typically when you solve a dependent system with substitution or elimination, you get 0 = 0. This indicates that many combinations of x and y will solve the system of equations…but how do you find those ordered pairs?

In the MathFAQ below, I demonstrate two strategies for writing out the solutions to dependent systems of linear equations.

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Section 2.4 Question 5

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,

2_4_5_01

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

2_4_5_02

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:

2_4_5_03

We can place a pivot in the second row and column by multiplying the row by the reciprocal of 0.75,

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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:

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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?

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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:

2_4_5_07

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

2_4_5_08

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:

2_4_5_09

To put a pivot in the second column, multiply the second row by the reciprocal of 0.20:

2_4_5_10

To put a 0 above the pivot,

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Writing this matrix as an a system of equations leads to

2_4_5_12

E85 appears in both equations so we’ll solve for the other variables in terms of E85,

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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

2_4_5_14

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

2_4_5_15

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

2_4_5_16

Here is a table of some of the possible solutions based on

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and their corresponding total costs:

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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.

 

Section 2.4 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 3 to solve the problem.

Example 6    Solve a System with More Variables Than Equations

Solve the system of equations

2_4_4_01

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 x1, the second column to x2, and the third column to x3, we can write the augmented matrix for this system as

2_4_4_02

To put this matrix into reduced row echelon form, we must transform it using row operations to

2_4_4_03

Since there is no third row, we have no hope of solving for a solution where x1, x2, and x3 are each a unique value. Instead, we’ll be able to solve for x1 and x2 in terms of x3.

The original augmented matrix already has a 1 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 -2, add it to the second row, and put the result in place of the second row:

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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 -1, add it to the first row, and place the result in the first row:

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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

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Notice that each of these equations contains x3 and it is easy to solve for x1 in the first equation and x2 in the second equation. If we solve for x1 and x2 we get

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Although we don’t have specific numbers for each variable, we do have a recipe for finding values. If we choose any value for x3, we can find the corresponding values for x1 and x2. For instance, if we choose x3 = 100, then

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If x3 = 0, then

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Since x3 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 x3 is chosen, the equations above must be used to calculate corresponding values for x1 and x2. This can be summarized by writing

2_4_4_10

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.


 

Example 7    Solve a System with More Equations Than Variables

Solve the system of equations

2_4_4_11

by transforming its augmented matrix to reduced row echelon form.

Solution The augmented matrix for this system is

2_4_4_12

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 1 so the pivot is already in place. To put zeros below the pivot,

2_4_4_13

In the second column and row, we need to transform the 12 to a 1 by multiplying the row by 1/12:

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Now place zeros in the rest of the column,

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If we convert this matrix back to a system of equations, we find that x1 = 4 and x2 = 1. 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, 1) into each equation:

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Since (x1, x2) = (4, 1) 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

2_4_4_17

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 and the second column to . The augmented matrix for this system is

2_4_4_18

To create a pivot in the first row and column, we have two possibilities. We could multiply the first row by 1/10  or interchange the first and third row. Since interchanging rows does not introduce and fractions into the matrix, we’ll do that to yield

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Now we’ll use row operations to fill the rest of the first column with zeros:

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To create a pivot in the second row and column, multiply the second row by 1/2:

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Now use row operations to place zeros above and below the pivot in the second column:

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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

2_4_4_23

The left side is simply 0. Since 0 cannot equal 27, 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 5. The system may be a system with infinitely many solutions like Example 6 or have no solutions like the inconsistent system in Example 8.