Coburn J.W., Herdlick J.D. College Algebra: Graphs and Models

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With , we ﬁrst substitute 1 for z in the second equation.

substitute 1 for

multiply

subtract 10

divide by ⫺3

We then substitute 1 for z and 1 for y in the ﬁrst equation.

substitute 1 for

and 1 for

simplify

subtract 3

The solution is (⫺7, 1, 1).

Now try Exercises 13 through 18

䊳

C. Solving a System Using Matrices

When a system of equations is written in augmented matrix form, we can solve the sys-

tem by applying the same operations to each row of the matrix, that would be applied

to the equations in the system. In this context, the operations are referred to as elemen-

tary row operations.

Elementary Row Operations

1. Any two rows in a matrix can be interchanged.

2. The elements of any row can be multiplied by a nonzero constant.

3. Any two rows can be added together, and the sum used to replace

one of the rows.

In this section, we’ll use these operations to triangularize the augmented matrix,

employing a solution method known as Gaussian elimination.A square matrix is said

to be in triangular form when all of the entries below the diagonal are zero. For example,

the matrix is in triangular form:

In system form we have meaning a matrix written in triangular

form can be used to solve the system using back-substitution. We’ll illustrate by

solving , using elimination to the left, and row operations on the

augmented matrix to the right. As before, R1 represents the ﬁrst equation in the system

and the ﬁrst row of the matrix, R2 represents equation 2 and row 2, and so on. The cal-

culations involved are shown for the ﬁrst stage only and are designed to offer a careful

comparison. In actual practice, the format shown in Example 4 is used.

•

1x ⫹ 4y ⫺ 1z ⫽ 4

2x ⫹ 5y ⫹ 8z ⫽ 15

1x ⫹ 3y ⫺ 3z ⫽ 1

•

x ⫹ 4y ⫺ z ⫽⫺10

⫺ 3y ⫹ 10z ⫽ 7

z ⫽ 1

14⫺1 ⫺10

0 ⫺310 7

00 1 1

§.£

14⫺1 ⫺10

0 ⫺310 7

00 1 1

x ⫽⫺7

x ⫹ 3 ⫽ 10

x ⫹ 4112⫺ 1 ⫽ 10

y ⫽ 1

⫺3y ⫽⫺3

⫺3y ⫹ 10 ⫽ 7

⫺3y ⫹ 10112⫽ 7

z ⫽ 1

•

1x ⫹ 4y ⫺ 1z ⫽⫺10

⫺ 3y ⫹ 10z ⫽ 7

1z ⫽ 1

14⫺1 ⫺10

0 ⫺310 7

00 1 1

640 CHAPTER 7 Matrices and Matrix Applications 7–4

College Algebra G&M—

B. You’ve just seen how

we can form the augmented

matrix of a system of

equations

cob19545_ch07_637-650.qxd 12/1/10 8:01 PM Page 640

WORTHY OF NOTE

The procedure outlined for solving

systems using matrices is virtually

identical to that for solving systems

by elimination. Using a 3  3

system for illustration, the “zeroes

below the ﬁrst diagonal entry”

indicates we’ve eliminated the

x-term from R2 and R3, the “zeroes

below the second entry” indicates

we’ve eliminated the y-term from

the subsystem, and the division

“to obtain a ‘1’ in the ﬁnal entry”

indicates we have just solved for z.

Elimination Row Operations

(System of Equations) (Augmented Matrix)

To eliminate the x-term in R2, we use Identical operations are

performed on the matrix, which begins the process of triangularizing the matrix.

System Form Matrix Form

For R3, the operations are

As always, we should look for opportunities to simplify any equation in the system

(and any row in the matrix). Note that 1R3 will make the coefﬁcients and related

matrix entries positive. Here is the new system and matrix.

New System New Matrix

On the left, we would ﬁnish by solving the subsystem using

. In matrix form, we eliminate the corresponding entry (third row,

second column) to triangularize the matrix.

Dividing R3 by 16 gives z  1 in the system, and entries of 0011 in the aug-

mented matrix. Completing the solution by back-substitution in the system gives the

ordered triple (1, 1, 1). For practice using these row operations, see Exercises 19

through 27.

The general solution process is summarized here.

Solving Systems by Triangularizing the Augmented Matrix

1. Write the system as an augmented matrix.

2. Use row operations to obtain zeroes below the ﬁrst diagonal entry.

3. Use row operations to obtain zeroes below the second diagonal entry.

4. Continue until the matrix is triangularized (entries below diagonal are zero).

5. Divide to obtain a “1” in the last diagonal entry (if it is nonzero), then convert

to equation form and solve using back-substitution.

Note: At each stage, look for opportunities to simplify row entries using multiplica-

tion or division. Also, to begin the process any equation with an x-coefﬁcient

of 1 can be made R1 by interchanging the equations.

1414

0 310 7

0 0 16 16

§ £

1414

0 3107

01 23

R2  3R3 S R3

2  2

1414

0 3107

01 23

§•

1x  4y  1z  4

3y  10z  7

1y  2z  3

1 4 1 4

1 3 3 1

0 1 2 3

1R1



New R3

1x  4y  1z 4

1x  3y  3z  1

 1y  2z 3

1R1



New R3

1R1  R3 S R3.

2 8 2 8

2 5 8 15

0 3 10 7

2R1



New R2

2x  8y  2z 8

2x  5y  8z  15

 3y  10z  7

2R1



New R2

2R1  R2 S R2.

1414

25815

1331

§•

1x  4y  1z  4

2x  5y  8z  15

1x  3y  3z  1

7–5 Section 7.1 Solving Linear Systems Using Matrices and Row Operations 641

College Algebra G&M—

R2  3R3 R3

¬¬¬¬¡

cob19545_ch07_637-650.qxd 10/21/10 8:40 AM Page 641

EXAMPLE 4 䊳 Solving Systems Using the Augmented Matrix

Solve by triangularizing the augmented matrix:

Solution 䊳 matrix form S R1 4 R2

2R1  R2 S R2 1R2 S R2

2R2  R3 S R3

Converting the augmented matrix back into equation form yields Back-substituting 1 for z

in the second equation gives , with showing . Back-substituting 1

for z and 1 for y in the ﬁrst equation gives , or . Subtracting 2 from

both sides shows . The solution is .

Now try Exercises 28 through 32

䊳

As mentioned, the process used in Example 4 is called Gaussian elimination

(Carl Friedrich Gauss, 1777–1855), with the last matrix written in row-echelon form.

It’s also possible to solve a system entirely using only the augmented matrix, by contin-

uing to use row operations until the diagonal entries are 1’s, with 0’s for all other entries

in the coefﬁcient matrix: . The process is then called Gauss-Jordan

elimination (Wilhelm Jordan, 1842–1899), with the ﬁnal matrix written in reduced

row-echelon form (see Appendix IV).

Note that with Gauss-Jordan elimination, our initial focus is less on getting 1’s

along the diagonal, and more on obtaining zeroes for all entries other than the diagonal

entries. This will enable us to work with integer values in the solution process, as

shown in Example 5.

EXAMPLE 5

䊳 Solving a System Using Gauss-Jordan Elimination

Solve using Gauss-Jordan elimination

standard form

Solution 䊳 matrix form S R1  R2 S R2

10 0 13 43

056 16

001 1

S R3

10 0 13 43

056 16

0 0 29 29

2R2  5R1 S R1

4R2  5R3 S R3

2 25 15

056 16

04 1 7

2 25 15

056 16

04 1 7

§£

2 2515

231 1

04 17

§•

2x  2y  5z  15

2x  3y  1z 1

0x  4y  1z 7

•

2x  5z  15  2y

2x  3y 1  z

4y  z 7

100a

010b

001c

13, 1, 12x 3

x  2 1x  1  1 1

y  1y  4  5y  4112 5

z  1.

1111

0145

0011

S R3

1111

0145

0077

§£

11 11

0145

0 2 1 3

11 11

0145

0 2 1 3

§£

11 11

0 1 4 5

0 2 1 3

§£

11 11

212 7

0 2 1 3

11 11

212 7

0 2 1 3

§£

212 7

11 11

0 2 1 3

§•

2x 

y  2z 7

x 

y  z 1

2y  z 3

•

2x 

y  2z 7

x 

y  z 1

2y  z 3

642 CHAPTER 7 Matrices and Matrix Applications 7–6

College Algebra G&M—

cob19545_ch07_637-650.qxd 10/21/10 8:40 AM Page 642

The ﬁnal matrix shows the solution is or .

Now try Exercises 33 through 36

䊳

D. Solving Systems of Equations Using Technology

Graphing calculators offer a very efﬁcient way to solve systems of equations using

matrices. Once the system has been written in matrix form, it can easily be entered into

a calculator and solved by having the calculator instantly perform the row operations

needed to produce the diagonal of 1’s (with zeroes for all other coefﬁcient entries). On

many calculators, pressing (MATRIX) gives a screen similar to the one

shown in Figure 7.1, where we begin by selecting the EDIT option (push the right ar-

row twice). Pressing places you on a screen where you can EDIT matrix A, chang-

ing the size as needed, then input the entries of the matrix. To use the 3 ⫻ 4 matrix

from Example 4, we press 3 and then 4 , giving the screen shown in Figure 7.2.

The dash marks to the right indicate that there is a fourth column that cannot be seen,

but that comes into view as you enter the elements of the matrix. Begin entering the

ﬁrst row of the matrix, which is . Press after each entry and the cur-

sor automatically goes to the next position in the matrix (note the calculator automati-

cally shifts left and right to allow all four columns to be entered). After entering the

second row and the third row the completed matrix

should look like the one shown in Figure 7.3 (the matrix is currently shifted to the

right, showing the fourth column).

50, ⫺2, ⫺1, ⫺3651, 1, 1, ⫺16

ENTER

52, 1, ⫺2, ⫺76

ENTERENTER

ENTER

-1

2nd

13, ⫺2, 12x ⫽ 3, y ⫽⫺2, and z ⫽ 1,

100 3

010⫺2

001 1

§£

1000 30

050⫺10

001 1

⫺13R3 ⫹ R1 S R1

6R3 ⫹ R2 S R2

10 0 13 43

05⫺6 ⫺16

001 1

7–7 Section 7.1 Solving Linear Systems Using Matrices and Row Operations 643

College Algebra G&M—

S R2

S R1

Figure 7.1 Figure 7.2 Figure 7.3

C. You’ve just seen how

we can solve a system

of equations using row

operations

Most graphing calculators have an rref function, which is short for reduced row

echelon form. To access this function, press (MATRIX) and select the

MATH option, then scroll upward (or downward) until you get to the B:rref option.

Pressing places this function on the home screen, where we must tell it to perform

the rref operation on matrix [A]. Pressing the once again returns us to the

matrix options, enabling us to select matrix NAMES

(notice that NAMES is automatically highlighted).

With the cursor over 1:[A] 3ⴛ4 we press and

matrix [A] is placed on the home screen as the object

of the rref function. After pressing and the

calculator quickly computes the reduced row echelon

form and displays it on the screen, as in Figure 7.4.

The solution is easily read as and

, just as we found in Example 4.z ⫽ 1

x ⫽⫺3, y ⫽ 1,

ENTER

)

ENTER

-1

2nd

ENTER

-1

2nd

Figure 7.4

cob19545_ch07_637-650.qxd 1/10/11 8:55 PM Page 643

Figure 7.5 Figure 7.6

D. You’ve just seen how

we can solve a system of

equations using technology

EXAMPLE 6

䊳

Solving a System of Equations Using Technology

Solve the following system using a graphing calculator.

Solution

䊳

Entering the system as matrix [A] produces the screen shown in Figure 7.5. Using

the rref([A]) command as shown above produces the screen in Figure 7.6. The

solution is the ordered triple (2, 1, 3).

Check by substituting 2 for x, 1 for y, and 3 for z in the original equations.

•

2x  3y  z 4

x  3y  2z  1

3x  y  4z 5

644 CHAPTER 7 Matrices and Matrix Applications 7–8

College Algebra G&M—

Now try Exercises 37 through 42

䊳

E. Inconsistent and Dependent Systems

Due to the strong link between a linear system and its augmented matrix, inconsistent

and dependent systems can be recognized just as in Sections 6.1 and 6.2. An inconsis-

tent system will yield an inconsistent or contradictory statement such as

meaning all entries in a row of the matrix of coefﬁcients are zero, but the constant is

not. A linearly dependent system will yield an identity statement such as

meaning all entries in one row of the matrix are zero. If the system has coincident

dependence, all rows will be zero except one.

EXAMPLE 7

䊳 Solving a Dependent System

Solve the system using Gauss-Jordan elimination:

Solution

䊳

standard form S matrix form S

Since all entries in the last row are zeroes and it’s the only row of zeroes, we conclude the system

is linearly dependent and equivalent to (a calculator solution is shown in the

ﬁgure). As in Chapter 6, we demonstrate this dependence by writing the (x, y, z) solution in terms of

 2z  1

y  3z  2

1021

0132

0000

1R2  R1 S R1

3R2  R3 S R3

1153

0132

0 396

R1  R2 S R2

2R1  R3 S R3

1153

10 21

2 1 10

1153

10 21

2 1 10

§•

x  y  5z  3

x  0y  2z 1

2x  y  z  0

•

x  y  5z  3

x 

2z 1

2x  y  z  0

•

x  y  5z  3

x 

2z 1

2x  y  z  0

0  0,

0 12,

cob19545_ch07_637-650.qxd 10/21/10 8:40 AM Page 644

a parameter. Solving for y in R2 gives y in terms of z:

Solving for x in R1 gives x in terms of z: . As written, the

solutions all depend on z: and

Selecting p as the parameter (or some other “neutral” variable), we

write the solution as Two of the inﬁnite number

of solutions would be (1, 2, 0) for and for

Test these triples in the original equations.

Now try Exercises 43 through 52

䊳

Since there was only one row of zeroes in the system from Example 7, we knew the

system was linearly dependent. As mentioned previously, for coincident dependence

the system will have only one nonzero row, with all other rows consisting entirely of

zeroes. Note that if we change the constant in the third equation of the original system

(the entry for row 3, column 4 was changed to a 5), an inconsistent system results as

one equation is no longer dependent on the others (see Figures 7.7 and 7.8).

p ⫽⫺1.

1⫺1, ⫺1, ⫺12p ⫽ 0,

12p ⫹ 1, 3p ⫹ 2, p2.

z ⫽ z.x ⫽ 2z ⫹ 1, y ⫽ 3z ⫹ 2,

x ⫽ 2z ⫹ 1

y ⫽ 3z ⫹ 2.

7–9 Section 7.1 Solving Linear Systems Using Matrices and Row Operations 645

College Algebra G&M—

Figure 7.7 Figure 7.8

E. You’ve just seen how

we can recognize inconsistent

and dependent systems

When the number of variables and equations in a system increases, solutions using

technology can be particularly effective. In addition, some applications generate sys-

tems where the number of equations and number of variables are unequal, which the

technology still handles efﬁciently.

EXAMPLE 8

䊳

Solving a System of Equations Using Technology

Solve the following system using a graphing calculator.

Solution

䊳

After entering the system as matrix [A], using the

rref([A]) command produces the screen shown in

the ﬁgure. Note that in this case, while the solution

contains a row of zeroes, the system is not linearly

dependent as there are more equations than

variables, and unique values have been found for x,

y, and z. The solution is (⫺1, 2, 1).

Now try Exercises 53 through 58

䊳

F. Solving Applications Using Matrices

As in other areas, solving applications using systems relies heavily on the ability to

mathematically model information given verbally or in context. As you work through

the exercises, read each problem carefully. Look for relationships that yield a system

of two equations in two variables, three equations in three variables, and so on.

x ⫹ 3y ⫺ z ⫽ 4

⫺2x ⫹ y ⫹ 3z ⫽ 7

⫺2y ⫹ 4z ⫽ 0

⫺x ⫹ 4y ⫹ 2z ⫽ 11

cob19545_ch07_637-650.qxd 1/8/11 3:39 PM Page 645

F. You’ve just seen how

we can solve applications

using matrix methods

EXAMPLE 9

䊳

Determining the Original Value of Collector’s Items

A museum purchases a famous painting, a ruby tiara, and a rare coin for its collection, spending

a total of $30,000. One year later, the painting has tripled in value, while the tiara and the coin

have doubled in value. The items now have a total value of $75,000. Find the purchase price of

each if the original price of the painting was $1000 more than twice the coin.

Solution 䊳 Let P represent the price of the painting, T the tiara, and C the coin.

Total spent was $30,000:

One year later:

Value of painting versus coin:

standard form S matrix form S

⫺1R2 ⫹ R3 S R3

From R3 of the triangularized form, directly. Since

R2 represents we ﬁnd the tiara was purchased

for Substituting these values into the ﬁrst equation

shows the painting was purchased for $15,000. The solution is

(15,000, 8000, 7000). The solution found using a graphing

calculator is also shown.

Now try Exercises 61 through 68

䊳

T ⫽ $8000.

T ⫹ C ⫽ 15,000,

C ⫽ $7000

1 1 1 30000

0 1 1 15000

0 0 1 7000

S R3

1 1 1 30000

0 1 1 15000

0 0 2 14000

§£

1 1 1 30000

0 1 1 15000

0 1 3 29000

1 1 1 30000

0 1 1 15000

0 1 3 29000

⫺1R2 S R2

⫺1R3 S R3

1 1 1 30000

0 ⫺1 ⫺1 ⫺15000

0 ⫺1 ⫺3 ⫺29000

⫺3R1 ⫹ R2 S R2

⫺1R1 ⫹ R3 S R3

1 1 1 30000

3 2 2 75000

10⫺2 1000

1 1 1 30000

3 2 2 75000

10⫺2 1000

§•

1P ⫹ 1T ⫹ 1C ⫽ 30,000

3P ⫹ 2T ⫹ 2C ⫽ 75,000

1P ⫹ 0T ⫺ 2C ⫽ 1000

•

P ⫹ T ⫹ C ⫽ 30,000

3P ⫹ 2T ⫹ 2C ⫽ 75,000

P ⫽ 2C ⫹ 1000

S P ⫽ 2C ⫹ 1000

S 3P ⫹ 2T ⫹ 2C ⫽ 75,000

S P ⫹ T ⫹ C ⫽ 30,000

646 CHAPTER 7 Matrices and Matrix Applications 7–10

College Algebra G&M—

7.1 EXERCISES

2. When the coefﬁcient matrix is used with the matrix

of constants, the result is a(n) matrix.

䊳

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A matrix with the same number of rows and

columns is called a(n) matrix.

3. Matrix is a by

matrix. The entry in the second row and third

column is .

A ⫽ c

24⫺3

1 ⫺21

4. Given matrix B shown

here, the diagonal entries

are , , and .

B ⫽ £

143

⫺152

3 ⫺21

5. The notation indicates that an

equivalent matrix is formed by performing what

operations/replacements?

⫺2R1 ⫹ R2 S R2

6. Describe how to tell an inconsistent system apart

from a dependent system when solving using

matrix methods (row reduction).

cob19545_ch07_637-650.qxd 10/22/10 12:25 AM Page 646

College Algebra G&M—

䊳

DEVELOPING YOUR SKILLS

Determine the size of each matrix and identify the third

row and second column entry. If the matrix given is a

square matrix, identify the diagonal entries.

7. 8.

Form the augmented matrix, then name the diagonal

entries of the coefﬁcient matrix.

10.

11.

12.

Write the system of equations for each matrix. Then use

back-substitution to ﬁnd its solution.

13.

14.

15.

16.

17.

18. £

1213

00 1

13429

01

00 1 3

10 7 5

01515

00 1 26

1210

0122

0013

1 5 15

0 1 2

145

•

2x  3y  z  5

2y  z  7

x  y  2z  5

•

x  2y  z  1

x 

z  3

2x  y  z  3

•

2x  3y  2z  7

x  y  2z 5

3x  2y  z  11

≥

10 4

137

5 12

2 39

10 4

137

5 12

§£

2.1 1

3 5.8

Perform the indicated row operation(s) in the order

given and write the new matrix.

19.

20.

21.

22.

23.

24.

What row operations would produce zeroes beneath the

ﬁrst entry in the diagonal?

25.

26.

27.

Solve each system by triangularizing the augmented

matrix and using back-substitution. Simplify by

clearing fractions or decimals before beginning.

28.

29.

30. 31.

32. •

2x  3y  2z  7

x  y  2z 5

3x  2y  z  11

•

x  2y  2z  7

2x  2y  z  5

3x  y  z  6



u 

v  1

u 

v  7

0.15g  0.35h 0.5

0.12g  0.25h  0.1

2y  5x  4

5x  2  4y

12010

512 6

4332

114 3

301 5

53 2 3

1302

24 11

3 1 29

211 3

31 1 0

43 2 3

3R1  2R2 S R2,

2R1  R3 S R3

31 1 8

6 1 110

4 2 322

2R1  R2 S R2,

4R1  3R3 S R3

3200

1126

4132

R1 4 R2,

4R1  R3 S R3

2104

5835

1 33 2

R1 4 R3,

5R1  R2 S R2

74 3

4 812

R2 S R2,

R1 4 R2

3 1

52 4

2R1 S R1,

5R1  R2 S R2

7–11 Section 7.1 Solving Linear Systems Using Matrices and Row Operations 647

cob19545_ch07_637-650.qxd 10/21/10 8:40 AM Page 647

Solve using Gauss-Jordan elimination.

33. 34.

35. 36.

Solve each system of equations using a graphing

calculator. Verify each solution on the home screen using

the keys, as in Chapter 6.

37.

38.

39.

40.

41.

42.

Solve each system by triangularizing the augmented

matrix and using back-substitution. If the system is

linearly dependent, give the solution in terms of a

parameter. If the system has coincident dependence,

answer in set notation as in Chapter 6.

43.

44.

45. •

x  3y  5z  20

2x  3y  4z  16

x  2y  3z  12

•

3x  y  z 2

x  2y  3z  1

2x  3y  5z  3

•

4x  8y  8z  24

2x  6y  3z  13

3x  4y  z 11

•

x 

y  z 

x 

y 

z 

x  y 

z 

•

x 

y  z 

x 

y 

z 

x  y 

z 

•

0.1x  2y  0.6z 36.4

3x  0.8y  0.2z 35

0.5x  y  0.4z  25

•

0.2x  0.5y  z  13

x  0.7y  0.4z 9

0.5x  y  0.8z 4.6

•

3x  5y  z  6

4x  7y  2z  24

x  10y  3z 40

•

x  2y  3z  9

2x  y  5z 20

3x  5y  z  14

ALPHA

•

x  y  2z 1

4x  y  3z  3

3x  2y  z  4

•

x  y  2z  2

x  y  z  1

2x  y  z  4

•

2x  3y  z  5

2y  z  7

x  y  2z  5

•

x  2y  z  1

x  z  3

2x  y  z  3

46.

47.

48.

49.

50.

51.

52.

Solve each system of equations using a graphing

calculator. Verify each solution on the home screen using

the keys, as in Chapter 6.

53.

54.

55.

56. μ

2x  3y  4

 2y  5z 19

x  4z  13

x  3y  4z  17

x  3y  z  1

x  2z  7

2y  3z 6

2x  3y  z  8

2x  5y 4

3x  4y  z  0

x  y  3z 8

5x  9y  z  4

2x  3y  z  0

x  2y  z  5

3x  2z  4

x  3y  z  4

ALPHA

•

2x  4y  3z  4

5x  6y  7z 12

x  2y  z 4

•

x  2y  z  4

3x  4y  z  4

6x  8y  2z  8

•

x  2y  3z  2

3x  4y  z  6

4x  2y  2z  7

•

2x  y  3z  1

2y  6z  2

x 

y 

z  5

•

2x  y  3z  1

4x  2y  6z  2

10x  5y  15z  5

•

3x  4y  2z 2

x  2y  z 1

6x  8y  4z  4

•

x  2y  3z 6

x  y  2z 4

3x  6y  9z  18

648 CHAPTER 7 Matrices and Matrix Applications 7–12

College Algebra G&M—

cob19545_ch07_637-650.qxd 10/21/10 8:40 AM Page 648

57. μ

x ⫹ 2y ⫺ 3z ⫹ w ⫽⫺13

⫺2x ⫹ 3y ⫹ z ⫺ 2w ⫽⫺3

⫺x ⫺ y ⫹ 2z ⫹ 3w ⫽ 4

3x ⫺ 2y ⫺ z ⫺ w ⫽ 5

58. μ

2x ⫹ 3y ⫺ 4z ⫹ w ⫽⫺10

x ⫺ 2y ⫺ 3w ⫽ 6

3x ⫹ 5z ⫹ 2w ⫽ 1

⫺x ⫹ y ⫹ 2z ⫺ 5w ⫽ 21

7–13 Section 7.1 Solving Linear Systems Using Matrices and Row Operations 649

College Algebra G&M—

䊳

WORKING WITH FORMULAS

Area of a triangle in the plane: (x

 x

 x

 x

 x

 x

)

The area of a triangle in the plane is given by the formula shown, where the vertices of the triangle are located at the

points (x

, y

), (x

, y

), and (x

, y

), and the sign is chosen to ensure a positive value.

A 

䊳

APPLICATIONS

Model each problem using a system of linear equations. Then solve using the augmented matrix.

Descriptive Translation

61. The distance (via air travel) from Los Angeles (LA), California, to

Saint Louis (STL), Missouri, to Cincinnati (CIN), Ohio, to New York

City (NYC), New York, is approximately 2480 mi. Find the length of

each ﬂight if the distance from LA to STL is 50 mi more than ﬁve

times the distance between STL and CIN, and 110 mi less than three

times the distance from CIN to NYC.

62. In the 2006 NBA Championship Series, Dwayne Wade of the Miami

Heat carried his team to the title after the ﬁrst two games were lost to

the Dallas Mavericks. If 187 points were scored in the title game and the

Heat won by 3 points, what was the ﬁnal score?

63. Moe is lecturing Larry and Curly once again (Moe, Larry, and Curly of The Three Stooges fame) claiming he is

twice as smart as Larry and three times as smart as Curly. If he is correct and the sum of their IQs is 165, what is

the IQ of each stooge?

New York City

St. Louis

Cincinnati

Los Angeles

59. Find the area of a triangle whose vertices are

(5, 2), and (1, 8).1⫺1, ⫺32,

60. Find the area of a triangle whose vertices are

, and 1⫺1, 72.16, ⫺22, 1⫺5, 42

64. A collector of rare books buys a handwritten,

autographed copy of Edgar Allan Poe’s Annabel

Lee, an original advance copy of L. Frank Baum’s

The Wonderful Wizard of Oz, and a ﬁrst print copy

of The Caine Mutiny by Herman Wouk, paying a

total of $100,000. Find the cost of each one, given

that the cost of Annabel Lee and twice the cost of

The Caine Mutiny sum to the price paid for The

Wonderful Wizard of Oz, and The Caine Mutiny

cost twice as much as Annabel Lee.

Geometry

65. A right triangle has a hypotenuse of 39 m. If the

perimeter is 90 m, and the longer leg is 6 m longer

than twice the shorter leg, ﬁnd the dimensions of

the triangle.

66. In triangle ABC, the sum of angles A and C is equal

to three times angle B. Angle C is 10 degrees more

than twice angle B. Find the measure of each angle.

Investment and Finance

67. Suppose $10,000 is invested in three different

investment vehicles paying 5%, 7%, and 9%

annual interest. Find the amount invested at each

rate if the interest earned after 1 yr is $760 and the

amount invested at 9% is equal to the sum of the

amounts invested at 5% and 7%.

68. The trustee of a union’s pension fund has

invested funds in three ways: a savings fund

paying 4% annual interest, a money market fund

paying 7%, and government bonds paying 8%.

Find the amount invested in each if the interest

earned after one year is $0.178 million and the

amount in government bonds is $0.3 million

more than twice the amount in money market

funds. The total amount invested is $2.5 million

dollars.

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