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6 Matrix Equations: Solving a Linear System

6 Matrix Equations: Solving a Linear System

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620



CHAPTER 7







Systems of Equations and Data in Categories



If A and B are n * n matrices, and if AB = BA = In, then we say that B is the

inverse of A, and we write B = A -1. The concept of the inverse of a matrix is analogous to that of the reciprocal of a real number.



Inverse of a Matrix

Let A be a square n * n matrix. If there exists an n * n matrix A -1 with the

property that

AA -1 = A -1A = In

then we say that A -1 is the inverse of A.



e x a m p l e 1 Verifying That a Matrix Is an Inverse

Verify that matrix B is the inverse of matrix A.

A = c



2

5



3

d

1



     



B = c



and



3 -1

d

-5

2



Solution

We perform the matrix multiplications to show that AB = I and BA = I.

c



2

5



c



3

-5



1

3

dc

3 -5

-1 2

dc

2 5



-1

2 # 3 + 11- 52

d = c #

2

5 3 + 31- 52



21- 12 + 1 # 2

1

d = c

#

51- 12 + 3 2

0



0

d

1



1

3 # 2 + 1- 125

d = c

3

1- 522 + 2 # 5



3 # 1 + 1- 123

1

d = c

#

1- 521 + 2 3

0



0

d

1







NOW TRY EXERCISE 3







The inverse of a matrix can be found by using a process that involves elementary row operations. This process is programmed into graphing calculators. On

TI-83 calculators, matrices are stored in memory by using names such as [A], [B],

[C], and so on. To find the inverse of [A], we key in

[A]



XϪ1



ENTER



e x a m p l e 2 Finding the Inverse of a Matrix

Find the inverse of matrix A.

1

A= £ 2

-3



-2

-3

6



-4

-6§

15



Solution

A graphing calculator gives the following output for the inverse of matrix A.



SECTION 7.6



Matrix Equations: Solving a Linear System







Calculator output



621



Inverse of matrix A



[A] -1 Frac

[[-3 2 0

]

[-4 1 -2/3]

[1

0 1/3 ]]



A



-1



-3

= £-4

1



2

1

0



0

- 23 §

1

3



We have used the ᭤Frac command to display the output in fraction form rather than

in decimal form.





2







NOW TRY EXERCISES 7 AND 13



■ Matrix Equations

A system of linear equations can be written as a single matrix equation. For example, the system

x - 2y - 4z = 7

c 2x - 3y - 6z = 5

- 3x + 6y + 15z = 0

is equivalent to the matrix equation

1

£ 2

-3



-4 x

7

-6§ £y§ = £5§

15 z

0



-2

-3

6

A



X



B



If we let

1

A= £ 2

-3



-2

-3

6



-4

-6§

15



     

x

X = £y§

z



7

B = £5§

0



then this matrix equation can be written as

AX = B

Solving the matrix equation AX = B

is very similar to solving the simple

real-number equation

3x = 12

which we do by multiplying each

side by the reciprocal (or inverse)

of 3:

1

3 13x 2



= 13 1122



x=4



Matrix A is called the coefficient matrix.

We solve this matrix equation by multiplying each side by the inverse of A (provided that this inverse exists).

AX = B



Given equation



A -1 1AX2 = A -1B

1A -1A2X = A-1B



I3 X = A -1B

X =A B

-1



Multiply on left by A -1

Associative Property

Property of inverses

Property of the identity matrix



622



CHAPTER 7







Systems of Equations and Data in Categories



We have proved that the matrix equation AX = B can be solved by the following

method.



Solving a Matrix Equation

If A is a square n * n matrix that has an inverse A -1, and if X is a variable

matrix and B a known matrix, both with n rows and one column, then the

solution of the matrix equation

AX = B

is given by

X = A -1B



e x a m p l e 3 Solving a System as a Matrix Equation

Solve the system of equations.

x - 2y - 4z = 7

c 2x - 3y - 6z = 5

- 3x + 6y + 15z = 0



Solution

The system is equivalent to the matrix equation AX = B described on page 621.

Using a calculator, we find that

A



-3

= £-4

1



-1



2

1

0



0



- 23 §

1

3



So because X = A -1B, we have

x

-3

£y§ = £-4

1

z

X



2

1

0



0



- 23 §

1

3



7

- 11

£ 5 § = £ - 23 §

0

7



A -1



=



B



So the solution of the system is 1- 11, - 23, 72 .





NOW TRY EXERCISES 19 AND 23



e x a m p l e 4 Solving a System Using a Matrix Inverse

A system of equations is given.

(a) Write the system of equations as a matrix equation.

(b) Solve the system by solving the matrix equation.

e



2x - 5y = 15

3x - 6y = 36







SECTION 7.6







Matrix Equations: Solving a Linear System



623



Solution

(a) We write the system as a matrix equation of the form AX = B.

c



-5 x

15

dc d = c d

-6 y

36



2

3



A



X =



B



(b) Using a graphing calculator to calculate the inverse, we get

A -1 = c



- 5 -1

-2

d = c

-6

-1



2

3



5

3

2d

3



So because X = A -1B, we have

x

-2

c d = c

y

-1

X



=



5

3

2d

3



c



A -1



15

30

d = c d

36

9

B



So the solution of the system is (30, 9).





NOW TRY EXERCISE 19







Not every square matrix has an inverse. A matrix that has no inverse is called

singular.



e x a m p l e 5 A Matrix That Does Not Have an Inverse

Find the inverse of the matrix.

2

£1

1



-3

2

1



-7



4



Solution

If we try to calculate the inverse of this matrix on the TI-83 calculator, we get the error message shown below. That means that this matrix has no inverse.

Calculator output

ERR:SINGULAR MAT

1:Quit

2:Goto







Inverse of matrix A

The matrix A does

not have an inverse.



NOW TRY EXERCISE 15







624



2



CHAPTER 7







Systems of Equations and Data in Categories



■ Modeling with Matrix Equations

Suppose we need to solve several systems of equations with the same coefficient matrix. Then expressing the systems as matrix equations provides an efficient way to

obtain the solutions, because we need to find the inverse of the coefficient matrix

only once. This procedure is particularly convenient if we use a graphing calculator

to perform the matrix operations, as in the next example.



e x a m p l e 6 Solving a System Using a Matrix Inverse

A pet store owner feeds his hamsters and gerbils different mixtures of three types of

rodent food: KayDee Food, Pet Pellets, and Rodent Chow. The protein, fat, and carbohydrates content (in mg) in one gram of each brand is given in the table below.









Hamsters need 340 mg of protein, 280 mg of fat, and 440 mg of carbohydrates each day.

Gerbils need 480 mg of protein, 360 mg of fat, and 680 mg of carbohydrates

each day.



The pet store owner wishes to feed his animals the correct amount of each brand to

satisfy their daily requirements exactly. How many grams of each food should the

storekeeper feed his hamsters and gerbils daily to satisfy their nutrient requirements?



KayDee Food



Pet Pellets



Rodent Chow



10

10

5



0

20

10



20

10

30



Protein (mg)

Fat (mg)

Carbohydrates (mg)



Solution

We let x, y, and z be the respective amounts (in grams) of KayDee Food, Pet Pellets,

and Rodent Chow that the hamsters should eat and r, s, and t be the corresponding

amounts for the gerbils. Then we want to solve the matrix equations

10

£ 10

5



0

20

10



20 x

340

10 § £ y § = £ 280 §

30 z

440



Hamster Equation



10

£ 10

5



0

20

10



20 r

480

10 § £ s § = £ 360 §

30 t

680



Gerbil Equation



Let

10

A = £ 10

5



0

20

10



20

10 §

30



  



340

B = £ 280 §

440



  



480

C = £ 360 §

680



     



Then we can write these matrix equations as

AX = B



Hamster Equation



AY = C



Gerbil Equation



x

X = £y§

z



r

Y = £s§

t



SECTION 7.6







Matrix Equations: Solving a Linear System



625



Solving for X and Y, we have

X = A -1B



Hamster Equation



Y = A -1C



Gerbil Equation



Using a graphing calculator, we find matrices X and Y.

[A]-1*[B]



[A]-1*[C]



[[10]

[3 ]

[12]]



[[8 ]

[4 ]

[20]]



Thus each hamster should be fed 10 g of KayDee Food, 3 g of Pet Pellets, and 12 g

of Rodent Chow, and each gerbil should be fed 8 g of KayDee Food, 4 g of Pet

Pellets, and 20 g of Rodent Chow daily.









NOW TRY EXERCISE 27



7.6 Exercises

CONCEPTS



Fundamentals

1. (a) The matrix I = c



1 0

d is called an _______ matrix.

0 1



(b) If A is a 2 * 2 matrix, then A * I = _______ and I * A = _______.

(c) If A and B are 2 * 2 matrices with AB = I, then B is the _______ of A.

2. (a) Write the following system as a matrix equation AX = B.



System



Matrix equation

A



e



c



5x + 3y = 4

3x + 2y = 3



(b) The inverse of A is A -1 = c



#



X



dc



=

d = c



B

d



d



(c) The solution of the matrix equation is X = A -1B.

X =

x

c d = c

y



A -1



#

dc



B

d = c



d



(d) The solution of the system is x = ______, y = ______.



SKILLS



3–6







Calculate the products AB and BA to verify that matrix B is the inverse of matrix A.



  

  



3. A = c



4

7



1

d

2



4. A = c



2

4



-3

d

-7



B= c



2

-7



-1

d

4



7



- 32

d

-1



B = c2

2



626



CHAPTER 7







Systems of Equations and Data in Categories



1

5. A = £ 1

-1

3

6. A = £ 1

2





7–18



2

1

1



  

  



-1



2



4

-6§

12



8

B = £-2

1



9

B = £ - 12

- 12



-3

1

0

- 10

14

1

2



4

-1§

1

-8

11 §

1

2



Use a calculator that can perform matrix operations to find the inverse of the

matrix, if it exists.



7. c



-3

2



-5

d

3



8. c



9. c



2

-5



5

d

- 13



10. c



11. c



6

-8



-3

d

4



12. c 2

5



4



1

-1§

0



5

14. £ 3

6



7

-1

7



2

13. £ - 1

1



4

1

4



3

7



4

d

9

4

d

-5



-7

8

1



1

3



d

4



5



1

15. Ê 4

1



2

5

-1



3

-1Đ

- 10



2

16. Ê 1

2



1

1

1



0



2



0

17. Ê 3

1



-2

1

-2



2



3



1

0

18.

0

1



2

1

1

2



0

1

0

0



1926



CONTEXTS



3

4

-3







3

1

Ơ

1

2



Solve the system of linear equations by expressing the system as a matrix

equation and using the inverse of the coefficient matrix. Use the inverses you

found in Exercises 7–10, and 13, 14, 17, and 18.



19. e



- 3x - 5y = 4

2x + 3y = 0



20. e



3x + 4y = 10

7x + 9y = 20



21. e



2x + 5y = 2

- 5x - 13y = 20



22. e



- 7x + 4y = 0

8x - 5y = 100



2x + 4y + z = 7

23. c- x + y - z = 0

x + 4y

= -2



5x + 7y + 4z = 1

24. c3x - y + 3z = 1

6x + 7y + 5z = 1



- 2y + 2z = 12

25. c3x + y + 3z = - 2

x - 2y + 3z = 8



x + 2y

ϩ 3w

y+z + w

26. μ

y

+ w

x + 2y

+ 2w



=

=

=

=



0

1

2

3



27. Sales Commissions An encyclopedia salesperson works for a company that offers

three different grades of bindings for its encyclopedias: standard, deluxe, and leather.

For each set that she sells, she earns a commission based on the set’s binding grade.

One week she sells one standard set, one deluxe set, and two leather sets and makes

$675 in commission. The next week she sells two standard sets, one deluxe set, and one

leather set for a $600 commission. The third week she sells one standard set, two deluxe

sets, and one leather set, earning $625 in commission.



CHAPTER 7







Review



627



(a) Let x, y, and z represent the commission she earns on standard, deluxe, and leather

sets, respectively. Translate the given information into a system of equations.

(b) Express the system of equations you found in part (a) as a matrix equation of the

form AX = B.

(c) Find the inverse of the coefficient matrix A, and use it to solve the matrix equation

in part (b). How much commission does the salesperson earn on a set of

encyclopedias in each grade of binding?

28. Nutrition A nutritionist is studying the effects of the nutrients folic acid, choline, and

inositol. He has three types of food available, and each type contains the amounts of

these nutrients per ounce shown in the table.



Folic acid (mg)

Choline (mg)

Inositol (mg)



Type A



Type B



Type C



3

4

3



1

2

2



3

4

4



Let A be the matrix

3

A = £4

3



1

2

2



3



4



(a) Find the inverse of matrix A.

(b) How many ounces of each food should the nutritionist feed his laboratory rats if he

wants each rat’s daily diet to contain 10 mg of folic acid, 14 mg of choline, and

13 mg of inositol?

(c) How many ounces of each food should the nutritionist feed his laboratory rats if he

wants each rat’s daily diet to contain 9 mg of folic acid, 12 mg of choline, and

10 mg of inositol?



CHAPTER 7 R E V I E W

CHAPTER 7



CONCEPT CHECK

Make sure you understand each of the ideas and concepts that you learned in this chapter,

as detailed below section by section. If you need to review any of these ideas, reread the

appropriate section, paying special attention to the examples.



7.1 Systems of Linear Equations in Two Variables

A system of equations is a set of equations in which each equation involves the same

variables. A solution of a system is a set of values for the variables that makes each

equation true.

A system of two equations in two variables can be solved by the substitution

method or the elimination method. (The details of these methods are described on

pages 569 and 570.)



628



CHAPTER 7







Systems of Equations and Data in Categories



The graph of a system of two linear equations in two variables is a pair of lines;

the solution of the system corresponds to the point of intersection of the lines. A twovariable linear system has either:











One solution if the lines have different slopes.

No solution if the lines are different but have the same slope. (That is, the

lines are parallel.)

Infinitely many solutions if the lines are the same.



7.2 Systems of Linear Equations in Several Variables

To solve a linear system involving several equations and variables, we use Gaussian

elimination to put the system in triangular form and then use back-substitution

to get the solution. The following elementary row operations are used in the

Gaussian elimination process:









Add a nonzero multiple of one equation to another.

Multiply an equation by a nonzero constant.

Interchange the positions of two equations.



A system of linear equations can have one solution, no solution, or infinitely

many solutions. A system with no solution is said to be inconsistent, and a system

with infinitely many solutions is said to be dependent. If we apply Gaussian elimination to a system and arrive at a false equation, then the system is inconsistent. The

solution of a dependent system can be described by using one or more parameters.



7.3 Using Matrices to Solve Systems of Equations

A matrix is a rectangular array of numbers. For instance, here is a matrix with three

rows and four columns:

6

£-5

9



-2

7

-4



4

1

1

2



0

-3§

8



A matrix with m rows and n columns is said to have dimension m * n; the

above matrix has dimension 3 * 4. The numbers in the matrix are its entries; the

(i, j) entry is the number in row i and column j. So in the above matrix the (2, 3) entry is 1, and the (3, 2) entry is - 4.

The augmented matrix of a system of linear equations is the matrix that is obtained by writing the coefficients of the variables and the constants in each equation

in matrix form, omitting the symbols for the variables and the equal signs. For example, the above matrix is the augmented matrix for the system

6x - 2y + 4z = 0

c- 5x + 7y + z = - 3

9x - 4y + 12 z = 8

To solve a linear system, we apply elementary row operations to the augmented

matrix of the system, transforming it into an equivalent system in row-echelon form.

A matrix is in row-echelon form if it satisfies the following conditions:





The first nonzero number (called the leading entry) is 1.



CHAPTER 7













Review



629



The leading entry in each row is to the right of the leading entry in the row

above it.

All rows that consist entirely of zeros are at the bottom of the matrix.



If the matrix also satisfies the following condition, then it is in reduced rowechelon form:





Every number above and below each leading entry is 0.



Graphing calculators have commands that put matrices into row-echelon and reduced row-echelon form (ref and rref on the TI-83 calculator).

If an augmented matrix in row-echelon form has a row equivalent to the equation 0 = 1, then the system is inconsistent; this means the system has no solutions.

If the augmented matrix in row-echelon form is not inconsistent but has fewer

nonzero rows than variables, then the system is dependent; this means that the system has infinitely many solutions.



7.4 Matrices and Data in Categories

Matrices can be used to organize and analyze data about two different categorical

characteristics of a population. For instance, if you gather data about the eye color

(brown, blue, or green) and the hair color (black, brown, blond, or red) of the members of your class, then the numbers can be conveniently presented as a 3 * 4 matrix in which the rows represent eye color and the columns represent hair color.

(For example, the (3, 4) entry would be the number of green-eyed redheads in the

class.)

To combine analogous data matrices for two different populations, we add the

matrices by adding corresponding entries.



7.5 Matrix Operations: Getting Information From Data

Suppose that A and B are two matrices of the same dimension m * n and that c is a

real number.









The sum A + B is the m * n matrix whose entries are the sums of the

corresponding entries of A and B.

The scalar multiple cA is the m * n matrix obtained by multiplying each

entry in A by the number c.



The product AB of two matrices A and B is defined if the number of columns

of A is the same as the number of rows of B. (The procedure for multiplying two matrices is described on pages 612–615.) Matrix products can be used to extract information from sets of categorical data.



7.6 Matrix Equations: Solving a Linear System

The identity matrix In is the square n * n matrix for which each main diagonal entry (from the top left corner to the bottom right corner) is a 1 and every other entry

is a 0. For instance, the 3 * 3 identity matrix is

1

I3 = £ 0

0



0

1

0



0



1



630



CHAPTER 7







Systems of Equations and Data in Categories



When a matrix is multiplied by an identity matrix I of the appropriate dimension, it remains unchanged:



     



AI = A



IB = B



and



If A is a square n * n matrix and if there exists an n * n matrix A -1 such that

A -1A = AA -1 = In

then we say that A -1 is the inverse of A.

A system of equations can be written as a matrix equation of the form AX = B.

To solve a matrix equation, we multiply both sides by A -1:

X = A -1B



C H A P T E R 7 REVIEW EXERCISES

SKILLS



1–4







Graph each linear system, either by hand or by using a graphing device. Use the

graph to determine whether the system has one solution, no solution, or infinitely

many solutions. If there is exactly one solution, use the graph to find it.



1. e



3x - y = 5

2x + y = 5



2. e



y = 2x + 6

y = -x + 3



3. e



6x - 8y = 15

- 32 x + 2y = - 4



4. e



2x - 7y = 28

y = 27 x - 4



5–8







Solve the linear system, or show that it has no solution. (Use either the elimination

or the substitution method.)



5. e



2x + 3y = 7

x - 2y = 0



6. e



2x + 5y = 9

- x + 3y = 1



7. e



3x - y = 12

- x + 13 y = - 4



8. e



6x + 9y = 5

4x + 6y = 3



9–16







Use Gaussian elimination to find the complete solution of the system, or show that

no solution exists.



x + y + 2z = 6

9. c2x

+ 5z = 12

x + 2y + 3z = 9



x - 2y + 3z = 1

10. c x - 3y - z = 0

2x

- 6z = 6



x + 2y + 2z = 6

11. c x - y

= -1

2x + y + 3z = 7



x- y + z =2

12. cx + y + 3z = 6

2y + 3z = 5



x-y+ z = 2

13. c x + y + 3z = 6

3x - y + 5z = 10



x + 2y - z = 1

14. c2x + 3y - 4z = - 3

3x + 6y - 3z = 4



x - 2y + 3z = - 2

15. c2x - y + z = 2

2x - 7y + 11z = - 9



x- y + z =0

16. c3x + 2y - z = 6

x + 4y - 3z = 3



17–18







The augmented matrix of a linear system is given. Use a graphing calculator to

put the matrix into reduced row-echelon form, and then find the complete solution

of the system. (Assume that the variables in the system are x, y, and z.)



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