6 Matrix Equations: Solving a Linear System
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620
CHAPTER 7
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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
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CHAPTER 7
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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
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
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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
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Systems of Equations and Data in Categories
1
5. A = £ 1
-1
3
6. A = £ 1
2
■
7–18
2
1
1
-1
0§
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
3Đ
5
1
15. Ê 4
1
2
5
-1
3
-1Đ
- 10
2
16. Ê 1
2
1
1
1
0
4Đ
2
0
17. Ê 3
1
-2
1
-2
2
3Đ
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§
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
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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
0§
1
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CHAPTER 7
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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.)