Download Matrix Solutions to Linear Systems

P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 805
Matrices and
Determinants
8
You are being drawn deeper into cyberspace, spending more time online each week. With
constantly improving high-resolution images, cyberspace is reshaping your life by nourishing
shared enthusiasms. The people who built your computer talk of bandwidth that will give you the
visual experience, in high-definition 3-D format, of being in the same room with a person who is
actually in another city. Rectangular arrays of numbers, called matrices, play a central role in
representing computer images and in the forthcoming technology of tele-immersion.
The use of rectangular arrays of numbers in the digital representation of images and the
manipulation of images on a computer screen is discussed in Examples 8 and 9 in Section 8.3.
805
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 806
806 Chapter 8 Matrices and Determinants
Section
8.1
Matrix Solutions to Linear Systems
Objectives
� Write the augmented matrix
�
�
�
for a linear system.
Perform matrix row
operations.
Use matrices and Gaussian
elimination to solve systems.
Use matrices and GaussJordan elimination to solve
systems.
he data below show that we spend a lot of time sprucing up.
T
Average Number of Minutes per Day Americans Spend on Grooming
Men
Ages
15–19
Ages
20–24
Ages
45–54
Ages
65ⴙ
Married
Single
37
37
34
28
31
34
59
49
46
46
44
50
B
Women
R
Source: Bureau of Labor Statistics’ American Time-Use Survey
The 12 numbers inside the brackets are arranged in two rows and six columns. This
rectangular array of 12 numbers, arranged in rows and columns and placed in brackets, is an example of a matrix (plural: matrices). The numbers inside the brackets are
called elements of the matrix. Matrices are used to display information and to solve
systems of linear equations. Because systems involving two equations in two
variables can easily be solved by substitution or addition, we will focus on matrix
solutions to systems of linear equations in three or more variables.
�
Write the augmented matrix for a
linear system.
Augmented Matrices
A matrix gives us a shortened way of writing a system of equations. The first step in
solving a system of linear equations using matrices is to write the augmented matrix.
An augmented matrix has a vertical bar separating the columns of the matrix into
two groups. The coefficients of each variable are placed to the left of the vertical line
and the constants are placed to the right. If any variable is missing, its coefficient is 0.
Here are two examples:
System of Linear Equations
3x + y + 2z = 31
c x + y + 2z = 19
x + 3y + 2z = 25
c
x + 2y - 5z = - 19
y + 3z =
9
z =
4
Augmented Matrix
3
C1
1
1
1
3
2 31
2 3 19 S
2 25
1
C0
0
2
1
0
-5 -19
3 3
9 S.
1
4
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 807
Section 8.1 Matrix Solutions to Linear Systems
807
Our goal in solving a system of linear equations in three variables using matrices
is to produce a matrix with 1s down the diagonal from upper left to lower right on the
left side of the vertical bar, called the main diagonal, and 0s below the 1s. In general,
the matrix will be of the form
1
C0
0
a
1
0
b c
d 3 e S,
1 f
where a through f represent real numbers. The third row of this matrix gives us the
value of one variable. The other variables can then be found by back-substitution.
�
Perform matrix row operations.
Matrix Row Operations
A matrix with 1s down the main diagonal and 0s below the 1s is said to be in rowechelon form. How do we produce a matrix in this form? We use row operations on
the augmented matrix. These row operations are just like what you did when solving
a linear system by the addition method. The difference is that we no longer write the
variables, usually represented by x, y, and z.
Matrix Row Operations
The following row operations produce matrices that represent systems with the
same solution set:
1. Two rows of a matrix may be interchanged. This is the same as interchanging
two equations in a linear system.
2. The elements in any row may be multiplied by a nonzero number. This is the
same as multiplying both sides of an equation by a nonzero number.
3. The elements in any row may be multiplied by a nonzero number, and these
products may be added to the corresponding elements in any other row. This
is the same as multiplying both sides of an equation by a nonzero number
and then adding equations to eliminate a variable.
Study Tip
When performing the row operation
Two matrices are row equivalent if one can be obtained from the other by a
sequence of row operations.
kRi + Rj
you use row i to find the products.
However, elements in row i do not
change. It is the elements in row j that
change: Add k times the elements in
row i to the corresponding elements
in row j. Replace elements in row j
by these sums.
Each matrix row operation in the preceding box can be expressed symbolically
as follows:
1. Interchange the elements in the ith and jth rows: Ri 4 Rj .
2. Multiply each element in the ith row by k: kRi .
3. Add k times the elements in row i to the corresponding elements in
row j: kRi + Rj .
EXAMPLE 1
Performing Matrix Row Operations
Use the matrix
3
C 1
-2
18
2
-3
-12 21
-3 3 5 S
4 -6
and perform each indicated row operation:
a. R1 4 R2
b.
1
3 R1
c. 2R2 + R3 .
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 808
808 Chapter 8 Matrices and Determinants
3
C 1
-2
18
2
-3
-12 21
-3 3 5 S
4 -6
Solution
a. The notation R1 4 R2 means to interchange the elements in row 1 and row 2.
This results in the row-equivalent matrix
This was row 2; now it’s row 1.
1 2 –3
5
C 3 18 –12 3 21 S .
–2 –3
4 –6
The given matrix (repeated)
This was row 1; now it’s row 2.
b. The notation 13 R1 means to multiply each element in row 1 by 13 . This results in
the row-equivalent matrix
1
3 132
C 1
-2
1
3 1182
1
3 1-122
2
-3
-3
4
1
3 1212
1
5 S = C 1
-6
-2
3
6
2
-3
-4
7
- 3 3 5 S.
4 -6
c. The notation 2R2 + R3 means to add 2 times the elements in row 2 to the
corresponding elements in row 3. Replace the elements in row 3 by these
sums. First, we find 2 times the elements in row 2, namely, 1, 2, -3, and 5:
2112 or 2,
2122 or 4,
21-32 or - 6,
2152 or 10.
Now we add these products to the corresponding elements in row 3. Although
we use row 2 to find the products, row 2 does not change. It is the elements in
row 3 that change, resulting in the row-equivalent matrix
Replace row 3 by the
sum of itself and
2 times row 2.
3
C
1
–2+2=0
18
2
–3+4=1
Check Point
1
–12
21
3
3
S = C1
–3
5
4+(–6)=–2 –6+10=4
0
18
2
1
–12 21
–3 3 5 S.
–2 4
Use the matrix
4 12 -20
8
C 1
6
-3 3 7 S
- 3 -2
1 -9
and perform each indicated row operation:
a. R1 4 R2
�
Use matrices and Gaussian
elimination to solve systems.
b.
1
4 R1
c. 3R2 + R3 .
Solving Linear Systems Using Gaussian Elimination
The process that we use to solve linear systems using matrix row operations is called
Gaussian elimination, after the German mathematician Carl Friedrich Gauss
(1777–1855). Here are the steps used in Gaussian elimination:
Solving Linear Systems of Three Equations with Three Variables Using Gaussian Elimination
1. Write the augmented matrix for the system.
2. Use matrix row operations to simplify the matrix to a row-equivalent matrix in row-echelon form, with 1s down
the main diagonal from upper left to lower right, and 0s below the 1s in the first and second columns.
*
1 * *
C* * * 3 * S
*
* * *
*
1 * *
C0 * * 3 * S
*
0 * *
Get 1 in the
upper lefthand corner.
Use the 1 in the
first column to
get 0s below it.
*
1 * *
C0 1 * 3 * S
*
0 * *
*
1 * *
C0 1 * 3 * S
*
0 0 *
Get 1 in the
second row, second
column position.
Use the 1 in the
second column to
get 0 below it.
*
1 * *
C0 1 * 3 * S
*
0 0 1
Get 1 in the
third row, third
column position.
3. Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find the
system’s solution.
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 809
Section 8.1 Matrix Solutions to Linear Systems
EXAMPLE 2
809
Gaussian Elimination with Back-Substitution
Use matrices to solve the system:
3x + y + 2z = 31
c x + y + 2z = 19
x + 3y + 2z = 25.
Solution
Step 1 Write the augmented matrix for the system.
Linear System
Augmented Matrix
3x + y + 2z = 31
c x + y + 2z = 19
x + 3y + 2z = 25
3
C1
1
1
1
3
2 31
2 3 19 S
2 25
Step 2 Use matrix row operations to simplify the matrix to row-echelon form, with
1s down the main diagonal from upper left to lower right, and 0s below the 1s in the
first and second columns. Our first step in achieving this goal is to get 1 in the top
position of the first column.
3
C1
1
We want 1 in
this position.
1
1
3
2 31
2 3 19 S
2 25
To get 1 in this position, we interchange row 1 and row 2: R1 4 R2 . (We could also
interchange row 1 and row 3 to attain our goal.)
1
C3
1
1
1
3
2 19
2 3 31 S
2 25
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.
Now we want to get 0s below the 1 in the first column.
1
C3
1
We want 0 in
these positions.
1
1
3
2 19
2 3 31 S
2 25
To get a 0 where there is now a 3, multiply the top row of numbers by -3 and add
these products to the second row of numbers: -3R1 + R2 . To get a 0 where there is
now a 1, multiply the top row of numbers by - 1 and add these products to the third
row of numbers: -1R1 + R3 . Although we are using row 1 to find the products, the
numbers in row 1 do not change.
Replace row 2 by
−3R1 + R2.
Replace row 3 by
−1R1 + R3.
1 1 2
19
1
1
2
19
C –3(1)+3 –3(1)+1 –3(2)+2 3 –3(19)+31 S = C 0 –2 –4 3 –26 S
0 2 0
6
–1(1)+1 –1(1)+3 –1(2)+2
–1(19)+25
We want 1 in this position.
We move on to the second column.To get 1 in the desired position, we multiply -2 by its
reciprocal, - 12 . Therefore, we multiply all the numbers in the second row by - 12 : - 12 R2 .
− 21 R2
1
1
2
C– 12 (0)
– 12 (–2)
– 12 (–4)
0
2
0
19
3
1 1 2 19
1 2 3 13 S .
0 2 0
6
– 12 (–26)S = C 0
6
We want 0 in this position.
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 810
810 Chapter 8 Matrices and Determinants
So far, our matrix row operations have resulted in the matrix that we repeated in the
margin. We are not yet done with the second column. The voice balloon shows that
we want to get a 0 where there is now a 2. If we multiply the second row of numbers
by -2 and add these products to the third row of numbers, we will get 0 in this position: -2R2 + R3 . Although we are using the numbers in row 2 to find the products,
the numbers in row 2 do not change.
1 1 2 19
C 0 1 2 3 13 S
0 2 0
6
We want 0 in this position.
The matrix from the bottom
of the previous page (repeated)
Replace row 3 by
−2R2 + R3.
1
1
2
19
1
C
3
S=C0
0
1
2
13
–2(0)+0 –2(1)+2 –2(2)+0 –2(13)+6
0
1 2
19
1 2 3 13 S
0 –4 –20
We want 1 in this position.
We move on to the third column.To get 1 in the desired position, we multiply - 4 by its
reciprocal, - 14 . Therefore, we multiply all the numbers in the third row by - 14 : - 14 R3 .
− 41 R3
1
2
1 1 2 19
1
19
C 0
3
1
2
13 S = C 0 1 2 3 13 S
0 0 1
5
– 14 (0) – 14 (0) – 14 (–4) – 14 (–20)
We now have the desired matrix in row-echelon form, with 1s down the main diagonal
and 0s below the 1s in the first and second columns.
Step 3 Write the system of linear equations corresponding to the matrix in step 2
and use back-substitution to find the system’s solution. The system represented by
the matrix in step 2 is
1
C0
0
1
1
0
2 19
1x + 1y + 2z = 19
3
2 13 S : c 0x + 1y + 2z = 13 or
1 5
0x + 0y + 1z = 5
x + y + 2z = 19
c
y + 2z = 13.
z = 5
(1)
(2)
(3)
We immediately see from equation (3) that the value for z is 5. To find y, we
back-substitute 5 for z in the second equation.
y + 2z
y + 2152
y + 10
y
=
=
=
=
13
13
13
3
Equation (2)
Substitute 5 for z.
Multiply.
Subtract 10 from both sides and solve for y.
Finally, back-substitute 3 for y and 5 for z in the first equation.
x + y + 2z
x + 3 + 2152
x + 13
x
Technology
Most graphing utilities can convert
an augmented matrix to row-echelon
form, with 1s down the main diagonal and 0s below the 1s. However,
row-echelon form is not unique. Your
graphing utility might give a rowechelon form different from the one
you obtained by hand. However, all
row-echelon forms for a given
system’s augmented matrix produce
the same solution to the system.
Enter the augmented matrix and
name it A. Then use the 冷 REF 冷
(row-echelon form) command on
matrix A.
=
=
=
=
19
19
19
6
Equation (1)
Substitute 3 for y and 5 for z.
Multiply and add.
Subtract 13 from both sides and solve for x.
With z = 5, y = 3, and x = 6, the solution set of the original system is 516, 3, 526.
Check to see that the solution satisfies all three equations in the given system.
Check Point
2
Use matrices to solve the system:
2x + y + 2z = 18
c x - y + 2z = 9
x + 2y - z = 6.
Modern supercomputers are capable of solving systems with more than 600,000
variables. The augmented matrices for such systems are huge, but the solution using
matrices is exactly like what we did in Example 2. Work with the augmented matrix,
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 811
Section 8.1 Matrix Solutions to Linear Systems
811
one column at a time. Get 1s down the main diagonal from upper left to lower right
and 0s below the 1s. Let’s see how this works for a linear system involving four
equations in four variables.
Gaussian Elimination with Back-Substitution
EXAMPLE 3
Use matrices to solve the system:
2w
w
d
w
-w
+
+
x
x
x
2x
+
+
-
3y
2y
y
2y
+
-
z
2z
z
z
= 6
= -1
= -4
= - 7.
Solution
Step 1 Write the augmented matrix for the system.
Linear System
2w
w
d
w
-w
+
+
x
x
x
2x
+
+
-
Augmented Matrix
3y
2y
y
2y
+
-
z
2z
z
z
= 6
= -1
= -4
= -7
2
1
D
1
-1
1
-1
-1
2
3
2
-1
-2
-1
6
-2 4 -1
T
1 -4
-1 -7
Step 2 Use matrix row operations to simplify the matrix to row-echelon form,
with 1s down the main diagonal from upper left to lower right, and 0s below the 1s
in the first, second, and third columns. Our first step in achieving this goal is to get
1 in the top position of the first column. To do this, we interchange row 1 and row
2: R1 4 R2 .
1
2
D
1
–1
We want
0s in these
positions.
–1
1
–1
2
2
3
–1
–2
–2 –1
–1
6
4
T
1 –4
–1 –7
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.
Now we use the 1 at the top of the first column to get 0s below it.
1
0
D
0
0
Use the previous matrix and:
Replace row 2 by −2R1 + R2.
Replace row 3 by −1R1 + R3.
Replace row 4 by 1R1 + R4.
–1
3
0
1
2
–1
–3
0
–2 –1
3
8
4
T
3 –3
–3 –8
We want
1 in this
position.
We move on to the second column. We can obtain 1 in the desired position by
multiplying the numbers in the second row by 13 , the reciprocal of 3.
1
1
3 (0)
D
0
0
–1
1
3 (3)
0
1
–2
–1
1
3 (–1)
2
1
3 (3)
1
3 (8)
–3
0
3
–3
4
1
0
T=D
–3
0
–8
0
–1
2
1 – 13
0 –3
1
0
We want 0s in these positions.
The top position already has a 0.
–2
–1
8
1
3
4
T
3
–3
–3
–8
1
3
R2
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 812
812 Chapter 8 Matrices and Determinants
1
0
D
0
0
–1
2
1 – 13
0 –3
1
0
So far, our matrix row operations have resulted in the matrix that we repeated in the
margin. Now we use the 1 in the second row, second column position to get 0s below it.
–2
–1
8
1
3
4
T
3
–3
–3
–8
Replace row 4 in the previous
matrix by −1R2 + R4.
We want 0s in these positions.
The top position already has a 0.
1
0
D
0
0
–1
1
0
0
2
– 13
–3
1
3
–2 –1
8
1
4 3T
3 –3
–4 – 323
We want
1 in this
position.
We move on to the third column.We can obtain 1 in the desired position by multiplying
the numbers in the third row by - 13 , the reciprocal of -3.
The matrix from the bottom
of the previous page (repeated)
1
–1
2
–2
–1
1
8
1
–3
0
1
1
0
3
D 1
4
T=D
0
– 3 (0) – 13 (0) – 13 (–3) – 13 (3) – 13 (–3)
32
1
0
0
–4
0
–3
3
–1
2
1
1 –3
0
1
1
0
3
–2
–1
8
1
3
4
T
–1
1
–4
– 323
− 31 R3
We want
0 in this
position.
Now we use the 1 in the third column to get 0 below it.
1
0
D
0
0
Replace row 4 in the previous
matrix by − 31 R3 + R4.
–1
1
0
0
2
–2
–1
8
1
–
3
4
T
1 –1
1
0 – 113 –11
1
3
We want
1 in this
position.
We move on to the fourth column. Because we want 1s down the diagonal from
upper left to lower right, we want 1 where there is now - 11
3 . We can obtain 1 in this
3
.
position by multiplying the numbers in the fourth row by - 11
1
–1
2
–2
–1
8
0
1
1
– 13
3
D
4
T
0
0
1
–1
1
– 113 (0) – 113 (0) – 113 (0) – 113 (–113) – 113 (–11)
1
0
=D
0
0
–1
2
1 – 13
0
1
0
0
–2
–1
8
1
3
4
T
–1
1
1
3
− 113 R4
We now have the desired matrix in row-echelon form, with 1s down the main
diagonal and 0 s below the 1s. An equivalent row-echelon matrix can be obtained
using a graphing utility and the 冷 REF 冷 command on the augmented matrix.
Step 3 Write the system of linear equations corresponding to the matrix in step 2
and use back-substitution to find the system’s solution. The system represented by
the matrix in step 2 is
1
0
D
0
0
-1
1
0
0
2
- 13
1
0
-2 -1
1w 0w +
1 4 83
T:d
1
0w +
-1
3
0w +
1
1x
1x
0x
0x
+
+
+
2y +
1y 0y +
1
3y
2z
1z
1z
1z
= -1
= 83
or
= 1
= 3
w - x + 2y - 2z = - 1
x - 13 y + z = 83
d
y - z = 1
z = 3.
We immediately see that the value for z is 3. We can now use back-substitution to
find the values for y, x, and w.
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 813
Section 8.1 Matrix Solutions to Linear Systems
These are the
four equations
from the last column.
z=3 y-z=1
9
y=4
1
8
y+z=
3
3
w-x+2y-2z=–1
1
8
(4)+3=
3
3
w-1+2(4)-2(3)=–1
x-
y-3=1 x-
813
9
x+
8 9
5
=
3
3
w-1+8-6=–1
x=1
w+1=–1
w=–2
Let’s agree to write the solution for the system in the alphabetical order of the
variables from left to right, namely 1w, x, y, z2. Thus, the solution set is
51 -2, 1, 4, 326. We can verify the solution by substituting the value for each variable
into the original system of equations and obtaining four true statements.
Check Point
3
Use matrices to solve the system:
w
2w
d
3w
5w
�
Use matrices and Gauss-Jordan
elimination to solve systems.
+
3x
7x
7x
x
+
2y
y
3y
4y
+
+
+
-
z
2z
3z
2z
= -3
= 1
= -5
= 18.
Solving Linear Systems Using Gauss-Jordan Elimination
Using Gaussian elimination, we obtain a matrix in row-echelon form, with 1s down
the main diagonal and 0s below the 1s. A second method, called Gauss-Jordan
elimination, after Carl Friedrich Gauss and Wilhelm Jordan (1842–1899), continues
the process until a matrix with 1s down the main diagonal and 0s in every position
above and below each 1 is found. Such a matrix is said to be in reduced row-echelon
form. For a system of three linear equations in three variables, x, y, and z, we must
get the augmented matrix into the form
1
C0
0
0
1
0
0 a
0 3 b S.
1 c
Based on this matrix, we conclude that x = a, y = b, and z = c.
Solving Linear Systems Using Gauss-Jordan Elimination
1. Write the augmented matrix for the system.
2. Use matrix row operations to simplify the matrix to a row-equivalent matrix
in reduced row-echelon form, with 1s down the main diagonal from upper
left to lower right, and 0s above and below the 1s.
a. Get 1 in the upper left-hand corner.
b. Use the 1 in the first column to get 0s below it.
c. Get 1 in the second row, second column.
d. Use the 1 in the second column to make the remaining entries in the second
column 0.
e. Get 1 in the third row, third column.
f. Use the 1 in the third column to make the remaining entries in the third
column 0.
g. Continue this procedure as far as possible.
3. Use the reduced row-echelon form of the matrix in step 2 to write the system’s
solution set. (Back-substitution is not necessary.)
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 814
814 Chapter 8 Matrices and Determinants
Technology
EXAMPLE 4
Most graphing utilities can convert a
matrix to reduced row-echelon form.
Enter the system’s augmented matrix
and name it A. Then use the 冷 RREF 冷
(reduced row-echelon form) command on matrix A.
Using Gauss-Jordan Elimination
Use Gauss-Jordan elimination to solve the system:
3x + y + 2z = 31
c x + y + 2z = 19
x + 3y + 2z = 25.
Solution In Example 2, we used Gaussian elimination to obtain the following matrix:
We want
0s in these
positions.
1
C0
0
1
1
0
2 19
2 3 13 S .
1
5
To use Gauss-Jordan elimination, we need 0s both above and below the 1s in the
main diagonal. We use the 1 in the second row, second column to get a 0 above it.
This is the
augmented matrix
for the system
in Example 4.
Replace row 1
in the previous
matrix by
−1R2 + R1.
1
C0
0
0
1
0
0
6
2 3 13 S
1
5
We want
0s in these
positions.
We use the 1 in the third column to get 0s above it.
1
C0
0
This is the matrix
in reduced rowechelon form we
obtained in
Example 4.
0
1
0
0 6
0 3 3S
1 5
Replace row 2 in the previous
matrix by −2R3 + R2.
This last matrix corresponds to
x = 6, y = 3, z = 5.
As we found in Example 2, the solution set is 516, 3, 526.
Check Point
4
Solve the system in Check Point 2 using Gauss-Jordan elimination.
Begin by working with the matrix that you obtained in Check Point 2.
Exercise Set 8.1
Practice Exercises
In Exercises 1–8, write the augmented matrix for each system of
linear equations.
2x + y + 2z = 2
1. c 3x - 5y - z = 4
x - 2y - 3z = - 6
2. c
3x - 2y + 5z = 31
x + 3y - 3z = - 12
- 2x - 5y + 3z = 11
3. c
x - y + z = 8
y - 12z = - 15
z = 1
4. c
x - 2y + 3z = 9
y + 3z = 5
z = 2
5. c
5x - 2y - 3z = 0
x + y = 5
2x - 3z = 4
6. c
x - 2y + z = 10
3x + y = 5
7x + 2z = 2
2w + 5x - 3y
3x
7. d
w - x
5w - 5x
+
+
+
-
z
y
5y
2y
=
=
=
=
2
4
9
1
4w + 7x - 8y + z
5x + y
8. d
w - x - y
2w - 2x + 11y
In Exercises 9–12, write the system of linear equations represented
by the augmented matrix. Use x, y, and z, or, if necessary, w, x, y,
and z, for the variables.
5
9. C 0
7
1
-1
11. D
2
0
0
1
2
3 - 11
- 4 3 12 S
0
3
1
1
0
0
4
-1
0
12
1 3
0 4 7
T
5 11
4 5
7
10. C 0
2
0
1
7
4
1
12. D
3
0
1
-1
0
0
4 - 13
- 5 3 11 S
0
6
5
0
0
11
1
-1
7
5
6
4 8T
4
3
In Exercises 13–18, perform each matrix row operation and write
the new matrix.
=
=
=
=
3
2
5
17 13. C 1
3
4
-6
5
0
4 10
-5 3 0 S
4 7
1
2 R1
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 815
Section 8.1 Matrix Solutions to Linear Systems
-12
-4
0
1
15. C 3
2
-3
1
-2
2 0
-1 3 7 S
1 3
1
16. C 3
1
-1
3
3
5 -6
- 1 3 10 S
2
5
1
0
17. D
2
5
-1
1
0
1
1
-2
3
2
1
0
18. D
3
-4
-5
1
0
1
6
4
7
1
3 R1
3
14. C 1
2
9
3 0S
4
2
-3
2
4
- 3R1 + R2
- 3R1 + R2
1 3
-1 4 0
T
4 11
4 6
- 2R1 + R3
- 5R1 + R4
4
-2
-1 4 0
T
-1
6
2 -3
- 3R1 + R3
4R1 + R4
In Exercises 19–20, a few steps in the process of simplifying the
given matrix to row-echelon form, with 1s down the diagonal from
upper left to lower right, and 0s below the 1s, are shown. Fill in the
missing numbers in the steps that are shown.
1
19. C 2
3
1
20. C 2
-3
-1
3
-2
-2
1
4
1
8
1
-1 3 -2 S : C 0
-9
9
0
-1
5
1
1 8
n 3 nS
n n
1
: C0
0
-1
1
1
1 8
n 3 nS
n n
815
x + 2y = z - 1
29. c x = 4 + y - z
x + y - 3z = - 2
2x + y = z + 1
30. c 2x = 1 + 3y - z
x + y + z = 4
3a - b - 4c = 3
31. c 2a - b + 2c = - 8
a + 2b - 3c = 9
3a + b - c = 0
32. c 2a + 3b - 5c = 1
a - 2b + 3c = - 4
2x + 2y + 7z = - 1
33. c 2x + y + 2z = 2
4x + 6y + z = 15
3x + 2y + 3z = 3
34. c 4x - 5y + 7z = 1
2x + 3y - 2z = 6
w
2w
35. d
w
3w
+
+
+
x
x
2x
2x
+
+
y
2y
y
y
+
+
z
z
2z
3z
= 4
= 0
= -2
= 4
w
w
36. d
w
2w
+
+
-
x
2x
3x
x
+
+
y
y
3y
2y
+
-
z
2z
z
z
= 5
= -1
= -1
= -2
3w
w
37. d
2w
-w
+
+
+
4x
x
x
2x
+
+
+
y
y
4y
y
+
-
z
z
2z
3z
=
=
=
=
2w
w
38. d
3w
w
+
y - 3z =
8
- x +
4z = - 10
+ 5x - y - z = 20
+ x - y - z =
6
9
0
3
3
Practice Plus
39. Find the quadratic function f1x2 = ax2 + bx + c for which
f1-22 = - 4, f112 = 2, and f122 = 0.
3
4
1
-4 3 3 S : C 0
-1 - 2
0
-2
5
-2
3 4
n 3 nS
n n
40. Find the quadratic function f1x2 = ax2 + bx + c for which
f1- 12 = 5, f112 = 3, and f122 = 5.
1
: C0
0
-2
1
-2
3 4
n 3 nS
n n
42. Find the cubic function f1x2 = ax3 + bx2 + cx + d for
which f1- 12 = 3, f112 = 1, f122 = 6, and f132 = 7.
41. Find the cubic function f1x2 = ax3 + bx2 + cx + d for
which f1- 12 = 0, f112 = 2, f122 = 3, and f132 = 12.
43. Solve the system:
In Exercises 21–38, solve each system of equations using matrices.
Use Gaussian elimination with back-substitution or Gauss-Jordan
elimination.
x + y - z = -2
21. c 2x - y + z = 5
- x + 2y + 2z = 1
x - 2y - z = 2
22. c 2x - y + z = 4
- x + y - 2z = - 4
x + 3y
= 0
23. c x + y + z = 1
3x - y - z = 11
24. c
2x - y - z = 4
25. c x + y - 5z = - 4
x - 2y
= 4
x
- 3z = - 2
26. c 2x + 2y + z = 4
3x + y - 2z = 5
x + y + z = 4
27. c x - y - z = 0
x - y + z = 2
3x + y - z = 0
28. c x + y + 2z = 6
2x + 2y + 3z = 10
3y - z = - 1
x + 5y - z = - 4
-3x + 6y + 2z = 11
2 ln w
4 ln w
d
ln w
ln w
+ ln x
+ 3 ln x
+ ln x
+ ln x
+ 3 ln y - 2 ln z = - 6
+ ln y - ln z = - 2
+ ln y + ln z = - 5
- ln y - ln z = 5.
(Hint: Let A = ln w, B = ln x, C = ln y, and D = ln z. Solve
the system for A, B, C, and D. Then use the logarithmic
equations to find w, x, y, and z.)
44. Solve the system:
ln w
-ln w
d
ln w
-ln w
+ ln x
+ 4 ln x
- 2 ln x
- 2 ln x
+
+
+
+
ln y
ln y
ln y
ln y
+ ln z
- ln z
- 2 ln z
+ 2 ln z
= -1
= 0
= 11
= - 3.
(Hint: Let A = ln w, B = ln x, C = ln y, and D = ln z. Solve
the system for A, B, C, and D. Then use the logarithmic
equations to find w, x, y, and z.)
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 816
816 Chapter 8 Matrices and Determinants
Application Exercises
45. A ball is thrown straight upward. A position function
s1t2 = 12 at2 + v0t + s0
can be used to describe the ball’s height, s1t2, in feet, after t
seconds.
Height above Ground
(feet)
s(t)
60
50
If a meal consisting of the three foods allows exactly
660 calories, 25 grams of protein, and 425 milligrams of vitamin C, how many ounces of each kind of food should be used?
48. A furniture company produces three types of desks: a children’s model, an office model, and a deluxe model. Each
desk is manufactured in three stages: cutting, construction,
and finishing. The time requirements for each model and
manufacturing stage are given in the following table.
(2, 48)
(1, 40)
Children’s
model
Office
model
Deluxe
model
Cutting
2 hr
3 hr
2 hr
Construction
2 hr
1 hr
3 hr
Finishing
1 hr
1 hr
2 hr
40
30
(3, 24)
20
10
1
2
3
4
Time (seconds)
5
t
a. Use the points labeled in the graph to find the values of
a, v0 , and s0 . Solve the system of linear equations involving
a, v0 , and s0 using matrices.
b. Find and interpret s13.52. Identify your solution as a
point on the graph shown.
Each week the company has available a maximum of 100
hours for cutting, 100 hours for construction, and 65 hours for
finishing. If all available time must be used, how many of
each type of desk should be produced each week?
49. Imagine the entire global population as a village of precisely
200 people. The bar graph shows some numeric observations
based on this scenario.
c. After how many seconds does the ball reach its maximum
height? What is its maximum height?
can be used to describe the ball’s height, s1t2, in feet, after t
seconds.
Height above Ground
(feet)
s(t)
300
(5, 246)
250
200
(2, 198)
100
(8, 6)
0
5
Time (seconds)
125
100
75
60
50
Asian
w
Under
age 15
African
Over
age 65
x
25
14
Unable
to read
or write
European
American
(U.S.)
y
32
z
Eat at
McDonald’s
each day
1
Source: Gary Rimmer, Number Freaking, The Disinformation Company Ltd., 2006
150
50
150
Number of People
46. A football is kicked straight upward. A position function
s1t2 = 12 at2 + v0t + s0
Earth’s Population as a Village of 200 People
10
t
a. Use the points labeled in the graph to find the values of
a, v0 , and s0 . Solve the system of linear equations involving
a, v0 , and s0 using matrices.
b. Find and interpret s172. Identify your solution as a point
on the graph shown.
c. After how many seconds does the ball reach its maximum
height? What is its maximum height?
Combined, there are 183 Asians, Africans, Europeans, and
Americans in the village. The number of Asians exceeds the
number of Africans and Europeans by 70. The difference
between the number of Europeans and Americans is 15. If
the number of Africans is doubled, their population exceeds
the number of Europeans and Americans by 23. Determine
the number of Asians, Africans, Europeans, and Americans in
the global village.
50. The bar graph shows the number of rooms, bathrooms,
fireplaces, and elevators in the U.S. White House.
The U.S. White House by the Numbers
150
w
Write a system of linear equations in three or four variables to
solve Exercises 47–50. Then use matrices to solve the system.
125
47. Three foods have the following nutritional content per ounce.
100
Calories
Protein
(in grams)
Vitamin C
(in milligrams)
Food A
40
5
Food B
200
2
10
Food C
400
4
300
30
Rooms
75
Bathrooms
50
x
25
Fireplaces
y
Elevators
z
Source: The White House
P-BLTZMC08_805-872-hr
21-11-2008
13:26
Page 817
Section 8.1 Matrix Solutions to Linear Systems
Combined, there are 198 rooms, bathrooms, fireplaces, and
elevators. The number of rooms exceeds the number of
bathrooms and fireplaces by 69. The difference between
the number of fireplaces and elevators is 25. If the number
of bathrooms is doubled, it exceeds the number of fireplaces and elevators by 39. Determine the number of
rooms, bathrooms, fireplaces, and elevators in the U.S.
White House.
Writing in Mathematics
817
61. When I use matrices to solve linear systems, I spend most of my
time using row operations to express the system’s augmented
matrix in row-echelon form.
62. Using row operations on an augmented matrix, I obtain a row
in which 0s appear to the left of the vertical bar, but 6 appears
on the right, so the system I’m working with has no solution.
In Exercises 63–66, determine whether each statement is true or false.
If the statement is false, make the necessary change(s) to produce a
true statement.
51. What is a matrix?
63. A matrix row operation such as - 45 R1 + R2 is not permitted
because of the negative fraction.
52. Describe what is meant by the augmented matrix of a system
of linear equations.
64. The augmented matrix for the system
x - 3y = 5
y - 2z = 7 is
2x + z = 4
53. In your own words, describe each of the three matrix row
operations. Give an example with each of the operations.
54. Describe how to use row operations and matrices to solve a
system of linear equations.
55. What is the difference between Gaussian elimination and
Gauss-Jordan elimination?
Technology Exercises
56. Most graphing utilities can perform row operations on
matrices. Consult the owner’s manual for your graphing
utility to learn proper keystrokes for performing these
operations. Then duplicate the row operations of any three
exercises that you solved from Exercises 13–18.
57. If your graphing utility has a
冷
REF
冷
(row-echelon form)
command or a 冷 RREF 冷 (reduced row-echelon form)
command, use this feature to verify your work with any five
systems that you solved from Exercises 21–38.
58. Solve using a graphing utility’s 冷 REF 冷 or 冷 RREF 冷 command:
2x1
x1
e x1
- x1
x1
+
+
+
-
2x2 + 3x3
2x2 - x3
x3
x2 - x3
x2 -
+
+
-
x4
2x4
x4
2x4
x4
+
x5
5x5
3x5
x5
= 12
= -7
= 1
= 0
= 4.
Critical Thinking Exercises
Make Sense? In Exercises 59–62, determine whether each
statement makes sense or does not make sense, and explain
your reasoning.
59. Matrix row operations remind me of what I did when solving
a linear system by the addition method, although I no longer
write the variables.
60. When I use matrices to solve linear systems, the only arithmetic
involves multiplication or a combination of multiplication and
addition.
1
C1
2
-3
-2
1
5
3 7 S.
4
65. In solving a linear system of three equations in three
variables, we begin with the augmented matrix and use row
operations to obtain a row-equivalent matrix with 0s down
the diagonal from left to right and 1s below each 0.
66. The row operation kRi + Rj indicates that it is the elements
in row i that change.
67. The table shows the daily production level and profit for a
business.
x (Number of Units
Produced Daily)
y (Daily Profit)
30
50
100
$5900
$7500
$4500
Use the quadratic function y = ax2 + bx + c to determine
the number of units that should be produced each day for
maximum profit. What is the maximum daily profit?
Preview Exercises
Exercises 68–70 will help you prepare for the material covered in
the next section. In each exercise, refer to the following system:
3x - 4y + 4z = 7
c x - y - 2z = 2
2x - 3y + 6z = 5.
68. Show that 112z + 1, 10z - 1, z2 satisfies the system for
z = 0.
69. Show that 112z + 1, 10z - 1, z2 satisfies the system for
z = 1.
70. a. Select a value for z other than 0 or 1 and show that
112z + 1, 10z - 1, z2 satisfies the system.
b. Based on your work in Exercises 68–70(a), how does this
system differ from those in Exercises 21–34?