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Matrices andDeterminants 8
805
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.
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 805
8.1
806 Chapter 8 Matrices and Determinants
Matrix Solutions to Linear SystemsObjectives
� Write the augmented matrixfor a linear system.
� Perform matrix rowoperations.
� Use matrices and Gaussianelimination to solve systems.
� Use matrices and Gauss-Jordan elimination to solvesystems.
S e c t i o n
Average Number of Minutes per Day Americans Spend on Grooming
Ages 15–19
Ages 20–24
Ages 45–54
Ages 65� Married Single
Men 37 37 34 28 31 34
Women 59 49 46 46 44 50
Source: Bureau of Labor Statistics’ American Time-Use Survey
System of Linear Equations Augmented Matrix
c 3x + y + 2z = 31x + y + 2z = 19x + 3y + 2z = 25
C3 1 21 1 21 3 2
3 311925S
c x + 2y - 5z = -19y + 3z = 9
z = 4C1 2 -5
0 1 30 0 1
3 -1994S .
The data below show that we spend a lot of time sprucing up.
RThe 12 numbers inside the brackets are arranged in two rows and six columns. Thisrectangular array of 12 numbers, arranged in rows and columns and placed in brack-ets, is an example of a matrix (plural: matrices).The numbers inside the brackets arecalled elements of the matrix. Matrices are used to display information and to solvesystems of linear equations. Because systems involving two equations in twovariables can easily be solved by substitution or addition, we will focus on matrixsolutions to systems of linear equations in three or more variables.
Augmented MatricesA matrix gives us a shortened way of writing a system of equations. The first step insolving 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 intotwo groups.The coefficients of each variable are placed to the left of the vertical lineand the constants are placed to the right. If any variable is missing, its coefficient is 0.Here are two examples:
� Write the augmented matrix for alinear system.
B
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 806
Study TipWhen performing the row operation
you use row to find the products.However, elements in row do notchange. It is the elements in row thatchange: Add times the elements inrow to the corresponding elementsin row Replace elements in row by these sums.
jj.i
kj
ii
kRi + Rj
Section 8.1 Matrix Solutions to Linear Systems 807
Our goal in solving a system of linear equations in three variables using matricesis to produce a matrix with 1s down the diagonal from upper left to lower right on theleft side of the vertical bar, called the main diagonal, and 0s below the 1s. In general,the matrix will be of the form
where through represent real numbers. The third row of this matrix gives us thevalue of one variable. The other variables can then be found by back-substitution.
Matrix Row OperationsA matrix with 1s down the main diagonal and 0s below the 1s is said to be in row-echelon form. How do we produce a matrix in this form? We use row operations onthe augmented matrix.These row operations are just like what you did when solvinga linear system by the addition method.The difference is that we no longer write thevariables, usually represented by and z.x, y,
fa
C1 a b
0 1 d
0 0 1 3 ce
f
S ,
� Perform matrix row operations.
Matrix Row OperationsThe following row operations produce matrices that represent systems with thesame solution set:
1. Two rows of a matrix may be interchanged. This is the same as interchangingtwo equations in a linear system.
2. The elements in any row may be multiplied by a nonzero number. This is thesame 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 theseproducts may be added to the corresponding elements in any other row. Thisis the same as multiplying both sides of an equation by a nonzero numberand then adding equations to eliminate a variable.
Two matrices are row equivalent if one can be obtained from the other by asequence of row operations.
Each matrix row operation in the preceding box can be expressed symbolicallyas follows:
1. Interchange the elements in the and rows:
2. Multiply each element in the row by
3. Add times the elements in row to the corresponding elements in row j: kRi + Rj .
ik
k: kRi .ith
Ri 4 Rj .jthith
Performing Matrix Row Operations
Use the matrix
and perform each indicated row operation:
a. b. c. 2R2 + R3 .13 R1R1 4 R2
C 3 18 -121 2 -3
-2 -3 4 3 21
5-6S
EXAMPLE 1
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 807
808 Chapter 8 Matrices and Determinants
Solution
a. The notation means to interchange the elements in row 1 and row 2.This results in the row-equivalent matrix
b. The notation means to multiply each element in row 1 by This results inthe row-equivalent matrix
c. The notation means to add 2 times the elements in row 2 to thecorresponding elements in row 3. Replace the elements in row 3 by thesesums. First, we find 2 times the elements in row 2, namely, 1, 2, , and 5:
Now we add these products to the corresponding elements in row 3. Althoughwe use row 2 to find the products, row 2 does not change. It is the elements inrow 3 that change, resulting in the row-equivalent matrix
2112 or 2, 2122 or 4, 21-32 or -6, 2152 or 10.
-3
2R2 + R3
C 13132
131182 1
31-1221 2 -3
-2 -3 4 3 131212
5-6S = C 1 6 -4
1 2 -3-2 -3 4
3 75
-6S .
13 .1
3 R1
C 3 S1
3
–2
2
18
–3
–3
–12
4
5
21
–6
This was row 2; now it’s row 1.
This was row 1; now it’s row 2..
R1 4 R2
� Use matrices and Gaussianelimination to solve systems.
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 downthe main diagonal from upper left to lower right, and 0s below the 1s in the first and second columns.
C 3 S = C 3 S .3
1
18
2
–12
–3
21
5
3
1
0
18
2
1
–12
–3
–2
21
5
4–2+2=0 –3+4=1 4+(–6)=–2 –6+10=4Replace row 3 by the
sum of itself and2 times row 2.
Check Point 1 Use the matrix
and perform each indicated row operation:
a. b. c.
Solving Linear Systems Using Gaussian EliminationThe process that we use to solve linear systems using matrix row operations is calledGaussian elimination, after the German mathematician Carl Friedrich Gauss(1777–1855). Here are the steps used in Gaussian elimination:
3R2 + R3 .14 R1R1 4 R2
C 4 12 -201 6 -3
-3 -2 1 3 8
7-9S
3. Write the system of linear equations corresponding to the matrix in step 2 and use back-substitution to find thesystem’s solution.
Get 1 in theupper left- hand corner.
Use the 1 in thefirst column toget 0s below it.
Use the 1 in thesecond column toget 0 below it.
Get 1 in thesecond row, secondcolumn position.
Get 1 in thethird row, thirdcolumn position.
1
**
***
***
***
C 3 S 1
0
0
***
***
***
C 3 S 1
0
0
*1*
***
***
C 3 S 1
0
0
*1
0
***
***
C 3 S 1
0
0
*1
0
**1
***
C 3 S
The given matrix (repeated)
C 3 18 -121 2 -3
-2 -3 4 3 21
5-6S
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 808
Section 8.1 Matrix Solutions to Linear Systems 809
Linear System Augmented Matrix
c 3x + y + 2z = 31x + y + 2z = 19x + 3y + 2z = 25
C3 1 21 1 21 3 2
3 311925S
Gaussian Elimination with Back-Substitution
Use matrices to solve the system:
SolutionStep 1 Write the augmented matrix for the system.
c 3x + y + 2z = 31x + y + 2z = 19x + 3y + 2z = 25.
EXAMPLE 2
Step 2 Use matrix row operations to simplify the matrix to row-echelon form, with1s down the main diagonal from upper left to lower right, and 0s below the 1s in thefirst and second columns. Our first step in achieving this goal is to get 1 in the topposition of the first column.
To get 1 in this position, we interchange row 1 and row 2: (We could alsointerchange row 1 and row 3 to attain our goal.)
Now we want to get 0s below the 1 in the first column.
To get a 0 where there is now a 3, multiply the top row of numbers by and addthese products to the second row of numbers: To get a 0 where there isnow a 1, multiply the top row of numbers by and add these products to the thirdrow of numbers: Although we are using row 1 to find the products, thenumbers in row 1 do not change.
-1R1 + R3 .-1
-3R1 + R2 .-3
We want 0 inthese positions. C 3 S1
3
1
1
1
3
2
2
2
19
31
25
C 3 S1
3
1
1
1
3
2
2
2
19
31
25
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.
R1 4 R2 .
We want 1 inthis position. C 3 S3
1
1
1
1
3
2
2
2
31
19
25
C C 33 S1
–3(1)+3
–1(1)+1
1
–3(1)+1
–1(1)+3
2
–3(2)+2
–1(2)+2
=19
–3(19)+31
–1(19)+25
S1
–2
2
2
–4
0
19
–26
6
1
0
0
Replace row 2 by−3R1 + R2.
Replace row 3 by−1R1 + R3.
We want 1 in this position.
We move on to the second column.To get 1 in the desired position,we multiply by itsreciprocal, Therefore, we multiply all the numbers in the second row by
C C3 3 SS0
(0)
112–
2
(–2)
1 2
2
0
19
13
6
1
1
2
1
0
0
12–
0
(–4)
212– = .
6
(–26)
1912–
12− R2
We want 0 in this position.
- 12 R2 .-
12 :-
12 .
-2
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 809
810 Chapter 8 Matrices and Determinants
So far, our matrix row operations have resulted in the matrix that we repeated in themargin. We are not yet done with the second column. The voice balloon shows thatwe want to get a 0 where there is now a 2. If we multiply the second row of numbersby and add these products to the third row of numbers, we will get 0 in this posi-tion: Although we are using the numbers in row 2 to find the products,the numbers in row 2 do not change.
-2R2 + R3 .-2
TechnologyMost graphing utilities can convertan augmented matrix to row-echelonform, with 1s down the main diago-nal and 0s below the 1s. However,row-echelon form is not unique.Yourgraphing utility might give a row-echelon form different from the oneyou obtained by hand. However, allrow-echelon forms for a givensystem’s augmented matrix producethe same solution to the system.Enter the augmented matrix andname it Then use the (row-echelon form) command onmatrix A.
� REF �A.
C C 33 S1
0
–2(0)+0
1
1
–2(1)+2
2
2
–2(2)+0
=19
13
–2(13)+6
S1
1
0
2
2
–4
19
13
–20
1
0
0Replace row 3 by
−2R2 + R3.
We want 1 in this position.
We move on to the third column.To get 1 in the desired position, we multiply by itsreciprocal, Therefore, we multiply all the numbers in the third row by
We now have the desired matrix in row-echelon form, with 1s down the main diagonaland 0s below the 1s in the first and second columns.
Step 3 Write the system of linear equations corresponding to the matrix in step 2and use back-substitution to find the system’s solution. The system represented bythe matrix in step 2 is
C1 1 20 1 20 0 1
3 19135S : c 1x + 1y + 2z = 19
0x + 1y + 2z = 130x + 0y + 1z = 5
or c x + y + 2z = 19y + 2z = 13.
z = 5
C C3 3 SS(0)
1
014– (0)
1
1
2
2
1
19
13
5
1
1
0
1
0
014– (–4)
2
214–
=(–20)
19
1314–
14− R3
- 14 R3 .-
14 :-
14 .
-4
The matrix from the bottom of the previous page (repeated)
C 3 S2
2
0
19
13
6
1
1
2
1
0
0
We want 0 in this position.
(1)(2)(3)
We immediately see from equation (3) that the value for is 5. To find weback-substitute 5 for in the second equation.
Equation (2)
Substitute 5 for
Multiply.
Subtract 10 from both sides and solve for
Finally, back-substitute 3 for and 5 for in the first equation.
Equation (1)
Substitute 3 for and 5 for
Multiply and add.Subtract 13 from both sides and solve for
With and the solution set of the original system is Check to see that the solution satisfies all three equations in the given system.
Check Point 2 Use matrices to solve the system:
Modern supercomputers are capable of solving systems with more than 600,000variables. The augmented matrices for such systems are huge, but the solution usingmatrices is exactly like what we did in Example 2. Work with the augmented matrix,
c 2x + y + 2z = 18x - y + 2z = 9x + 2y - z = 6.
516, 3, 526.x = 6,z = 5, y = 3,
x. x = 6 x + 13 = 19
z.y x + 3 + 2152 = 19
x + y + 2z = 19
zy
y. y = 3
y + 10 = 13
z. y + 2152 = 13
y + 2z = 13
zy,z
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 810
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 rightand 0s below the 1s. Let’s see how this works for a linear system involving fourequations in four variables.
Gaussian Elimination with Back-Substitution
Use matrices to solve the system:
SolutionStep 1 Write the augmented matrix for the system.
d2w + x + 3y - z = 6
w - x + 2y - 2z = -1w - x - y + z = -4
-w + 2x - 2y - z = -7.
EXAMPLE 3
Linear System Augmented Matrix
d2w + x + 3y - z = 6
w - x + 2y - 2z = -1w - x - y + z = -4
-w + 2x - 2y - z = -7
D2 1 3 -11 -1 2 -21 -1 -1 1
-1 2 -2 -1
4 6
-1-4-7
T
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 1sin the first, second, and third columns. Our first step in achieving this goal is to get1 in the top position of the first column. To do this, we interchange row 1 and row2:
Now we use the 1 at the top of the first column to get 0s below it.
We move on to the second column. We can obtain 1 in the desired position bymultiplying the numbers in the second row by the reciprocal of 3.
4 41
0
0
=
–1
–3
–8
D DT T
We want 0s in these positions.The top position already has a 0.
(0)13
–1
0
1
(3)13
–2
3
–3
1
0
0
0
–1
1
0
1
2
–3
0
–2
1
3
–3
–1
–3
–8
(3)13 (8)1
3
2
–3
0
(–1)13 – 1
383
13 R2
13 ,
41
0
0
0
–1
3
0
1
2
–1
–3
0
–2
3
3
–3
–1
8
–3
–8
D TUse the previous matrix and:Replace row 2 by −2R1 + R2.Replace row 3 by −1R1 + R3.Replace row 4 by 1R1 + R4.
We want1 in thisposition.
41
2
1
–1
–1
1
–1
2
2
3
–1
–2
–2
–1
1
–1
–1
6
–4
–7
D TWe want0s in thesepositions.
This was row 2; now it’s row 1.
This was row 1; now it’s row 2.
R1 4 R2 .
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 811
812 Chapter 8 Matrices and Determinants
So far, our matrix row operations have resulted in the matrix that we repeated in themargin. Now we use the 1 in the second row, second column position to get 0s below it.
We move on to the third column.We can obtain 1 in the desired position by multiplyingthe numbers in the third row by the reciprocal of -3.-
13 ,
41
0
0
0
–1
1
0
0
2
–3
–2
1
3
–4
–1
–3D T
Replace row 4 in the previousmatrix by −1R2 + R4.
– 13
– 323
13
83
We want1 in thisposition.
We immediately see that the value for is 3. We can now use back-substitution tofind the values for and w.y, x,
z
D1 -1 2 -20 1 -
13 1
0 0 1 -10 0 0 1
4 -1
83
13
T : d1w - 1x + 2y - 2z = -10w + 1x -
13 y + 1z =
83
0w + 0x + 1y - 1z = 10w + 0x + 0y + 1z = 3
or dw - x + 2y - 2z = -1
x -13 y + z =
83
y - z = 1z = 3.
4 41
0
0
=
–1
D DT TWe want0 in thisposition.
(0)13
–1
1
0
–2
1
–4
1
0
0
0
–1
1
0
0
2
1
–2
1
–1
–4
–1
113
83
83
2
(–3)1
3– (0)13– – (–3)1
3–323– 32
3–
(3)13–
13
13
–13–
13 R3−
Now we use the 1 in the third column to get 0 below it.
We move on to the fourth column. Because we want 1s down the diagonal fromupper left to lower right, we want 1 where there is now We can obtain 1 in this position by multiplying the numbers in the fourth row by
We now have the desired matrix in row-echelon form, with 1s down the maindiagonal and 0 s below the 1s. An equivalent row-echelon matrix can be obtained
using a graphing utility and the command on the augmented matrix.
Step 3 Write the system of linear equations corresponding to the matrix in step 2and use back-substitution to find the system’s solution. The system represented bythe matrix in step 2 is
� REF �
4
4
1
0
0
=
–1
1D
D
T
T(0)3
11
–1
1
0
–2
1
–1
1
0
0
0
–1
1
0
0
2
1
0
–2
1
–1
1
–1
1
3
83
2
1
– (0)311– (0)3
11– (–11)311–( )3
11– 113–
– 13
– 13
83
− R4311
- 311 .
- 113 .
41
0
0
0
–1
1
0
0
2
1
0
–2
1
–1
–1
1
–11
D T– 13
83
We want1 in thisposition.
Replace row 4 in the previousmatrix by − R3 + R4.
13 –11
3
The matrix from the bottom of the previous page (repeated)
4D T
We want 0s in these positions.The top position already has a 0.
1
0
0
0
–1
1
0
1
2
–3
0
–2
1
3
–3
–1
–3
–8
– 13
83
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 812
Section 8.1 Matrix Solutions to Linear Systems 813
� Use matrices and Gauss-Jordanelimination to solve systems.
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 matrixin reduced row-echelon form, with 1s down the main diagonal from upperleft 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 secondcolumn 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 thirdcolumn 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’ssolution set. (Back-substitution is not necessary.)
13
83
z=3 x- y+z=y-z=1 w-x+2y-2z=–1
13
83
x- (4)+3=y-3=1 w-1+2(4)-2(3)=–1
53
83
x+ = w-1+8-6=–1
x=1
y=4
w+1=–1
w=–2
9 9 9
These are thefour equations
from the last column.
Let’s agree to write the solution for the system in the alphabetical order of thevariables from left to right, namely Thus, the solution set is
We can verify the solution by substituting the value for each variableinto the original system of equations and obtaining four true statements.
Check Point 3 Use matrices to solve the system:
Solving Linear Systems Using Gauss-Jordan EliminationUsing Gaussian elimination, we obtain a matrix in row-echelon form, with 1s downthe main diagonal and 0s below the 1s. A second method, called Gauss-Jordanelimination, after Carl Friedrich Gauss and Wilhelm Jordan (1842–1899), continuesthe process until a matrix with 1s down the main diagonal and 0s in every positionabove and below each 1 is found. Such a matrix is said to be in reduced row-echelonform. For a system of three linear equations in three variables, and we mustget the augmented matrix into the form
Based on this matrix, we conclude that and z = c.x = a, y = b,
C1 0 00 1 00 0 1
3 abc
S .
z,x, y,
dw - 3x - 2y + z = -3
2w - 7x - y + 2z = 13w - 7x - 3y + 3z = -55w + x + 4y - 2z = 18.
51-2, 1, 4, 326.1w, x, y, z2.
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 813
814 Chapter 8 Matrices and Determinants
Using Gauss-Jordan Elimination
Use Gauss-Jordan elimination to solve the system:
Solution In Example 2,we used Gaussian elimination to obtain the following matrix:
To use Gauss-Jordan elimination, we need 0s both above and below the 1s in themain diagonal. We use the 1 in the second row, second column to get a 0 above it.
We use the 1 in the third column to get 0s above it.
This last matrix corresponds to
As we found in Example 2, the solution set is
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.
516, 3, 526.
x = 6, y = 3, z = 5.
Replace row 2 in the previousmatrix by −2R3 + R2.
1
0
0
0
1
0
6
3
5
0
0
1C 3 S
C 3 S1
0
0
0
1
0
0
2
1
6
13
5
Replace row 1in the previous
matrix by−1R2 + R1.
We want0s in thesepositions.
C 3 S .1
0
0
1
1
0
2
2
1
19
13
5
We want0s in thesepositions.
c 3x + y + 2z = 31x + y + 2z = 19x + 3y + 2z = 25.
EXAMPLE 4
Exercise Set 8.1
Practice ExercisesIn Exercises 1–8, write the augmented matrix for each system oflinear equations.
1. 2.
3. 4.
5. 6. c x - 2y + z = 103x + y = 57x + 2z = 2
c 5x - 2y - 3z = 0x + y = 5
2x - 3z = 4
c x - 2y + 3z = 9y + 3z = 5
z = 2c x - y + z = 8
y - 12z = -15z = 1
c 3x - 2y + 5z = 31x + 3y - 3z = -12
-2x - 5y + 3z = 11c 2x + y + 2z = 2
3x - 5y - z = 4x - 2y - 3z = -6
In Exercises 9–12, write the system of linear equations representedby the augmented matrix. Use or, if necessary,and for the variables.
9. 10.
11. 12.
In Exercises 13–18, perform each matrix row operation and writethe new matrix.
13. C2 -6 41 5 -53 0 4
3 1007S 1
2 R1
D4 1 5 11 -1 0 -13 0 0 70 0 11 5
4 6843
TD1 1 4 1
-1 1 -1 02 0 0 50 0 12 4
4 37
115
T
C7 0 40 1 -52 7 0
3 -13116SC5 0 3
0 1 -47 2 0
3 -11123S
z,w, x, y,x, y, and z,
7. 8. d4w + 7x - 8y + z = 3
5x + y = 5w - x - y = 17
2w - 2x + 11y = 4
d2w + 5x - 3y + z = 2
3x + y = 4w - x + 5y = 9
5w - 5x - 2y = 1
TechnologyMost graphing utilities can convert amatrix to reduced row-echelon form.Enter the system’s augmented matrixand name it Then use the (reduced row-echelon form) com-mand on matrix
This is the matrixin reduced row-echelon form weobtained inExample 4.
This is theaugmented matrixfor the systemin Example 4.
A.
� RREF �A.
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 814
Section 8.1 Matrix Solutions to Linear Systems 815
14.
15.
16.
17.
18.
In Exercises 19–20, a few steps in the process of simplifying thegiven matrix to row-echelon form, with 1s down the diagonal fromupper left to lower right, and 0s below the 1s, are shown. Fill in themissing numbers in the steps that are shown.
19.
20.
In Exercises 21–38, solve each system of equations using matrices.Use Gaussian elimination with back-substitution or Gauss-Jordanelimination.
21. 22.
23. 24.
25. 26.
27. 28. c 3x + y - z = 0x + y + 2z = 6
2x + 2y + 3z = 10c x + y + z = 4
x - y - z = 0x - y + z = 2
c x - 3z = -22x + 2y + z = 43x + y - 2z = 5
c 2x - y - z = 4x + y - 5z = -4x - 2y = 4
c 3y - z = -1x + 5y - z = -4
-3x + 6y + 2z = 11
c x + 3y = 0x + y + z = 1
3x - y - z = 11
c x - 2y - z = 22x - y + z = 4-x + y - 2z = -4
c x + y - z = -22x - y + z = 5-x + 2y + 2z = 1
: C1 -2 30 1 n
0 -2 n
3 4n
n
SC 1 -2 3
2 1 -4-3 4 -1
3 43
-2S : C1 -2 3
0 5 n
0 -2 n
3 4n
n
S: C1 -1 1
0 1 n
0 1 n
3 8n
n
SC 1 -1 1
2 3 -13 -2 -9
3 8-2
9S : C1 -1 1
0 5 n
0 1 n
3 8n
n
S
D1 -5 2 -20 1 -3 -13 0 2 -1
-4 1 4 2
4 406
-3
T
-3R1 + R3
4R1 + R4
D1 -1 1 10 1 -2 -12 0 3 45 1 2 4
4 30
116
T
-2R1 + R3
-5R1 + R4
C 1 -1 53 3 -11 3 2
3 -6105S -3R1 + R2
C 1 -3 23 1 -12 -2 1
3 073S -3R1 + R2
C 3 -12 61 -4 42 0 7
3 904S 1
3 R1
29. 30.
31. 32.
33. 34.
35.
36.
37.
38.
Practice Plus39. Find the quadratic function for which
and
40. Find the quadratic function for whichand
41. Find the cubic function forwhich and
42. Find the cubic function forwhich and
43. Solve the system:
(Hint: Let and Solvethe system for and Then use the logarithmicequations to find and )
44. Solve the system:
(Hint: Let and Solvethe system for and Then use the logarithmicequations to find and )z.w, x, y,
D.A, B, C,D = ln z.A = ln w, B = ln x, C = ln y,
dln w + ln x + ln y + ln z = -1
- ln w + 4 ln x + ln y - ln z = 0ln w - 2 ln x + ln y - 2 ln z = 11
- ln w - 2 ln x + ln y + 2 ln z = -3.
z.w, x, y,D.A, B, C,
D = ln z.A = ln w, B = ln x, C = ln y,
d2 ln w + ln x + 3 ln y - 2 ln z = -64 ln w + 3 ln x + ln y - ln z = -2
ln w + ln x + ln y + ln z = -5 ln w + ln x - ln y - ln z = 5.
f132 = 7.f1-12 = 3, f112 = 1, f122 = 6,f1x2 = ax3
+ bx2+ cx + d
f132 = 12.f1-12 = 0, f112 = 2, f122 = 3,f1x2 = ax3
+ bx2+ cx + d
f122 = 5.f1-12 = 5, f112 = 3,f1x2 = ax2
+ bx + c
f122 = 0.f1-22 = -4, f112 = 2,f1x2 = ax2
+ bx + c
d2w + y - 3z = 8
w - x + 4z = -103w + 5x - y - z = 20
w + x - y - z = 6
d3w - 4x + y + z = 9
w + x - y - z = 02w + x + 4y - 2z = 3-w + 2x + y - 3z = 3
dw + x + y + z = 5w + 2x - y - 2z = -1w - 3x - 3y - z = -1
2w - x + 2y - z = -2
dw + x + y + z = 4
2w + x - 2y - z = 0w - 2x - y - 2z = -2
3w + 2x + y + 3z = 4
c 3x + 2y + 3z = 34x - 5y + 7z = 12x + 3y - 2z = 6
c 2x + 2y + 7z = -12x + y + 2z = 24x + 6y + z = 15
c 3a + b - c = 02a + 3b - 5c = 1
a - 2b + 3c = -4c 3a - b - 4c = 3
2a - b + 2c = -8a + 2b - 3c = 9
c 2x + y = z + 12x = 1 + 3y - z
x + y + z = 4c x + 2y = z - 1
x = 4 + y - z
x + y - 3z = -2
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 815
816 Chapter 8 Matrices and Determinants
Application Exercises45. A ball is thrown straight upward. A position function
can be used to describe the ball’s height, in feet, after seconds.
a. Use the points labeled in the graph to find the values ofand Solve the system of linear equations involvingand using matrices.
b. Find and interpret Identify your solution as apoint on the graph shown.
c. After how many seconds does the ball reach its maximumheight? What is its maximum height?
46. A football is kicked straight upward. A position function
can be used to describe the ball’s height, in feet, after seconds.
a. Use the points labeled in the graph to find the values ofand Solve the system of linear equations involvingand using matrices.
b. Find and interpret Identify your solution as a pointon the graph shown.
c. After how many seconds does the ball reach its maximumheight? What is its maximum height?
Write a system of linear equations in three or four variables tosolve Exercises 47–50. Then use matrices to solve the system.
47. Three foods have the following nutritional content per ounce.
s172.s0a, v0 ,s0 .a, v0 ,
1050
s(t)
t
Time (seconds)
50
100
150
200
250
300
Hei
ght a
bove
Gro
und
(fee
t)
(5, 246)
(2, 198)
(8, 6)
ts1t2,
s1t2 =12 at2
+ v0t + s0
s13.52.
s0a, v0 ,s0 .a, v0 ,
60
50
40
30
20
54321
10
s(t)
t
Time (seconds)
Hei
ght a
bove
Gro
und
(fee
t)
(3, 24)
(2, 48)(1, 40)
ts1t2,
s1t2 =12 at2
+ v0t + s0
If a meal consisting of the three foods allows exactly660 calories, 25 grams of protein, and 425 milligrams of vita-min C, how many ounces of each kind of food should be used?
48. A furniture company produces three types of desks: a chil-dren’s model, an office model, and a deluxe model. Eachdesk is manufactured in three stages: cutting, construction,and finishing. The time requirements for each model andmanufacturing stage are given in the following table.
CaloriesProtein
(in grams)Vitamin C
(in milligrams)
Food A 40 5 30
Food B 200 2 10
Food C 400 4 300
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
Earth’s Population as a Village of 200 People
150
50
75
100
125
60
14
w
x yz
32
125N
umbe
r of
Peo
ple Under
age 15
Overage 65
AfricanEuropean
American(U.S.)
Unableto reador write
Eat atMcDonald’seach day
Asian
Source: Gary Rimmer, Number Freaking,The Disinformation Company Ltd., 2006
Each week the company has available a maximum of 100hours for cutting, 100 hours for construction, and 65 hours forfinishing. If all available time must be used, how many ofeach type of desk should be produced each week?
49. Imagine the entire global population as a village of precisely200 people. The bar graph shows some numeric observationsbased on this scenario.
150
125
100
75
50
The U.S. White House by the Numbers
w
25
xy
z
Bathrooms Fireplaces
Elevators
Rooms
Source: The White House
Combined, there are 183 Asians, Africans, Europeans, andAmericans in the village. The number of Asians exceeds thenumber of Africans and Europeans by 70. The differencebetween the number of Europeans and Americans is 15. Ifthe number of Africans is doubled, their population exceedsthe number of Europeans and Americans by 23. Determinethe number of Asians,Africans, Europeans, and Americans inthe global village.
50. The bar graph shows the number of rooms, bathrooms,fireplaces, and elevators in the U.S. White House.
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 816
Section 8.1 Matrix Solutions to Linear Systems 817
Combined, there are 198 rooms, bathrooms, fireplaces, andelevators. The number of rooms exceeds the number ofbathrooms and fireplaces by 69. The difference betweenthe number of fireplaces and elevators is 25. If the numberof bathrooms is doubled, it exceeds the number of fire-places and elevators by 39. Determine the number ofrooms, bathrooms, fireplaces, and elevators in the U.S.White House.
Writing in Mathematics51. What is a matrix?
52. Describe what is meant by the augmented matrix of a systemof linear equations.
53. In your own words, describe each of the three matrix rowoperations. Give an example with each of the operations.
54. Describe how to use row operations and matrices to solve asystem of linear equations.
55. What is the difference between Gaussian elimination andGauss-Jordan elimination?
Technology Exercises56. Most graphing utilities can perform row operations on
matrices. Consult the owner’s manual for your graphingutility to learn proper keystrokes for performing theseoperations. Then duplicate the row operations of any threeexercises that you solved from Exercises 13–18.
57. If your graphing utility has a (row-echelon form)
command or a (reduced row-echelon form) command, use this feature to verify your work with any fivesystems that you solved from Exercises 21–38.
58. Solve using a graphing utility’s or command:
Critical Thinking ExercisesMake Sense? In Exercises 59–62, determine whether eachstatement makes sense or does not make sense, and explainyour reasoning.
59. Matrix row operations remind me of what I did when solvinga linear system by the addition method, although I no longerwrite the variables.
60. When I use matrices to solve linear systems, the only arithmeticinvolves multiplication or a combination of multiplication andaddition.
e2x1 - 2x2 + 3x3 - x4 = 12x1 + 2x2 - x3 + 2x4 - x5 = -7x1 + x3 + x4 - 5x5 = 1
-x1 + x2 - x3 - 2x4 - 3x5 = 0x1 - x2 - x4 + x5 = 4.
� RREF �� REF �
� RREF �� REF �
61. When I use matrices to solve linear systems, I spend most of mytime using row operations to express the system’s augmentedmatrix in row-echelon form.
62. Using row operations on an augmented matrix, I obtain a rowin which 0s appear to the left of the vertical bar, but 6 appearson 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 atrue statement.
63. A matrix row operation such as is not permittedbecause of the negative fraction.
64. The augmented matrix for the system
65. In solving a linear system of three equations in threevariables, we begin with the augmented matrix and use rowoperations to obtain a row-equivalent matrix with 0s downthe diagonal from left to right and 1s below each 0.
66. The row operation indicates that it is the elementsin row that change.
67. The table shows the daily production level and profit for abusiness.
ikRi + Rj
x - 3y = 5y - 2z = 7
2x + z = 4 is C1 -3
1 -22 1
3 574S .
- 45 R1 + R2
(Number of Units Produced Daily)x
30 50 100
(Daily Profit)y $5900 $7500 $4500
Use the quadratic function to determinethe number of units that should be produced each day formaximum profit. What is the maximum daily profit?
Preview Exercises
Exercises 68–70 will help you prepare for the material covered inthe next section. In each exercise, refer to the following system:
68. Show that satisfies the system for
69. Show that satisfies the system for
70. a. Select a value for other than 0 or 1 and show thatsatisfies the system.
b. Based on your work in Exercises 68–70(a), how does thissystem differ from those in Exercises 21–34?
112z + 1, 10z - 1, z2z
z = 1.112z + 1, 10z - 1, z2
z = 0.112z + 1, 10z - 1, z2
c 3x - 4y + 4z = 7x - y - 2z = 2
2x - 3y + 6z = 5.
y = ax2+ bx + c
P-BLTZMC08_805-872-hr 21-11-2008 13:26 Page 817
