CHAPTER 1 Systems of Linear Equations - Cengage...CHAPTER 1 Systems of Linear Equations Section 1.1 Introduction to Systems of Linear Equations 1. Because the equation is in the form
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Adding 2 times the first equation to the second equation produces a new second equation.
10 7
x y− =
=
Because the second equation is a false statement, you can conclude that the original system of equations has no solution. Geometrically, the two lines are parallel.
21. 3 5 72 9
x yx y− =
+ =
Adding the first equation to 5 times the second equation produces a new second equation, 13 52,x = or 4.x = So, ( )2 4 9,y+ = or 1,y = and the solution is: 4,x =
1.y = This is the point where the two lines intersect.
23. 2 55 11
x yx y− =
− =
Subtracting the first equation from the second equation produces a new second equation, 3 6,x = or 2.x = So, ( )2 2 5,y− = or 1,y = − and the solution is: 2,x =
1.y = − This is the point where the two lines intersect.
25. 3 1 14 3
2 12
x y
x y
+ −+ =
− =
Multiplying the first equation by 12 produces a new first equation.
3 4 72 12
x yx y+ =
− =
Adding the first equation to 4 times the second equation produces a new second equation, 11 55,x = or
5.x = So, ( )2 5 12,y− = or 2,y = − and the solution
is: 5,x = 2.y = − This is the point where the two lines intersect.
27. 0.05 0.03 0.070.07 0.02 0.16
x yx y− =
+ =
Multiplying the first equation by 200 and the second equation by 300 produces new equations.
10 6 1421 6 48
x yx y− =
+ =
Adding the first equation to the second equation produces a new second equation, 31 62,x = or 2.x = So, ( )10 2 6 14,y− = or 1,y = and the solution is:
2,x = 1.y = This is the point where the two lines intersect.
83. To begin, reduce the system to row-echelon form.
( )2 6
8 3 14
x y kz
k z
+ + =
− = −
This system will have no solution if 8 3 0,k− = that is, 83.k =
85. Reducing the system to row-echelon form, you have
( ) ( )( ) ( )2
3
1 1 1
1 1 1 3
x y kz
k y k z
k y k z k
+ + =
− + − = −
− + − = −
( ) ( )( )2
3
1 1 1
.2 3
x y kz
k y k z
k k z k
+ + =
− + − = −
− − + = −
If 2 2 0,k k− − + = then there is no solution. So, if 1k = or 2,k = − there is not a unique solution.
87. (a) All three of the lines will intersect in exactly one point (corresponding to the solution point). (b) All three of the lines will coincide (every point on
these lines is a solution point). (c) The three lines have no common point.
89. Answers vary. ( :Hint Choose three different values for x
and solve the resulting system of linear equations in the variables a, b, and ).c
91. 4 35 6 13
x yx y− = −
− =
4 314 28
x yy
− = −
=
4 32
x yy
− = −
=
52
xy=
=
At each step, the lines always intersect at ( )5, 2 , which is
the solution to the system of equations.
93. Solve each equation for y.
1100199
2
2
y x
y x
= +
= −
The graphs are misleading because, while they appear parallel, when the equations are solved for y they have slightly different slopes.
Section 1.2 Gaussian Elimination and Gauss-Jordan Elimination
1. Because the matrix has 3 rows and 3 columns, it has size 3 3.×
3. Because the matrix has 2 rows and 4 columns, it has size 2 4.×
5. Because the matrix has 1 row and 5 columns, it has size 1 5.×
7. Because the matrix has 4 rows and 5 columns, it has size 4 5.×
9. The matrix satisfies all three conditions in the definition of row-echelon form. Moreover, because each column that has a leading 1 (columns one and two) has zeros elsewhere, the matrix is in reduced row-echelon form.
11. Because the matrix has two non-zero rows without leading 1’s, it is not in row-echelon form.
13. Because the matrix has a non-zero row without a leading 1, it is not in row-echelon form.
15. Because the matrix is in reduced row-echelon form, convert back to a system of linear equations
1 1 0 2 1 0 0 10 1 0 1 0 1 0 10 0 1 1 0 0 1 10 0 0 0 0 0a b c a b c
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
+ + + +⎣ ⎦ ⎣ ⎦
Converting back to a system of linear equations 1
11
0 .
xyz
a b c
==== + +
The system (a) will have a unique solution if 0,a b c+ + = (b) will have no solution if 0,a b c+ + ≠ and (c) cannot have an infinite number of solutions.
For each of these, examine the possible second rows.
0 0 0 1 1 0 1
, , ,0 0 0 0 0 1 0 0
k⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦
These represent all possible 2 2× reduced row-echelon matrices.
57. (a) True. In the notation ,m n m× is the number of rows of the matrix. So, a 6 3× matrix has six rows.
(b) True. On page 19, after Example 4, the sentence reads, “It can be shown that every matrix is row-equivalent to a matrix in row-echelon form.”
(c) False. Consider the row-echelon form
1 0 0 0 00 1 0 0 10 0 1 0 20 0 0 1 3
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
which gives the solution 1 2 30, 1, 2,x x x= = =
4and 3.x =
(d) True. Theorem 1.1 states that if a homogeneous system has fewer equations than variables, then it must have an infinite number of solutions.
59. First, a and c cannot both be zero. So, assume 0,a ≠ and use row reduction as follows.
00
a ba b a bcbc d ad bcda
⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⇒ ⇒−⎢ ⎥ ⎢ ⎥⎢ ⎥ −+⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦
So, 0.ad bc− ≠ Similarly, if 0,c ≠ interchange rows and proceed as above. So the original matrix is row equivalent to the identity if and only if 0.ad bc− ≠
61. Form the augmented matrix for this system
2 1 0
1 2 0λ
λ−⎡ ⎤
⎢ ⎥−⎣ ⎦
and reduce the system using elementary row operations.
2
1 2 0 1 2 02 1 0 0 4 3 0
λ λλ λ λ
− −⎡ ⎤ ⎡ ⎤⇒⎢ ⎥ ⎢ ⎥− − +⎣ ⎦ ⎣ ⎦
To have a nontrivial solution you must have
( )( )
2 4 3 0
1 3 0.
λ λ
λ λ
− + =
− − =
So, if 1λ = or 3,λ = the system will have nontrivial solutions.
63. To show that it is possible you need give only one example, such as
1 2 3
1 2 3
01
x x xx x x
+ + =
+ + =
which has fewer equations than variables and obviously has no solution.
Section 1.3 Applications of Systems of Linear Equations 15
The rows have been interchanged. In general, the second and third elementary row operations can be used in this manner to interchange two rows of a matrix. So, the first elementary row operation is, in fact, redundant.
67. When a system of linear equations is inconsistent, the row-echelon form of the corresponding augmented matrix will have a row that is all zeros except for the last entry.
69. A matrix in reduced row-echelon form has zeros above each row’s leading 1, which may not be true for a matrix in row-echelon form.
Section 1.3 Applications of Systems of Linear Equations
1. (a) Because there are three points, choose a second- degree polynomial, ( ) 2
0 1 2 .p x a a x a x= + + Then
substitute 2, 3,x = and 4 into ( )p x and equate the
results to 5, 2,y = and 5, respectively.
( ) ( )( ) ( )( ) ( )
20 1 2 0 1 2
20 1 2 0 1 2
20 1 2 0 1 2
2 2 2 4 5
3 3 3 9 2
4 4 4 16 5
a a a a a a
a a a a a a
a a a a a a
+ + = + + =
+ + = + + =
+ + = + + =
Form the augmented matrix
1 2 4 51 3 9 21 4 16 5
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix
1 0 0 290 1 0 18 .0 0 1 3
⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
So, ( ) 229 18 3 .p x x x= − +
(b)
3. (a) Because there are three points, choose a second- degree polynomial, ( ) 2
0 1 2 .p x a a x a x= + + Then
substitute 2, 3,x = and 5 into ( )p x and equate the
results to 4, 6,y = and 10, respectively.
( ) ( )( ) ( )( ) ( )
20 1 2 0 1 2
20 1 2 0 1 2
20 1 2 0 1 2
2 2 2 4 4
3 3 3 9 6
5 5 5 25 10
a a a a a a
a a a a a a
a a a a a a
+ + = + + =
+ + = + + =
+ + = + + =
Use Gauss-Jordan elimination on the augmented matrix for this system.
5. (a) Using the translation 2007,z x= − the points ( ),z y are ( ) ( )1, 5 , 0, 7 ,− and ( )1, 12 . Because there are three points,
choose a second-degree polynomial ( ) 20 1 2 .p z a a z a z= + + Then substitute 1, 0,z = − and 1 into ( )p z and equate the
results to 5, 7,y = and 12 respectively.
( ) ( )( ) ( )( ) ( )
20 1 2 0 1 2
20 1 2 0
20 1 2 0 1 2
1 1 5
0 0 7
1 9 1 12
a a a a a a
a a a a
a a a a a
+ − + − = − + =
+ + = =
+ + = + + =
Use Gauss-Jordan elimination on the augmented matrix for this system.
7232
1 1 1 5 1 0 0 71 0 0 7 0 1 01 1 1 12 0 0 1
⎡ ⎤−⎡ ⎤⎢ ⎥⎢ ⎥ ⇒ ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦
So, ( ) 27 32 27 .p z z z= + +
Letting 2007,z x= − you have ( ) ( ) ( )27 32 27 2007 2007 .p x x x= + − + −
(b)
7. Choose a fourth-degree polynomial and substitute 1, 2, 3,x = and 4 into
( ) 2 3 40 1 2 3 4 .p x a a x a x a x a x= + + + + However,
when you substitute 3x = into ( )p x and equate it to
2y = and 3y = you get the contradictory equations
0 1 2 3 4
0 1 2 3 4
3 9 27 81 23 9 27 81 3
a a a a aa a a a a
+ + + + =
+ + + + =
and must conclude that the system containing these two equations will have no solution. Also, y is not a function of x because the x-value of 3 is repeated. By similar reasoning, you cannot choose ( ) 2 3 4
0 1 2 3 4p y b b y b y b y b y= + + + + because
1y = corresponds to both 1x = and 2.x =
9. Letting ( ) 20 1 2 ,p x a a x a x= + + substitute 0, 2,x =
and 4 into ( )p x and equate the results to 1, 3,y = and
5, respectively.
( ) ( )( ) ( )( ) ( )
20 1 2 0
20 1 2 0 1 2
20 1 2 0 1 2
0 0 1
2 2 2 4 3
4 4 4 16 5
a a a a
a a a a a a
a a a a a a
+ + = =
+ + = + + =
+ + = + + =
Use Gauss-Jordan elimination on the augmented matrix for this system.
1 0 0 1 1 0 0 11 2 4 3 0 1 0 11 4 16 5 0 0 1 0
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
So, ( ) 1 .p x x= +
The graphs of ( ) ( )1 1 1y p x x= = + and that of the
function 27 115 151y x x= − + are shown below.
z
3
9
12
−1 1(2006) (2008)(2007)
(−1, 5)
(0, 7)
(1, 12)
y
−1 2 4
3
3
4
5
x(0, 1)
2, 4,1 13 5
y = − + 1 x2 7x15 15
) )) )
y
1
y = 11 + x
Section 1.3 Applications of Systems of Linear Equations 17
So, ( ) 2 311,041.25 1606.5 613.25 45 .p z z z z= + + −
Letting 2000,z x= − ( ) ( ) ( ) ( )2 311,041.25 1606.5 2000 613.25 2000 45 2000 .p x x x x= + − + − − −
Because the actual net profit increased each year from 2000 2007− except for 2006 and the predicted values decrease each year after 2008, the solution does not produce a reasonable model for predicting future net profits.
17. Choosing a second-degree polynomial approximation,
( ) 20 1 2 ,p x a a x a x= + + substitute 0, ,
2x π= and
π into ( )p x and equate the results to 0, 1, and 0,y =
respectively.
0
22
0 1
20 1 2
0
12 4
0
a
a a a
a a a
π π
π π
=
+ + =
+ + =
Then form the augmented matrix,
2
2
1 0 0 0
1 12 4
1 0
π π
π π
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix
2
1 0 0 040 1 0 .
40 0 1
π
π
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥
−⎢ ⎥⎢ ⎥⎣ ⎦
So, ( ) ( )2 22 2
4 4 4 .p x x x x xππ π π
= − = −
Furthermore,
sin3π
≈2
2
2
2
43 3 3
4 2 8 0.889.9 9
p π π πππ
ππ
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦
⎡ ⎤= = ≈⎢ ⎥
⎣ ⎦
Note that sin 3 0.866π = to three significant digits.
19. (i) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
20 1 2 0 1 2
20 1 2 0
20 1 2 0 1 2
1 1 1
0 0 0
1 1 1
p a a a a a a
p a a a a
p a a a a a a
− = + − + − = − +
= + + =
= + + = + +
(ii) 0 1 2
0
0 1 2
000
a a aaa a a
− + =
=
+ + =
(iii) From the augmented matrix
1 1 1 01 0 0 01 1 1 0
−⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
and use Gauss-Jordan elimination to obtain the equivalent reduced row-echelon matrix
1 0 0 00 1 0 0 .0 0 1 0
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
So, 0 10, 0,a a= = and 2 0.a =
Section 1.3 Applications of Systems of Linear Equations 19
which is equivalent to the reduced row-echelon matrix
1 1 0
.0 0 1
−⎡ ⎤⎢ ⎥⎣ ⎦
Because the second row corresponds to 0 1,= which is a false statement, you can conclude that the system has no solution.
19. Multiplying both equations by 100 and forming the augmented matrix produces
20 30 14
.40 50 20⎡ ⎤⎢ ⎥⎣ ⎦
Use Gauss-Jordan elimination as shown below.
3 7 13 7 3 7
22 102 10 2 104455
1 011 10 10 140 50 20 0 10 8
⎡ ⎤ ⎡ ⎤−⎡ ⎤ ⎡ ⎤⇒ ⇒ ⇒⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥
− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦
So, the solution is: 112x = − and 2
45.x =
21. Expanding the second equation, 3 2 0,x y+ = the augmented matrix for this system is
1 12 3 03 2 0
⎡ ⎤−⎢ ⎥⎢ ⎥⎣ ⎦
which is equivalent to the reduced row-echelon matrix
1 0 0
.0 1 0⎡ ⎤⎢ ⎥⎣ ⎦
So, the solution is: 0x = and 0.y =
23. Because the matrix has 2 rows and 3 columns, it has size 2 3.×
25. This matrix has the characteristic stair step pattern of leading 1’s so it is in row-echelon form. However, the leading 1 in row three of column four has 1’s above it, so the matrix is not in reduced row-echelon form.
27. Because the first row begins with 1,− this matrix is not in row-echelon form.
29. This matrix corresponds to the system 1 2
3
2 00.
x xx
+ =
=
Choosing 2x t= as the free variable you can describe the solution as 1 2 ,x t= − 2 ,x t= and 3 0,x = where t is a real number.
31. The augmented matrix for this system is
1 1 2 12 3 1 25 4 2 4
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
which is equivalent to the reduced row-echelon matrix
1 0 0 20 1 0 3 .0 0 1 3
⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
So, the solution is: 2, 3,x y= = − and 3.z =
33. Use Gauss-Jordan elimination on the augmented matrix.
1213
2 3 3 3 1 0 06 6 12 13 0 1 0
12 9 1 2 0 0 1 1
⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥ ⇒ −⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥−⎣ ⎦ ⎣ ⎦
So, 12,x = 1
3,y = − and 1.z =
35. The augmented matrix for this system is
1 2 1 62 3 0 71 3 3 11
− −⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥− −⎣ ⎦
which is equivalent to the reduced row-echelon matrix
1 0 3 40 1 2 5 .0 0 0 0
−⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦
Choosing z t= as the free variable you find that the solution set can be described by 4 3 ,x t= +
(a) There will be no solution if 2 0b a− = and 9 3 0.a− − ≠ That is if 2b a= and 3.a = − (b) There will be exactly one solution if 2 .b a≠ (c) There will be an infinite number of solutions if 2b a= and 3.a = − That is, if 3a = − and 6.b = −
55. You can show that two matrices of the same size are row equivalent if they both row reduce to the same matrix. The two given matrices are row equivalent because each is row equivalent to the identity matrix.
57. Adding a multiple of row one to each row yields the following matrix.