Solution manual For Elementary Linear Algebra 12e Anton

, Solution manual For Elementary Linear
Algebra 12e Anton
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, 1.1 Introduction to Systems of Linear Equations 1




1.1 Introduction to Systems of Linear Equations

1. (a) This is a linear equation in x1 , x2 , and x3 .

(b) This is not a linear equation in x1 , x2 , and x3 because of the term x1 x3 .

(c) We can rewrite this equation in the form x1  7 x2  3 x3  0 therefore it is a linear equation in x1 , x2 , and x3 .

(d) This is not a linear equation in x1 , x2 , and x3 because of the term x12 .

(e) This is not a linear equation in x1 , x2 , and x3 because of the term x13/5 .

(f) This is a linear equation in x1 , x2 , and x3 .

2. (a) This is a linear equation in x and y .

(b) This is not a linear equation in x and y because of the terms 2x1/3 and 3 y .

(c) This is a linear equation in x and y .

(d) This is not a linear equation in x and y because of the term 7
cos x .

(e) This is not a linear equation in x and y because of the term xy .

(f) We can rewrite this equation in the form  x  y  7 thus it is a linear equation in x and y .

3. (a) a11 x1  a12 x2  b1
a21 x1  a22 x2  b2

(b) a11 x1  a12 x2  a13 x3  b1
a21 x1  a22 x2  a23 x3  b2
a31 x1  a32 x2  a33 x3  b3

(c) a11 x1  a12 x2  a13 x3  a14 x4  b1
a21 x1  a22 x2  a23 x3  a24 x4  b2



4. (a) (b) (c)
 a11 a12 b1   a11 a12 a13 b1   a11 a12 a13 a14 b1 
a b2  a
 21 a22
 21 a22 a23 b2  a
 21 a22 a23 a24 b2 
 a31 a32 a33 b3 

, 1.1 Introduction to Systems of Linear Equations 2


5. (a) (b)
2 x1  0 3 x1  2 x3  5
3 x1  4 x2  0 7 x1  x2  4 x3  3
x2  1  2 x2  x3  7


6. (a) (b)
3 x2  x3  x4  1 3 x1  x3  4 x4  3
5 x1  2 x2  3 x4  6 4 x1  4 x3  x4  3
 x1  3 x2  2 x4  9
 x4  2


7. (a) (b) (c)
 2 6  6 1 3 4   0 2 0 3 1 0
 3 8 0  3 1 1 0 0 1
   5 1 1   
 9 3   6 2 1 2 3 6 



8. (a) (b) (c)
 3 2 1  2 0 2 1 1 0 0 1 
4  3 1 4 7  0 1 0 2 
 5 3    
 7 3 2   6 1 1 0  0 0 1 3 



9. The values in (a), (d), and (e) satisfy all three equations – these 3-tuples are solutions of the system.
The 3-tuples in (b) and (c) are not solutions of the system.
10. The values in (b), (d), and (e) satisfy all three equations – these 3-tuples are solutions of the system.
The 3-tuples in (a) and (c) are not solutions of the system.
11. (a) We can eliminate x from the second equation by adding 2 times the first equation to the second. This yields
the system

3x  2 y  4
0  1

The second equation is contradictory, so the original system has no solutions. The lines represented by the
equations in that system have no points of intersection (the lines are parallel and distinct).
(b) We can eliminate x from the second equation by adding 2 times the first equation to the second. This yields
the system

2x  4y  1
0  0

, 1.1 Introduction to Systems of Linear Equations 3


The second equation does not impose any restriction on x and y therefore we can omit it. The lines
represented by the original system have infinitely many points of intersection. Solving the first equation for x
we obtain x  12  2 y . This allows us to represent the solution using parametric equations

1
x  2t , yt
2

where the parameter t is an arbitrary real number.

(c) We can eliminate x from the second equation by adding 1 times the first equation to the second. This yields
the system

x  2y  0
 2y  8

From the second equation we obtain y  4 . Substituting 4 for y into the first equation results in x  8 .
Therefore, the original system has the unique solution
x  8, y  4

The represented by the equations in that system have one point of intersection:  8, 4  .

12. We can eliminate x from the second equation by adding 2 times the first equation to the second. This yields
the system

2 x  3y  a
0  b  2a

If b  2 a  0 (i.e., b  2 a ) then the second equation imposes no restriction on x and y ; consequently, the
system has infinitely many solutions.
If b  2 a  0 (i.e., b  2 a ) then the second equation becomes contradictory thus the system has no solutions.
There are no values of a and b for which the system has one solution.

13. (a) Solving the equation for x we obtain x  37  75 y therefore the solution set of the original equation can be
described by the parametric equations

3 5
x  t, yt
7 7

where the parameter t is an arbitrary real number.

(b) Solving the equation for x1 we obtain x1  37  35 x2  34 x3 therefore the solution set of the original equation can
be described by the parametric equations

7 5 4
x1   r  s, x2  r , x3  s
3 3 3

where the parameters r and s are arbitrary real numbers.

, 1.1 Introduction to Systems of Linear Equations 4


(c) Solving the equation for x1 we obtain x1   81  14 x2  85 x3  43 x4 therefore the solution set of the original
equation can be described by the parametric equations

1 1 5 3
x1    r  s  t , x2  r , x3  s, x4  t
8 4 8 4

where the parameters r , s , and t are arbitrary real numbers.

(d) Solving the equation for v we obtain v  83 w  23 x  13 y  43 z therefore the solution set of the original equation
can be described by the parametric equations

8 2 1 4
v  t1  t2  t3  t 4 , w  t1 , x  t2 , y  t3 , z  t4
3 3 3 3

where the parameters t1 , t2 , t3 , and t4 are arbitrary real numbers.

14. (a) Solving the equation for x we obtain x  2  10 y therefore the solution set of the original equation can be
described by the parametric equations
x  2  10t, yt

where the parameter t is an arbitrary real number.

(b) Solving the equation for x1 we obtain x1  3  3 x2  12 x3 therefore the solution set of the original equation can
be described by the parametric equations
x1  3  3r  12 s, x2  r , x3  s

where the parameters r and s are arbitrary real numbers.

(c) Solving the equation for x1 we obtain x1  5  12 x2  34 x3  14 x4 therefore the solution set of the original
equation can be described by the parametric equations

1 3 1
x1  5  r  s  t, x2  r , y  s, zt
2 4 4

where the parameters r , s , and t are arbitrary real numbers.
(d) Solving the equation for v we obtain v  w  x  5 y  7z therefore the solution set of the original equation
can be described by the parametric equations
v  t1  t2  5t3  7t 4 , w  t1 , x  t2 , y  t3 , z  t4

where the parameters t1 , t2 , t3 , and t4 are arbitrary real numbers.

15. (a) We can eliminate x from the second equation by adding 3 times the first equation to the second. This yields
the system

2 x  3y  1
0  0

, 1.1 Introduction to Systems of Linear Equations 5


The second equation does not impose any restriction on x and y therefore we can omit it. Solving the first
equation for x we obtain x  12  23 y . This allows us to represent the solution using parametric equations

1 3
x  t, yt
2 2

where the parameter t is an arbitrary real number.
(b) We can see that the second and the third equation are multiples of the first: adding 3 times the first equation
to the second, then adding the first equation to the third yields the system
x1  3 x2  x3  4
00
00

The last two equations do not impose any restriction on the unknowns therefore we can omit them. Solving the
first equation for x1 we obtain x1  4  3 x2  x3 . This allows us to represent the solution using parametric
equations
x1  4  3r  s, x2  r , x3  s

where the parameters r and s are arbitrary real numbers.

16. (a) We can eliminate x1 from the first equation by adding 2 times the second equation to the first. This yields
the system
00

3 x1  x2  4

The first equation does not impose any restriction on x1 and x2 therefore we can omit it. Solving the second
equation for x1 we obtain x1   43  13 x2 . This allows us to represent the solution using parametric equations

4 1
x1    t, x2  t
3 3

where the parameter t is an arbitrary real number.
(b) We can see that the second and the third equation are multiples of the first: adding 3 times the first equation
to the second, then adding 2 times the first equation to the third yields the system
2 x  y  2 z  4

00

00

The last two equations do not impose any restriction on the unknowns therefore we can omit them. Solving the
first equation for x we obtain x  2  12 y  z . This allows us to represent the solution using parametric
equations

, 1.1 Introduction to Systems of Linear Equations 6


1
x  2  r  s, y  r, zs
2

where the parameters r and s are arbitrary real numbers.

 1 7 8 8 
17. (a) Add 2 times the second row to the first to obtain 2 3 3 2  .
0 2 3 1

 1 3 8 3
(b) Add the third row to the first to obtain 2 9 3 2 
 1 4 3 3

 1 4 3 3 
(another solution: interchange the first row and the third row to obtain 2 9 3 2  ).
0 1 5 0 

 1 2 3 4 
18. (a) Multiply the first row by 1
2
to obtain  7 1 4 3 .
 5 4 2 7 

 1 1 3 6 
(b) Add the third row to the first to obtain  3 1 8 1
 6 3 1 4 

 1 2 18 0 
(another solution: add 2 times the second row to the first to obtain  3 1 8 1 ).
 6 3 1 4 

1 k 4 
19. (a) Add 4 times the first row to the second to obtain  which corresponds to the system
0 8  4k 18 

x ky  4

8  4k  y  18
If k  2 then the second equation becomes 0  18 , which is contradictory thus the system becomes
inconsistent.
If k  2 then we can solve the second equation for y and proceed to substitute this value into the first equation
and solve for x .
Consequently, for all values of k  2 the given augmented matrix corresponds to a consistent linear system.

1 k 1
(b) Add 4 times the first row to the second to obtain  which corresponds to the system
0 8  4 k 0 

, 1.1 Introduction to Systems of Linear Equations 7


x ky  1

8  4k  y  0
If k  2 then the second equation becomes 0  0 , which does not impose any restriction on x and y therefore
we can omit it and proceed to determine the solution set using the first equation. There are infinitely many
solutions in this set.
If k  2 then the second equation yields y  0 and the first equation becomes x  1 .

Consequently, for all values of k the given augmented matrix corresponds to a consistent linear system.

 3 4 k 
20. (a) Add 2 times the first row to the second to obtain  which corresponds to the system
0 0 2 k  5 

3x  4 y  k

0  2k  5

If k   25 then the second equation becomes 0  0 , which does not impose any restriction on x and y
therefore we can omit it and proceed to determine the solution set using the first equation. There are infinitely
many solutions in this set.
If k   25 then the second equation is contradictory thus the system becomes inconsistent.

Consequently, the given augmented matrix corresponds to a consistent linear system only when k   25 .

 k 1 2 
(b) Add the first row to the second to obtain  which corresponds to the system
4  k 0 0 

kx  y  2
4  k  x  0

If k  4 then the second equation becomes 0  0 , which does not impose any restriction on x and y
therefore we can omit it and proceed to determine the solution set using the first equation. There are infinitely
many solutions in this set.
If k  4 then the second equation yields x  0 and the first equation becomes y  2 .

Consequently, for all values of k the given augmented matrix corresponds to a consistent linear system.
21. Substituting the coordinates of the first point into the equation of the curve we obtain

y1  ax12  bx1  c

Repeating this for the other two points and rearranging the three equations yields

x12 a  x1b  c  y1

x22 a  x2 b  c  y2

x32 a  x3 b  c  y3

, 1.1 Introduction to Systems of Linear Equations 8


 x12 x1 1 y1 
 
This is a linear system in the unknowns a , b , and c . Its augmented matrix is  x22 x2 1 y2  .
 x32 x3 1 y3 


23. Solving the first equation for x1 we obtain x1  c  kx2 therefore the solution set of the original equation can be
described by the parametric equations

x1  c  kt, x2  t

where the parameter t is an arbitrary real number.
Substituting these into the second equation yields
c  kt  lt  d

which can be rewritten as
c  kt  d  lt

This equation must hold true for all real values t , which requires that the coefficients associated with the same power
of t on both sides must be equal. Consequently, c  d and k  l .
24. (a) The system has no solutions if either
 at least two of the three lines are parallel and distinct or
 each pair of lines intersects at a different point (without any lines being parallel)
(b) The system has exactly one solution if either
 two lines coincide and the third one intersects them or
 all three lines intersect at a single point (without any lines being parallel)
(c) The system has infinitely many solutions if all three lines coincide.

25. 2 x  3y  z  7
2 x  y  3z  9
4 x  2 y  5z  16

26. We set up the linear system as discussed in Exercise 21:

12 a  1b  c  1 a  b  c  1
2 2 a  2b  c  4 i.e. 4 a  2b  c  4
 1
2
a  1b  c  1 a  b  c  1

One solution is expected, since exactly one parabola passes through any three given points  x1 , y1  ,  x2 , y2  ,  x3 , y3 
if x1 , x2 , and x3 are distinct.

27. x  y  z  12
2x  y  2z  5
x  z  1