Elementary Linear Algebra Applications Version, Enhanced eText, 12th Edition , Howard Anton , Chris

Elementary Linear Algebra Applications Version, Enhanced eText, 12th Edition By Howard Anton, Chris Rorres , Anton Kaul

Email: Richard@qwconsultancy.com

1.1 Introduction to Systems of Linear Equations 1

CHAPTER 1: SYSTEMS OF LINEAR EQUATIONS AND MATRICES 1.1 Introduction to Systems of Linear Equations 1.

(a)

This is a linear equation in x1 , x2 , and x 3 .

(b)

This is not a linear equation in x1 , x2 , and x 3 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 x 3 .

3.

(d)

This is not a linear equation in x1 , x2 , and x 3 because of the term x12 .

(e)

This is not a linear equation in x1 , x2 , and x 3 because of the term x13/5 .

(f)

This is a linear equation in x1 , x2 , and x 3 .

(a)

a11 x1 a21 x1

 a12 x2  a22 x2

 b1  b2

(b)

a11 x1

 a12 x2

 a13 x3

 b1

a21 x1 a31 x1

 a22 x2  a32 x2

 a23 x3  a33 x3

 b2  b3

a11 x1 a21 x1

 a12 x2  a22 x2

 a13 x3  a23 x3

 a14 x4  a24 x4

(c)

5.

(a)

(b)

2 x1 3 x1

7.

 b1  b2

 4 x2 x2

(a) 6  2  3 8    9 3 

 0

3 x1

 0  1

7 x1

 x2  2 x2

 2 x3



 4 x3  x3

 3  7

5

(b)

(c)

 6 1 3 4   0 5 1 1  

1 0  0 2 0 3  3 1 1 0 0 1    6 2 1 2 3 6 

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.

11.

(a)

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

1.1 Introduction to Systems of Linear Equations 2 3x

 2y 0

 4  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

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 x

1  2t , 2

yt

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  . 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 x

3 5  t, 7 7

yt

where the parameter t is an arbitrary real number. (b)

Solving the equation for x1 we obtain x1  73  35 x2  43 x3 therefore the solution set of the original equation can be described by the parametric equations x1 

7 5 4  r  s, 3 3 3

x2  r ,

x3  s

where the parameters r and s are arbitrary real numbers. (c)

Solving the equation for x1 we obtain x1   81  14 x2  85 x3  34 x4 therefore the solution set of the original equation can be described by the parametric equations

1.1 Introduction to Systems of Linear Equations 3 1 1 5 3 x1    r  s  t , 8 4 8 4

x2  r ,

x3  s,

x4  t

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 , 3 3 3 3

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 2x

 3y

 1

0

 0

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  32 y . This allows us to represent the solution using parametric equations x

1 3  t, 2 2

yt

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.

17.

(a)

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

(b)

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

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

19.

(a)

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

k 1 Add 4 times the first row to the second to obtain  0 8  4 k x

1 which corresponds to the system 0 

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. 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

 x12  This is a linear system in the unknowns a , b , and c . Its augmented matrix is  x22  x32  23.

x1 1 y1   x2 1 y2  . x3 1 y3 

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