Chapter 1 Solutions to Review Problems

Chapter 1 Solutions to Review Problems

Chapter 1

Exercise 42

Which of the following equations are not linear and why:

(a) x21 + 3x2 - 2x3 = 5.

(b) x1 + x1x2 + 2x3 = 1.

(c)

x1

+

2 x2

+

x3

=

5.

Solution. (a) The given equation is linear by (??). (b) The equation is not linear because of the term x1x2. (c) The equation is nonlinear because x2 has a negative power

Exercise 43 Show that (2s + 12t + 13, s, -s - 3t - 3, t) is a solution to the system

2x1 + 5x2 + 9x3 + 3x4 = -1

x1 + 2x2 + 4x3

=1

Solution. Substituting these values for x1, x2, x3, and x4 in each equation.

2x1 + 5x2 + 9x3 + 3x4 = 2(2s + 12t + 13) + 5s + 9(-s - 3t - 3) + 3t = -1 x1 + 2x2 + 4x3 = (2s + 12t + 13) + 2s + 4(-s - 3t - 3) = 1.

Since both equations are satisfied, then it is a solution for all s and t

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CHAPTER 1. SOLUTIONS TO REVIEW PROBLEMS

Exercise 44 Solve each of the following systems using the method of elimination: (a)

4x1 - 3x2 = 0 2x1 + 3x2 = 18

(b) 4x1 - 6x2 = 10 6x1 - 9x2 = 15

(c) 2x1 + x2 = 3 2x1 + x2 = 1

Which of the above systems is consistent and which is inconsistent?

Solution. (a) Adding the two equations to obtain 6x1 = 18 or x1 = 3. Substituting this value for x1 in one of the given equations and then solving for x2 we find x2 = 4. So system is consistent.

(b) The augmented matrix of the given system is

4 -6 10 6 -9 15

Divide the first row by 4 to obtain

1

-

3 2

5 2

6 -9 15

Now, add to the second row -6 times the first row to obtain

1

-

3 2

5 2

000

Hence,

x2

=

s

is

a

free

variable.

Solving

for

x1

we

find

x1

=

5+3t 2

.The

system

is consistent.

(c) Note that according to the given equation 1 = 3 which is impossible. So the given system is inconsistent

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Exercise 45 Find the general solution of the linear system

x1 - 2x2 + 3x3 + x4 = -3 2x1 - x2 + 3x3 - x4 = 0

Solution. The augmented matrix of the given system is

1 -2 3 1 -3 2 -1 3 -1 0

A corresponding row-echelon matrix is obtained by adding negative two times the first row to the second row.

1 -2 3 1 -3 0 3 -3 -3 6

Thus x3 = s and x4 = t are free variables. Solving for the leading variables one finds x1 = 1 - s + t and x2 = 2 + s + t

Exercise 46 Find a, b, and c so that the system

x1

+ ax2 + cx3 = 0

bx1 + cx2 - 3x3 = 1

ax1 + 2x2 + bx3 = 5

has the solution x1 = 3, x2 = -1, x3 = 2.

Solution. Simply substitute these values into the given system to obtain

-a

+ 2c = -3 3b - c = 7

3a + 2b

=7

The augmented matrix of the system is

-1 0 2 -3 0 3 -1 7

32 0 7

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CHAPTER 1. SOLUTIONS TO REVIEW PROBLEMS

A row-echelon form of this matrix is obtained as follows.

Step 1: r1 -r1

1 0 -2 3

0 3 -1 7

32 0 7

Step 2: r3 r3 - 3r1

1 0 -2 3 0 3 -1 7

0 2 6 -2

Step 3: r2 r2 - r3

1 0 -2 3 0 1 -7 9

0 2 6 -2

Step 4: r3 r3 - 2r2

1 0 -2 3 0 1 -7 9

0 0 20 -20

Step

5:

r3

1 20

r3

1 0 -2 0 1 -7

00 1

3 9 -1

The corresponding system is

a

- 2c = 3

b - 7c = 9

c = -1

Using back substitution we find the solution a = 1, b = 2, c = -1

Exercise 47 Find a relationship between a,b,c so that the following system is consistent

x1

+ x2 + 2x3 = a

x1

+ x3 = b

2x1 + x2 + 3x3 = c

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Solution. The augmented matrix of the system is

1 1 2 a 1 0 1 b

213c

We reduce this matrix into row-echelon form as follows.

Step 1: r2 r2 - r1 and r3 r3 - 2r1

1 1 2 a 0 -1 -1 b - a

0 -1 -1 c - 2a

Step 2: r2 -r2

1 1 2 a 0 1 1 a-b

0 -1 -1 c - 2a

Step 3: r3 r3 + r2

1 1 2 a 0 1 1 a-b

0 0 0 c-a-b

The system is consistent provided that a + b - c = 0

Exercise 48 For which values of a will the following system have (a) no solutions? (b) exactly one solution? (c) infinitely many solutions?

x1

+ 2x2 -

3x3

=4

3x1 - x2 +

5x3

=2

4x1 + x2 + (a2 - 14)x3 = a + 2

Solution. The augmented matrix is

1 2 -3

4

3 -1 5

2

4 1 a2 - 14 a + 2

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