# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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10. Start with the given matrix augmented by the identity

matrix.

4 . 1

0

Add 2 times the first row to the second row.

1

4

1

0

0 11 . 2 1 V 7

Multiply the second row by (1/11).

^-4 2 6j

1

^ -2 3 . 0 1 )

Section 7.2

125

' 14. 1 0 ^

v0 1 . 2/11 1/11,

Add (-4) times the second row to the first row.

'1 0 . 3/11 -4/11^

v0 1 . 2/11 1/11 ,

Since we have performed the same operation on the given matrix and the identity matrix, the 2 x 2 matric appearing on the right side of this augmented matrix is the desired inverse matrix. The answer can be checked by multiplying it by the given matrix; the result should be the indentity matrix.

12. The augmented matrix in this case is:

' 1 2 3 1 0 0 N

2 4 5 0 1 0

V 3 5 6 0 0 1,

Add (-2) times the first row to the second row and (-3) times the first row to the third row.

' 1 2 ro 1 0 0 N

0 0 -1 -2 1 0

V 0 -1 -3 -3 0 1 ,

Multiply the second and third rows by (-1) and interchange them.

' 1 2 ro 1 0 0 N

0 1 3 3 0 -1

0 V 0 1 2 -1 0 /

Add (-3) times the third row to the first and second

126

Section 7.2

' 1 2 0 -5 3 0

rows. 0 1 0 .-3 3 -1

v 0 0 1 . 2 -1 0

Add (-2) times the second row to th

' 1 0 0 1 -3 2 N

0 1 0 -3 3 -1

0 V 0 1 2 -1 0 )

The desired answer appears on the r

augmented matrix.

14. Again, start with the given matrix augmented by the

"1 2 1 . 1 0 0 "

identity matrix.

-2 1 8 . 0 1 0

Add (2) times the first row to the second row and add(-1) times the first row to the third row.

' 1 2 1 . 1 0 0N

0 5 10 . 2 1 0

V 0 CO 1 1 1 0 1,

the second row to the third

' 1 2 1 . 1 0 0

0 5 10 . 2 1 0

v0 0 0 .3/5 4/5 0y

Since the third row of the left matrix is all zeros, no further reduction can be performed, and the given matrix is singular.

^1 -2 -7 . 0 0 1)

Section 7.3

127

22. x' =

2e

2t

2t

; and

' 3 -2 ' ' 3 -2 ' ' 4 ' 2t ' 12-4" 2t CO

x = e = e =

v2 -2, v2 -2, V 2 y i CO V 4 ,

2t

25. ^ =

r -, -3t _ 2t ^

-3e 2e

V12e 3t 2e2ty

1 1

—3t

ee

—3t

2t

—4e

2t

e

Section 7.3, Page 366

1. Form the augmented matrix, as in Example 1, and use row reduction.

' 1 0 T—1 1 0 N

CO 1 1 1

T—1 1 1 2 2 ,

Add (-3) times the first row to the second and add the first row to the third.

' 1 0 -1 . 0 N

vv 0 1 4 . 1

V 0 1 1 . 2 ,

Add (-1) times the second row to the third.

' 1 0 -1 . 0 N

vv 0 1 4 . 1

V 0 0 -3 . 1,

The third row is equivalent to - 3x3 = 1 or x3 = - 1/3. Likewise the second row is equivalent to x2 + 4x3 = 1, so x2 = 7/3. Finally, from the first row, x3 - x3 = 0, so x3 = - 1/3. The answer can be checked by substituting into the original equations.

8

4

c4 —2 )

128

Section 7.3

2 . The augmented matrix is

' 1 2 -1 1N

2 1 1 1

t—1 l—1 1 2 t—1

Row reduction then

' 1 2 -1 . 1N

yields 0 -3 3 . -1

V 0 0 0 . 1,

The last row corresponds to the equation

0x1 + 0x2 + 0x3 = 1, and there is no choice of x3, x2, and x3 that satisfies this equation. Hence the given system of equations has no solution.

3. Form the augmented matrix and use row reduction.

' 1 2 l—1 1 2 N

2 1 1 . 1

t—1 l—1 1 2 . t—1 1

Add (-2) times the first row to the second and add (-1) times the first row to the third.

' 1 2 1—1 1 2 N

0 ro 1 ro ro 1

V 0 ro 1 ro ro 1

Add (-1) times the second row to the third row and then multiply the second row by (-1/3).

' 1 2 1—1 1 2 N

0 1 -1 . 1

V 0 0 0 . 0,

Sectuib 7.3

129

Since the last row has only zero entries, it may be

dropped. The second row corresponds to the equation

x2 - x3 = 1. We can assign an arbitrary value to either x2

or x3 and use this equation to solve for the other. For

example, let x3 = c, where c is arbitrary. Then

x2 = 1 + c. The first row corresponds to the equation

xi + 2x2 - x3 = 2, so

x1 = 2 - 2x2 + x3 = 2 - 2(1+c)+c = -c.

6. To determine whether the given set of vectors is linearly independent we must solve the system

c3x(1) + c2x(2) + c3x(3) = 0 for c3, c2, and c3. Writing this in scalar form, we have c1 + c3 = 0

c1 + c2 = 0, so the

c2 + c3 = 0

augmented matrix is

1 1 0

0 1 1

Row reduction yields

1 0 1

1 -1

From the third row we have c3 = 0. Then from the second row, c2 - c3 = 0, so c2 = 0. Finally from the first row

c1 + c3 = 0, so c1 = 0. Since c1 = c2 = c3 = 0, we

conclude that the given vectors are linearly independent.

1

0

1

0

0

0

0

0

0

0

0

2

0

8. As in Problem 6 we wish to solve the system

(1) (2) (3) (4) n -p ^

c3x + c2x + c3x + c4x = 0 for c1, c2, c3, and

c4. Form the augmented matrix and use row reduction.

130

Section 7.3

14.

' 1 T—1 1 -2 ro 1 0 N

2 0 -1 0 . 0

2 ro 1 -1 . 0

ro 1 0 ro 0,

Add (-2) times the first row to the second, add (-2) times the first row to the third, and add (-3) times the first row to the fourth.

' 1 -1 -2 -3 . 0 ^

0 2 3 6 . 0

0 5 5 5 . 0

V 0 4 6 12 . 0,

Multiply the second row by

the second row to the third

row to the fourth.

' 1 -1 -2 -3 . 0 ^

0 1 3/2 3 . 0

0 0 -5/2 O o T—1 1

V 0 0 0 o o

The third row is equivalent to the equation c3 + 4c4 = 0. One way to satisfy this equation is by choosing c4 = -1; then c3 = 4. From the second row we then have c2 = - (3/2)c3 - 3c4 = - 6 + 3 = -3. Then, from the first row, c4 = c2 + 2c3 + 3c4 = -3 + 8 - 3 = 2. Hence the given vectors are linearly dependent, and satisfy 2x(1) - 3x(2) + 4x(3) - x(4) = 0.

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