IM Chapter1 7e

8/11/2019 IM Chapter1 7e

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Chapter 1

Exercise Set 1.1

1. (a) 3 x 3 (b) 3 x 2 (c) 2 x 4 (d) 3 x 1 (e) 3 x 5 (f) 1 x 4

2. 1, 4, 9, - 1, 3, 8 3. 4, 5, 6, 7, 2, 3

4.

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

5. (a)

1 3

2 - 5 and

1 3 7

2 - 5 - 3(b)

5 2 - 4

1 3 6

4 6 - 9 and

5 2 - 4 8

1 3 6 4

4 6 - 9 7

(c)

- 1 3 - 5

2 - 2 4

1 3 0

and

- 1 3 - 5 - 3

2 - 2 4 8

1 3 0 6

(d)

5 4

2 - 8

1 2 and

5 4 9

2 - 8 - 4

1 2 3

(e)

5 2 - 4

0 4 31 0 - 1

and

5 2 - 4 8

0 4 3 01 0 - 1 7

(f)

- 1 3 - 9

1 0 - 4

1 8 0

and

- 1 3 - 9 - 4

1 0 - 4 11

1 8 0 1

(g)

1 0 0

0 1 0

0 0 1and

1 0 0 - 3

0 1 0 12

0 0 1 8

(h)

- 4 2 - 9 1

1 6 - 8 - 70 - 1 3 - 5

and

- 4 2 - 9 1 - 1

1 6 - 8 - 7 150 - 1 3 - 5 0

6. (a) x1 + 2x2 = 3

4x1

+ 5x2

= 6

(b) 7x1

+ 9x2

= 8

6x1

+ 4x2

= - 3

(c) x1

+ 9x2

= - 3

5x1

= 2


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(d) 8x1

+ 7x2

+ 5x3

= - 1

4x1

+ 6x2

+ 2x3

= 4

9x1

+ 3x2

+ 7x3

= 6

(e) 2x1

- 3x2

+ 6x3

= 4

7x1

- 5x2

- 2x3

= 3

2x2

+ 4x3

= 0

(f) - 2x2

= 4

5x1

+ 7x2

= - 3

6x1

= 8

(g) x1

= 3

x2

= 8

x3

= 4

(h) x1

+ 2x2

- x3

= 6

x2

+ 4x3

= 5

x3

= - 2

7. (a)

1 3 - 2 0

1 2 - 3 6

8 3 2 5

(b)

2 7 5 1

0 - 8 4 3

3 - 5 8 9 (c)

1 2 3 - 1

0 3 10 0

0 - 8 - 1 - 1

(d)

1 0 - 1 - 60 1 2 1

0 0 11 - 1 (e)

1 0 0 - 230 1 0 17

0 0 1 5 (f)

1 0 2 70 1 5 3

0 0 1 4

8. (a) Create zeros below the leading 1 in the first column.x

1is eliminated from all equations except the first.

(b) Normalize the (2,2) element, i.e., make the (2,2) element 1. This becomes a leading 1.It is now possible to have x

2 in the second equation with coefficient 1.

(c) Need to have the leading 1 in row 2 to the left of leading 1 in row 3.The second equation now contains an x

2 term.

(d) Create zeros above and below the leading 1 in row 2.x

2 is eliminated from all equations except the second.

9. (a) Create zeros above the leading 1 in column 3.x

3 is eliminated from all equations except the third.

(b) Need to have the leading 1 in row 1 to the left of leading 1s in other rows.

It is now possible to have x1 in Equation 1 with leading coefficient 1.

(c) Normalize the (3,3) element, i.e., make the (3,3) element 1. This becomes a leading1.The coefficient of x

3 in the third equation becomes 1.

(d) Create zeros above the leading 1 in column 3.


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x3 is eliminated from all equations except the third.

10. (a)1 2 8

2 3 11

R2 (2)R1

1 2 8

0 1 5

R1(2)R2

1 0 2

0 1 5

,

so the solution is x1 = 2 and x

2 = 5.

(b)2 2 4

3 2 3

(1/2)R1

1 1 2

3 2 3

R2 (3)R1

1 1 2

0 1 3

(1)R2

1 1 2

0 1 3

R1(1)R2

1 0 1

0 1 3

, so the solution is x1 = -1, x2 = 3.

(c)

1 0 1 3

0 2 2 40 1 2 5

(1/2)R2

1 0 1 3

0 1 1 20 1 2 5

R3 (1)R2

1 0 1 3

0 1 1 20 0 1 7

(1)R3

1 0 1 3

0 1 1 2

0 0 1 7

R1(1)R3

R2 R3

1 0 0 10

0 1 0 9

0 0 1 7

,

so the solution is x1 = 10, x

2 = - 9, x

3 = - 7.

(d)

1 1 3 6

1 2 4 9

2 1 6 11

R R

R R

R R

R R

R R

R R

, so the solution is x

1 = -1, x

2 = 1, x

3 = 2.

(e)1 1 3 32 1 2 2

3 1 2 3

R R

R R

R R

R R2

R

RR

R R


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2 = 4, x

3 = 2.

(f)

1 1 1 2

3 1 1 10

4 2 3 14

(1)R1

1 1 1 2

3 1 1 10

4 2 3 14

R2 (3)R1

R3 (4)R1

1 1 1 2

0 4 2 4

0 6 1 6

(1/4)R2

1 1 1 2

0 1 1/2 1

0 6 1 6

R1R2

R3 (6)R2

1 0 1/2 3

0 1 1/2 1

0 0 2 0

(1/2)R3

1 0 1/2 3

0 1 1/2 1

0 0 1 0

R1(1/2)R3

R2 (1/2)R3


1 = 3, x

2 = 1, x

3 = 0.

11. (a)

1 2 3 14

2 5 8 36

1 1 0 4

R2 (2)R1

R3 (1)R1

1 2 3 14

0 1 2 8

0 3 3 18

R1(2)R2

R3 (3)R2

1 0 1 2

0 1 2 8

0 0 3 6

(1/3)R3

1 0 1 2

0 1 2 8

0 0 1 2

R1R3

R2 (2)R3

1 0 0 0

0 1 0 4

0 0 1 2

,


2 = 4, x

3 = 2.

(b)

1 1 1 1

2 6 10 14

2 1 6 9

R2 (2)R1

R3 (2)R1

1 1 1 1

0 4 8 12

0 3 8 11

(1/4)R2

1 1 1 1

0 1 2 3

0 3 8 11

R1R2

R3 (3)R2

1 0 1 2

0 1 2 3

0 0 2 2

(1/2)R3

1 0 1 2

0 1 2 3

0 0 1 1

R1(1)R3

R2 (2)R3

1 0 0 1

0 1 0 1

0 0 1 1

,


2 = 1, x

3 = 1.

(c)

2 2 4 14

3 1 1 8

2 1 2 1

(1/2)R1

1 1 2 7

3 1 1 8

2 1 2 1

R2 (3)R1

R3 (2)R1

1 1 2 7

0 2 7 13

0 3 6 15


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Let us swap rows 2 and rows 2 and 3 to avoid awkward fractions at the next step.

R2R3

1 1 2 7

0 3 6 15

0 2 7 13

(1/3)R2

1 1 2 7

0 1 2 5

0 2 7 13

R1(1)R2

R3 (2)R2

1 0 0 2

0 1 2 5

0 0 3 3

(1/3)R3

1 0 0 2

0 1 2 5

0 0 1 1

R2 (2)R3

1 0 0 2

0 1 0 3

0 0 1 1

,


2 = 3, x

3 = - 1.

(d)

0 2 4 8

2 2 0 61 1 1 5

R1 R2

2 2 0 6

0 2 4 81 1 1 5

(1/2)R1

1 1 0 3

0 2 4 81 1 1 5

R3 (1)R1

1 1 0 3

0 2 4 8

0 0 1 2

(1/2)R2

1 1 0 3

0 1 2 4

0 0 1 2

R1(1)R2

1 0 2 1

0 1 2 4

0 0 1 2

R1 (2)R3

R2 (2)R3


1 = 3, x

2 = 0, x

3 = 2.

(e)

1 0 1 3

1 0 2 8

3 1 1 0

R2 R1

R2 (3)R1

1 0 1 3

0 0 1 5

0 1 2 9

R2R3

1 0 1 3

0 1 2 9

0 0 1 5

R1R3

R2 (2)R3


1 = -2, x

2 = 1, x

3 = -5.

12 (a)3/2 0 3 151 7 9 45

2 0 5 22

(2/3)R1

1 0 2 101 7 9 45

2 0 5 22

R2 R1

R3 (2)R1

1 0 2 100 7 7 35

0 0 1 2


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(1/7)R2

1 0 2 10

0 1 1 5

0 0 1 2

R1(2)R3

R2 R3

1 0 0 6

0 1 0 3

0 0 1 2

,


2 = - 3, x

3 = 2.

(b)

3 6 15 3

2 3 9 1

4 7 17 4

(1/3)R1

1 2 5 1

2 3 9 1

4 7 17 4

R2 (2)R1

R3 (4)R1

1 2 5 1

0 1 1 1

0 1 3 0

(1)R2

1 2 5 1

0 1 1 1

0 1 3 0

R1(2)R2

R3 (1)R2

1 0 3 1

0 1 1 1

0 0 2 1

(1/2)R3

1 0 3 1

0 1 1 1

0 0 1 1/2

R1(3)R3

R2 (1)R3

1 0 0 1/2

0 1 0 3/2

0 0 1 1/2


1 = 1/2, x

2 = 3/2, x

3 = -1/2.

(c)

3 6 0 3 3

1 3 1 4 12

1 1 1 2 8

2 3 0 0 8

(1/3)R1

1 2 0 1 1

1 3 1 4 12

1 1 1 2 8

2 3 0 0 8

R2 (1)R1

R3 (1)R1

R4 (2)R1

1 2 0 1 10 1 1 3 13

0 3 1 3 7

0 1 0 2 6

R1(2)R2

R3 (3)R2

R4R2

1 0 2 5 270 1 1 3 13

0 0 2 6 32

0 0 1 1 7

(1/2)R3

1 0 2 5 27

0 1 1 3 13

0 0 1 3 16

0 0 1 1 7

R1(2)R3

R2 R3

R4R3

1 0 0 1 5

0 1 0 0 3

0 0 1 3 16

0 0 0 2 9

(1/2)R4

1 0 0 1 5

0 1 0 0 3

0 0 1 3 16

0 0 0 1 9 /2

R1R4

R3 (3)R4

1 0 0 0 1/2

0 1 0 0 3

0 0 1 0 5 /2

0 0 0 1 9 /2

,


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so the solution is x1 = - 1/2, x

2 = 3, x

3 = 5/2, x

4= 9/2.

(d)

1 2 2 5 11

2 4 2 8 14

1 3 4 8 19

1 1 1 0 2

R R

R R

R R

1 5 9

R2R3

RR

R R

1 0 2 1 5

0 1 2 3 8

0 0 2 2 8

0 0 5 4 15

R

RR

R R

R R

1 0 0 1 3

0 1 0 1 0

0 0 1 1 4

0 0 0 1 5

R

R R

R R

R R

1 0 0 0 2

0 1 0 0 5

0 0 1 0 1

0 0 0 1 5

,

so the solution is x1 = - 2, x

2 = - 5, x

3 = - 1, x

4= 5.

(e)

1 1 2 6 11

2 3 6 19 36

0 3 4 15 28

1 1 1 6 12

R R

R R

2 23

R1(1)R2

R3 (3)R2

R4 (2)R2

1 0 0 1 30 1 2 7 14

0 0 2 6 14

0 0 1 2 5

(1/2)R3

1 0 0 1 30 1 2 7 14

0 0 1 3 7

0 0 1 2 5


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R2 (2)R3

R4(1)R3

1 0 0 1 3

0 1 0 1 0

0 0 1 3 7

0 0 0 1 2

(1)R4

1 0 0 1 3

0 1 0 1 0

0 0 1 3 7

0 0 0 1 2

R1R4

R2 (1)R4

R3(3)R4

1 0 0 0 1

0 1 0 0 2

0 0 1 0 1

0 0 0 1 2

, so the solution is x1 = - 1, x

2 = - 2, x

3 = 1, x

4=

2.

13. (a)1 2 3 4 3

3 5 8 9 7

R2 (3)R1

1 2 3 4 3

0 1 1 3 2

(1)R2

1 2 3 4 3

0 1 1 3 2

R1 2R2

1 0 1 2 1

0 1 1 3 2

, so x1 = 1, x2 = 1; x1 = - 2, x2 = 3; and x1 = - 1, x2 = 2.

(b)1 1 0 5 1

2 3 1 13 2

R2 (2)R1

1 1 0 5 1

0 1 1 3 0

R1(1)R2

1 0 1 2 1

0 1 1 3 0

,

so the solutions are in turn x1 = - 1, x

2 = 1; x

1 = 2, x

2 = 3; and x

1 = 1, x

2 =

0.

(c)

1 2 3 6 5 4

1 1 2 5 3 3

2 3 6 14 8 9

R2 (1)R1

R3(2)R1

1 2 3 6 5 4

0 1 1 1 2 1

0 1 0 2 2 1

R1 (2)R2

R3 (1)R2

1 0 1 4 1 2

0 1 1 1 2 1

0 0 1 3 0 2

R1(1)R3

R2 R3

1 0 0 1 1 0

0 1 0 2 2 1

0 0 1 3 0 2

,

so the solutions are in turn x1 = 1, x2 = 2, x3 = 3; x1 = - 1, x2 = 2, x3 = 0; and x

1 = 0, x

2 = 1, x

3 = 2.

(d)

1 2 1 1 6 0

1 1 1 1 4 2

3 7 1 1 18 4

R2 R1

R3 (3)R1

1 2 1 1 6 0

0 1 0 0 2 2

0 1 2 2 0 4


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R1(2)R2

R3 (1)R2

1 0 1 1 2 4

0 1 0 0 2 2

0 0 2 2 2 2

(1/2)R3

1 0 1 1 2 4

0 1 0 0 2 2

0 0 1 1 1 1

1 0 0 0 1 3

0 1 0 0 2 2

0 0 1 1 1 1

,

so the solutions are in turn x1 = 0, x

2 = 0, x

3 = 1; x

1 = 1, x

2 = 2, x

3 = - 1;

and x1 = 3, x

2 = - 2, x

3 = - 1.

Exercise Set 1.2

1. (a) Yes. (b) Yes.

(c) No. The second column contains a leading 1, so other elements in that columnshould be zero.

(d) No. The second row does not have 1 as the first nonzero number.

(e) Yes. (f) Yes. (g) Yes.

(h) No. The second row does not have 1 as the first nonzero number. (i) Yes.

2. (a) No. The leading 1 in row 3 is not to the right of the leading 1 in row 2.

(b) Yes. (c) Yes.

(d) No. The fourth and fifth columns contain leading 1s, so the other numbers in thosecolumns should be zeros.

(e) No. The row containing all zeros should be at the bottom of the matrix.

(f) Yes.

(g) No. The leading 1 in row 3 is not to the right of the leading 1 in row 2. Also, since column 3 contains a leading 1, all other numbers in that column should be zero.

(h) No. The leading 1 in row 3 is not to the right of the leading 1s in rows 1 and 2.

(i) Yes.


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3. (a) x1 = 2, x

2 = 4, x

3 = - 3. (b) x

1 = 3r + 4, x

2 = - 2r + 8, x

3 = r.

(c) x1 = - 3r + 6, x2 = r, x3 = - 2. (d) There is no solution. The last row gives 0 = 1.

(e) x1 = - 5r + 3, x

2 = - 6r- 2, x

3 = - 2r- 4, x

4 = r.

(f) x1 = - 3r + 2, x

2 = r , x

3 = 4, x

4 = 5.

4. (a) x1 = - 2r- 4s + 1, x

2 = 3r- 5s - 6, x

3 = r, x

4 = s.

(b) x1 = 3r- 2s + 4, x2 = r, x3 = s, x4 = - 7.

(c) x1 = 2r- 3s + 4, x

2 = r, x

3 = - 2s + 9, x

4 = s, x

5 = 8.

(d) x1 = - 2r- 3s + 6, x

2 = - 5r- 4s + 7, x

3 = r, x

4 = - 9s - 3, x

5 = s.

5. (a)

1 4 3 1

2 8 11 7

1 6 7 3

R2 (2)R1

R3 (1)R1

1 4 3 1

0 0 5 5

0 2 4 2

R2R3

1 4 3 1

0 2 4 2

0 0 5 5

(1/2)R2

1 4 3 1

0 1 2 1

0 0 5 5

R1(4)R2

1 0 5 3

0 1 2 1

0 0 5 5

(1/5)R3

1 0 5 3

0 1 2 1

0 0 1 1

R1 (5)R3

R2 (2)R3

1 0 0 2

0 1 0 1

0 0 1 1


1 = 2, x

2 = -1, x

3 = 1.

(b)

1 2 4 15

2 4 9 33

1 3 5 20

R2 (2)R1

R3 (1)R1

1 2 4 15

0 0 1 3

0 1 1 5

R2R3

1 2 4 15

0 1 1 5

0 0 1 3


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2 = -2, x

3 = 1.

6. (a)

3 6 3 6

2 4 3 13 6 2 10

(1/3)R1

1 2 1 2

2 4 3 13 6 2 10

R2 (2)R1R3 (3)R1

1 2 1 2

0 0 5 30 0 1 4

It is now clear that there is no solution. The last two rows give - 5x3 = 3 and x

3 =

4.

(b)

1 2 1 7

1 2 2 11

2 4 3 18

R2 (1)R1

R3 (2)R1

1 2 1 7

0 0 1 4

0 0 1 4

R1 (1)R2

R3 (1)R2

1 2 0 3

0 0 1 4

0 0 0 0

so x1 + 2x

2 = 3 and x

3 = 4. Thus the general solution is

x1 = 3 - 2r, x

2 = r, x

3 = 4.

(c)

1 2 1 3

2 4 2 6

3 6 2 1

R2 (2)R1

R3 (3)R1

1 2 1 3

0 0 0 0

0 0 5 10

R2R3

1 2 1 3

0 0 5 10

0 0 0 0

(1/5)R2

1 2 1 30 0 1 2

0 0 0 0

R1R2

1 2 0 10 0 1 2

0 0 0 0

, so x1+ 2x 2= 1, x3 = - 2.

Thus the general solution is x1 = 1 - 2r, x

2 = r, x

3 = - 2.

(d)

1 2 3 8

3 7 9 26

2 0 6 11

R2 (3)R1

R3(2)R1

1 2 3 8

0 1 0 2

0 4 0 5

, so there is no solution since the

last two rows give x2 = 2 and - 4 x

2 = - 5.


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(e)

0 1 2 5

1 2 5 13

1 0 2 4

R1 R2

1 2 5 13

0 1 2 5

1 0 2 4

R3 (1)R1

1 2 5 13

0 1 2 5

0 2 3 9

R1(2)R2

R3 (2)R2

1 0 1 3

0 1 2 5

0 0 1 1

R1 (1)R3

R2 (2)R3

1 0 0 2

0 1 0 3

0 0 1 1

,


2 = 3, x

3 = 1.

(f)

1 2 8 7

2 4 16 14

0 1 3 4

R2 (2)R1

1 2 8 7

0 0 0 0

0 1 3 4

R2 R3

1 2 8 7

0 1 3 4

0 0 0 0

R1(2)R2

1 0 2 1

0 1 3 4

0 0 0 0

, so x

1 + 2 x

3 = -1, x

2 + 3x

3 = 4.

Thus the general solution is x1 = - 1 - 2r, x

2 = 4 - 3r, x

3 = r.

7. (a)1 1 3 10

3 2 4 24

R2 (3)R1

1 1 3 10

0 1 5 6

R1 (1)R2

1 0 2 4

0 1 5 6

,

so x1 + 2x

3 = 4 and x

2- 5x


x1 = 4 - 2r, x

2 = 6 + 5r, x

3 = r.

(b)2 6 14 38

3 7 15 37

(1/2)R1

1 3 7 19

3 7 15 37

R2 (3)R1

1 3 7 19

0 2 6 20

(1/2)R2

1 3 7 19

0 1 3 10

R1(3)R2

1 0 2 11

0 1 3 10

,

so x1 + 2x

3 = - 11 and x

2 + 3x

3 = - 10. Thus the general solution is

x1 = - 11- 2r, x

2 = - 10 - 3r, x

3 = r.


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(c)

1 2 1 1 0

1 2 0 1 4

1 2 2 4 5

R2 (1)R1

R3 R1

1 2 1 1 0

0 0 1 2 4

0 0 1 3 5

R1R2

R3 (1)R2

1 2 0 1 4

0 0 1 2 4

0 0 0 1 1

R1(1)R3

R2 (2)R3

1 2 0 0 3

0 0 1 0 2

0 0 0 1 1

, so x

1 + 2 x

2 = 3 and x

3 = 2, and x

4 = 1.

Thus the general solution is x1 = 3 - 2r, x

2 = r, x

3 = 2, and x

4 = 1.

(d)1 2 0 4 0

2 4 3 2 0

R2 (2)R1

1 2 0 4 0

0 0 3 6 0

(1/3)R2

1 2 0 4 0

0 0 1 2 0

,

so x1 + 2x

2 + 4x

4 = 0 and x

3 + 2x


x1 = - 2r- 4s, x2 = r, x3 = - 2s, x4 = s.

(e)

0 1 3 1 0

1 1 1 4 0

2 2 2 8 0

R1 R2

1 1 1 4 0

0 1 3 1 0

2 2 2 8 0

R3 (2)R1

1 1 1 4 0

0 1 3 1 0

0 0 0 0 0

R1 (1)R2

1 0 2 3 0

0 1 3 1 0

0 0 0 0 0

,

so x1 + 2x

3 + 3x

4 = 0 and x

2- 3x

3 + x


x1 = - 2r- 3s, x

2 = 3r- s, x

3 = r, x

4 = s.

8. (a)

1 1 1 1 3

2 3 1 5 9

1 3 1 6 7

1 1 1 0 1

R2 (2)R1

R3 (1)R1

R4R1

1 1 1 1 3

0 1 1 3 3

0 2 2 5 4

0 0 0 1 2

R1(1)R2

R3(2)R2

1 0 2 2 0

0 1 1 3 3

0 0 0 1 2

0 0 0 1 2

R1 (2)R3

R2 (3)R3

R4R3

1 0 2 0 4

0 1 1 0 3

0 0 0 1 2

0 0 0 0 0

,

so x1 + 2x

3 = -4, x

2 - x

3 = 3, x

4 = 2.


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The general solution is x1 = -2r- 4, x

2 = r + 3, x

3 = r, x

4 = 2.

(b)

0 1 2 7

1 2 6 18

1 1 2 52 5 15 46

R1 R2

1 2 6 18

0 1 2 7

1 1 2 52 5 15 46

R3 R1

R4 (2)R1

1 2 6 18

0 1 2 7

0 3 8 23

0 1 3 10

R1(2)R2

R3 (3)R2

R4 R2

1 0 2 4

0 1 2 7

0 0 2 2

0 0 1 3

, and there are no

solutions because the last two rows of the matrix give, respectively,x

3 = 1 and x

3 = 3.

(c)

2 4 16 14 10

1 5 17 19 2

1 3 11 11 4

3 4 18 13 17

(1/2)R1

1 2 8 7 5

1 5 17 19 2

1 3 11 11 4

3 4 18 13 17

R2 R1

R3(1)R1

R4 (3)R1

1 2 8 7 5

0 3 9 12 3

0 1 3 4 1

0 2 6 8 2

(1/3)R2

1 2 8 7 5

0 1 3 4 1

0 1 3 4 1

0 2 6 8 2

R1 (2)R2

R3 R2

R4 (2)R2

1 0 2 1 7

0 1 3 4 1

0 0 0 0 0

0 0 0 0 0

, so x1+2x3+x4= 7 and x2-3x3+4x4=1.

Thus the general solution is x1 = 7 - 2r- s, x

2 = 1 + 3r- 4s, x

3 = r, x

4 = s.

(d)

1 1 2 0 7

2 2 2 4 12

1 1 1 2 4

3 1 8 10 29

R2 (2)R1

R3 R1

R4 (3)R1

1 1 2 0 7

0 0 2 4 2

0 0 1 2 3

0 2 2 10 8


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16

R2 R4

1 1 2 0 7

0 2 2 10 8

0 0 1 2 3

0 0 2 4 2

(1/2)R2

1 1 2 0 7

0 1 1 5 4

0 0 1 2 3

0 0 2 4 2

R1R2

1 0 3 5 11

0 1 1 5 4

0 0 1 2 3

0 0 2 4 2

R1(3)R3

R2 (1)R3

R4 (2R3)

1 0 0 1 2

0 1 0 3 1

0 0 1 2 3

0 0 0 0 4

The last row gives 0 = 4, so there is no solution.

(e)

1 6 1 4 0

2 12 5 17 0

3 18 1 6 0

R2 (2)R1

R3 (3)R1

1 6 1 4 0

0 0 3 9 0

0 0 2 6 0

(1/3)R2

1 6 1 4 0

0 0 1 3 0

0 0 2 6 0

R1R2

R3 (2)R2

1 6 0 1 0

0 0 1 3 0

0 0 0 0 0

,

so x1 + 6x

2- x

4 = 0 and x

3 + 3x

4 = 0.

Thus the general solution is x1 = - 6r + s, x

2 = r , x

3 = - 3s, x

4 = s.

(f)

4 8 12 28

1 2 3 72 4 8 16

3 6 9 21

(1/4)R1

1 2 3 7

1 2 3 72 4 8 16

3 6 9 21

R2 R1

R3(2)R1

R4 (3)R1

1 2 3 7

0 0 0 0

0 0 2 2

0 0 0 0

R2R3

1 2 3 7

0 0 2 2

0 0 0 0

0 0 0 0

(1/2)R2

1 2 3 70 0 1 1

0 0 0 0

0 0 0 0

R1 (3)R2

1 2 0 40 0 1 1

0 0 0 0

0 0 0 0

, so x1 + 2x

2 = 4 and x

3 = - 1.

Thus the general solution is x1 = 4 - 2r , x

2 = r , x

3 = - 1.


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17

(g)

1 1 2

2 3 3

1 3 0

1 2 1

R2 (2)R1

R3 (1)R1

R4 (1)R1

1 1 2

0 1 1

0 2 2

0 1 1

R1 (1)R2

R3 (2)R2

R4 (1)R2

1 0 3

0 1 1

0 0 0

0 0 0

,

so x1 = 3, x

2 = - 1.

9. (a) The system of equations

3x1

+ 2x2

- x3

+ x4

= 4

3x1

+ 2x2

- x3

+ x4

= 1

clearly has no solution, since the equations are inconsistent. To make a system

that is less obvious, add another equation to the system and perform an elementarytransformation on this new system of three equations. For example, replace thesecond equation by the sum of the second equation and some multiple (2 in theexample below) of the third equation:

3x1

+ 2x2

- x3

+ x4

= 4

5x1

+ 4x2

- x3

- x4

= 1

x1

+ x2

- x4

= 0

(b) Choose a solution, e.g., x1 = 1, x

2 = 2. Now make up equations thinking of

x1 as 1 and x

2 as 2:

x1 + x2 = 3x

1+ 2x

2= 5

x1

- 2x2

= - 3

An easy way to ensure that there are no additional solutions is to include x1 = 1 or

x2 = 2 as an equation in the system.

10. (a)

1 0 00 1 0

0 0 1

1 0 *0 1 *

0 0 0

1 * 00 0 1

0 0 0

1 * *0 0 0

0 0 0no unique no many

solution solution solution solutions


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(b)

1 0 0 *

0 1 0 *

0 0 1 *

1 0 * 0

0 1 * 0

0 0 0 1

1 0 * *

0 1 * *

0 0 0 0

1 * 0 0

0 0 1 0

0 0 0 1unique no many no

solution solution solutions solution

1 * 0 *

0 0 1 *

0 0 0 0

1 * * 0

0 0 0 1

0 0 0 0

1 * * *

0 0 0 0

0 0 0 0 many no many solutions solution solutions

11. (a) If ax0 + by

0 = 0 then k(ax

0 + by

0) = 0 so that a(kx

0) + b(ky

0) = 0.

Thus x = kx0 , y = ky

0 is a solution. Likewise for the equation cx + dy = 0.

(b) If ax0 + by

0 = 0 and ax

1 + by

1 = 0 then ax

0 + by

0 + ax

1 + by

1 = 0 + 0 = 0.

But ax0 + by

0 + ax

1 + by

1 = ax

0 + ax

1 + by

0 + by

1 = a(x

0 + x

1)+ b(y

0 + y

1) so

that a(x0 + x

1) + b(y

0 + y

1) = 0. Thus x = x

0 + x

1, y = y

0 + y

1 is a solution.

Likewise for the equation cx + dy = 0.

12. a(0) + b(0) = 0 and c(0) + d(0) = 0, so x = 0, y = 0 is a solution. Multiply 1

stequation by c, 2

ndby a to eliminate x. Get

cax+cby=0 and acx+ady=0. Subtract, ady-cby=0, (ad-bc)y=0. Similarly (ad-bc)x=0.

If ad-bc0, x=0,y=0. If ad-bc=0 the x and y can be anything; thus many solutions. Therefore x=0,y=0 is the only solution if and only if ad-bc0.

13. (a) and (b), No. If the first system of equations has a unique solution, then the reducedechelon form of the matrix [A:B

1] will be [I

3:X]. The reduced echelon form of [A:B

2

]must therefore be [I

3:Y]. So the second system must also have a unique solution.

(c) If the first system of equations has many solutions, then at least one row of thereduced echelon form of [A:B

1] will consist entirely of zeros. Therefore the

corresponding row(s) of the reduced echelon form of [A:B2] will have zeros in the

first three columns. If any such row has a nonzero number in the fourth column, thesystem will have no solution.


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19

14. (a)

1 1 5 2 3

1 2 8 5 2

2 4 16 10 4

R2 (1)R1

R3 (2)R1

1 1 5 2 3

0 1 3 3 1

0 2 6 6 2

R1 (1)R2

R3 (2)R2

1 0 2 1 4

0 1 3 3 1

0 0 0 0 0

,

so the general solution to the first system is x1 = - 1 - 2r, x

2 = 3 - 3r, x

3 = r, and

the general solution to the second system is x1 = 4 - 2r, x

2 = - 1 - 3r, x

3 = r.

(b)

1 2 4 8 5

1 1 2 5 3

2 3 6 13 11

R2 (1)R1

R3 (2)R1

1 2 4 8 5

0 1 2 3 2

0 1 2 3 1

(1)R2

1 2 4 8 5

0 1 2 3 2

0 1 2 3 1

,

R1(2)R2

R3 R2

1 0 0 2 1

0 1 2 3 2

0 0 0 0 3

, so the general solution to the first system is

x1 = 2, x

2 = 3 - 2r, x

3 = r, and the second system has no solution.

15. A 3x3 matrix represents the equations of three lines in a plane. In order for there to be aunique solution, the three lines would have to meet in a point. For there to be manysolutions, the three lines would all have to be the same. It is far more likely that the lineswill meet in pairs (or that one pair will be parallel), i.e., that there will be no solution, thesituation represented by the reduced echelon form I

3 .

16. A 3x4 matrix represents the equations of three planes. In order for there to be manysolutions, the three planes must have at least one line in common. For there to be nosolutions, either at least two of the three planes must be parallel or the line of intersection

of two of the planes must lie in a plane that is parallel to the third plane. It is more likely that the three planes will meet in a single point, i.e., that there will be a unique solution. The reduced echelon form therefore will be [I

3:X].

17. The difference between no solution and at least one solution is the presence of anonzero number in the last position of a row that otherwise consists entirely of zeros.Round-off error is more likely to produce a nonzero number when there should be a

zero than the reverse. Thus the answer is (b). Thinking geometrically, a small move by

one or more of the linear surfaces (round-off error) may destroy a solution if there is one,but probably won't produce a solution if there is none.

Exercise Set 1.3


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20

3. (a) (b)

(c)


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21

4. (a) (b)

5. (a) 3(1,4) = (3,12). (b) - 2(- 1,3) = (2,- 6). (c) (1/2)(2,6) = (1,3).

(d) (- 1/2)(2,4,2) = (- 1,- 2,- 1). (e) 3(- 1,2,3) = (- 3,6,9).

(f) 4(- 1,2,3,- 2) = (- 4,8,12,- 8). (g) - 5(1,- 4,3,- 2,5) = (- 5,20,- 15,10,- 25).

(h) 3(3,0,4,2,- 1) = (9,0,12,6,- 3).

6. (a) u+w = (1,2) + (- 3,5) = (- 2,7), (b) u+ 3v= (1,2) + 3(4,- 1) = (13,- 1).

(c) v+w = (4,- 1) + (- 3,5) = (1,4).

(d) 2u+ 3v - w= 2(1,2) + 3(4,- 1) - (- 3,5) = (17,- 4).

(e) - 3u+ 4v - 2w= - 3(1,2) + 4(4,- 1) - 2(- 3,5) = (19,- 20).

7. (a) u+w = (2,1,3) + (2,4,- 2) = (4,5,1). (b) 2u+v= 2(2,1,3) + (- 1,3,2) = (3,5,8).

(c) u+ 3w = (2,1,3) + 3(2,4,- 2) = (8,13,- 3).

(d) 5u - 2v+ 6w= 5(2,1,3) - 2(- 1,3,2) + 6(2,4,- 2) = (24,23,- 1).

(e) 2u - 3v - 4w= 2(2,1,3) - 3(- 1,3,2) - 4(2,4,- 2) = (- 1,- 23,8).

8. (a) u+v=

2

3 +

- 1

- 4 =

1

- 1. (b) 2v - 3w= 2

- 1

- 4- 3

4

- 6 =

- 14

10.


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22

(c) 2u+ 4v - w= 2

2

3 + 4

- 1

- 4-

4

- 6 =

- 4

- 4.

(d) - 3u - 2v+ 4w= - 3

2

3- 2

- 1

- 4 + 4

4

- 6 =

12

- 25 .

9. (a) u+ 2v=

1

2

- 1 + 2

3

0

1 =

7

2

1. (b) - 4v+ 3w= - 4

3

0

1 + 3

- 1

0

5 =

- 15

0

11.

(c) 3u - 2v+ 4w= 3

1

2

- 1- 2

3

0

1 + 4

- 1

0

5 =

- 7

6

15.

2u+ 3v - 8w= 2

12

- 1 + 3

30

1- 8

- 10

5 =

194

- 39.

10. (a) a(1,2) + b(-1,3) = (1,7). a-b=1, 2a+3b=7. Unique solution a=2, b=1. (1,7)=2(1,2) + (-1,3).

(1,7) is a linear combination of (1,2) and (-1,3).

(b) a(1,1) + b(3,2) = (1,2). a+3b=1, a+2b=2. Unique solution a=4,b=-1. (1,2)=4(1,1) - (3,2). (1,2) is a linear combination of (1,1) and (3,2).

(c) a(1,-3) + b(-2,6) = (3,5). a-2b=3, -3a+6b=5. No solution. (3,5) is not a linear combination of (1,-3) and (-2,6).

(d) a(2,4) + b(-4,-8) = (6,2). 2a-4b=6, 4a-8b=2. No solution. (6,2) is not a linear combination of (2,4) and (-4,-8).

11. (a) a(1,1,2) + b(1,2,1) + c(2,3,4) = (7,9,15). a+b+2c = 7, a+2b+3c = 9, 2a+b+4c = 15. Unique solution, a=2, b=-1, c=3. Is a linear combination.

(b) a(1,1,1) + b(1,2,4) + c(0,1,-3) = (6,13,9). a+b = 6, a+2b+c = 13, a+4b-3c = 9. Unique solution, a=2, b=4, c=3. Is a linear combination.

(c) a(1,2,0) + b(-1,-1,2) + c(1,3,2) = (1,2,-1). a-b+c = 1, 2a-b+3c = 2, 2b+2c = -1. No solution . Not a linear combination.

(d) a(1,2,3) + b(2,5,7) + c(0,0,1) = (-1,-1,2). a+2b = -1, 2a+5b = -1, 3a+7b+c = 2. Unique solution, a=-3, b=1, c=4. Is a linear combination.

(e) a(1,1,1) + b(1,2,-1) + c(5,7,1) = (5,8,1). a+b+5c = 5, a+2b+7c = 8, a-b+c = 1.


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23

No solution . Not a linear combination.

(f) a(1,1,2) + b(2,2,4) + c(1,-1,1) = (5,-1,7). a+2b+c = 5, a+2b-c = -1, 2a+4b+c = 7. Many solutions, a=-2r+2, b=r, c=3. (5,-1,7) = (-2r+2)(1,1,2) + r(2,2,4) + 3(1,-1,1). There are many linear combinations.

For example, when r=1 we get (5,-1,7) = 0(1,1,2) + (2,2,4) + 3(1,-1,1). When r=2, (5,-1,7) = -2(1,1,2) + 2(2,2,4) + 3(1,-1,1).

12. (a) u+ (v+w) = (u1

, u2

, . . . , un

) + ((v1

, v2

, . . . , vn

) + (w1

, w2

, . . . , wn

))

= (u1

, u2

, . . . , un

) + (v1

+w1

, v2

+w2

, . . . , vn

+wn

)

= (u1

+(v1

+w1

), u2

+(v2

+w2

), . . . , un

+(vn

+wn

))

= ((u1

+v1

)+w1

, (u2

+v2

)+w2

, . . . , (un

+vn

)+wn

)

= ((u1

+v1

), (u2

+v2

) , . . . , (un

+vn

))+ (w1

, w2

, . . . , wn

)

= ((u1

, u2

, . . . , un

) + (v1

, v2

, . . . , vn

)) + (w1

, w2

, . . . , wn

) = (u+v) +w .

(b) u+ (- u) = (u1

, u2

, . . . , un

) + (- 1)(u1

, u2

, . . . , un

)

= (u1

, u2

, . . . , un

) + (- u1

, - u2

, . . . , - un

) = (u1

- u1

,u2

- u2

, . . . , un

- un

)

= (0,0, . . . ,0) = 0.

(c) (c+d)u= (c+d)(u1

, u2

, . . . , un

) = ((c+d)u1

,(c+d)u2

, . . . , (c+d)un

)

= (cu1

+du1, cu

2+du

2, . . . , cu

n+du

n)

= (cu1

, cu2

, . . . , cun

) + (du1

, du2

, . . . , dun

)

= c(u1

, u2

, . . . , un

) + d(u1

, u2

, . . . , un

) = cu+ du.

(d) 1u= 1(u1

, u2

, . . . , un

) = (1xu1

,1xu2

, . . . , 1xun

) = (u1

,u2

, . . . ,un

) =u.

Exercise Set 1.4

1. (a) Let W be the subset of vectors of the form (a,3a). Let u=(a,3a),v=(b,3b) and k be a scalar. Then u+v=((a+b), 3(a+b)) and ku=(ka,3ka). The second component of bothu+vand kuis 3 times the first component. Thus W is closed under addition and scalar multiplication - it is a subspace.

(b) Let W be the subset of vectors of the form (a,-a). Let u=(a,-a), v=(b,-b) and k be a scalar. Then u+v=((a+b), -(a+b)) and ku=(ka,-ka). The second component of bothu+vand kuis minus the first component. Thus W is closed under addition and scalar multiplication - it is a subspace.

(c) Let W be the subset of vectors of the form (a,0). Let u=(a,0),v=(b,0) and k be a scalar. Then u+v=((a+b), 0) and ku=(ka,0). The second component of bothu+vand kuis zero. Thus W is closed


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24

under addition and scalar multiplication - it is a subspace.

(d) Let W be the subset of vectors of the form (2a,3a). Letu=(2a,3a), v=(2b,3b) and k be a scalar. Then u+v=(2(a+b), 3(a+b)) and ku=(2ka,3ka). The second component of bothu+vand kuis 3/2 times the first component. Thus W is

closed under addition and scalar multiplication - it is a subspace.

2. (a) Let W be the subset of vectors of the form (a,b,b). Let u=(a,b,b), v=(c,d,d) and k be a scalar. Then u+v=(a+c, b+d,b+d) and ku=(ka,kb,kb). The second and third components ofu+vare the same; so are those of ku. Thus W is

closed under addition and scalar multiplication - it is a subspace.

(b) Let W be the subset of vectors of the form (a,-a,b). Letu=(a,-a,b),v=(c,-c,d) and k be a scalar. Then u+v=(a+c,-(a+c),b+d) and ku=(ka,-ka,kb). The second component ofu+vequals minus the first component; and same for ku. Thus W is closed under addition and scalar multiplication - it is a subspace.

(c) Let W be the subset of vectors of the form (a,2a,-a). Let u=(a,2a,-a), v=(b,2b,-b) and kbe a scalar. Thenu+v=(a+b,2(a+b),-(a+b)) and ku=(ka,2ka,-ka).

The second component ofu+vis twice the first, and the third component is minus the first; and same for ku. Thus W is closed under addition and scalar multiplication - it is a subspace.

(d) Let W be the subset of vectors of the form (a,a,b,b). Letu=(a,a,b,b), v=(c,c,d,d) and kbe a scalar. Then u+v=(a+c,a+c,b+d,b+d) and ku=(ka,ka,kb,kb).

The 1stand 2

ndcomponents ofu+vare the same, so are 3

rdand 4

th; same for ku. Thus

W is closed under addition and scalar multiplication - it is a subspace.

3. (a) Let W be the subset of vectors of the form (a,b,2a+3b). Letu=(a,b,2a+3b),v=(c,d,2c+3d) and k be a scalar. Then u+v=(a+c, b+d,2(a+c)+3(b+d)) and

ku=(ka,kb,2ka+3kb).The third component ofu+vis twice the first plus three times the second; same for ku. Thus W is closed under addition and scalar multiplication - it is a subspace.

(b) Let W be the subset of vectors of the form (a,b,3). Letu=(a,b,3), v=(c,d,3) and k be ascalar. Thenu+v=(a+c,b+d,6). The third component is not 6. Thus u+vis not in W. W is

not a subspace. Let us check closure under scalar multiplication. ku=(ka,kb,3k). Thus

unless k=1, kuis not in W. W is not closed under scalar multiplication either.

(c) Let W be the subset of vectors of the form (a,a+2,b). Let u=(a,a+2,b),v=(c,c+2,d) and k be a scalar. Then u+v=(a+c,a+c+4,b+d). The second component is not the first plus 2. Thusu+vis not in W. W is not a subspace. Let us check closure under scalar multiplication. ku=(ka,ka+2k,kb). Thus unless k=1 kuis not in W. W is not closed under scalar multiplication either.


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25

(d) Let W be the subset of vectors of the form (a,-a,0). Letu=(a,-a,0),v=(b,-b,0) and k be a scalar. Then u+v=(a+b,-(a+b),0) and

ku=(ka,-ka,0).The second component ofu+vis minus the first and the last component is zero; same for ku. Thus W is closed under addition and scalar multiplication - it is a subspace.

4.

1 3 1 0

0 1 1 0

2 7 3 0

1 3 1 0

0 1 1 0

0 1 1 0

1 0 2 0

0 1 1 0

0 0 0 0

. General solution (2r,-r,r).

Let W be the subset of vectors of the form (2r,-r,r). Let u=(2r,-r,r),v=(2s,-s,s) and k be a scalar. Then u+v=(2(r+s),-(r+s),r+s) and ku=(2kr, -kr, kr).The first component ofu+vis twice the last component, and the second component is minus the last component; same for ku. Thus W is closed under addition and scalar multiplication - it is a subspace. Line defined by vector (2, -1, 1).

5.

1 1 7 0

0 1 4 0

1 0 3 0

1 1 7 0

0 1 4 0

0 1 4 0

1 0 3 0

0 1 4 0

0 0 0 0

. General solution (3r,4r,r).

Let W be the subset of vectors of the form (3r,4r,r). Let u=(3r,4r,r),v=(3s,4s,s) and k bea

scalar. Then u+v=(3(r+s),4(r+s),r+s) and ku=(3kr, 4kr, kr).The first component ofu+vis three times the last component, and the second component is four times the last component; same for ku. Thus W is closed under addition and scalar multiplication - it

is a subspace. It is a line defined by vector (3,4,1).

6.

1 2 3 0

1 2 1 0

1 2 4 0

1 2 3 0

0 0 4 0

0 0 1 0

1 2 0 0

0 0 1 0

0 0 0 0

. General solution (2r, r, 0).

Let W be the subset of vectors of the form (2r,r,0). Let u=(2r,r,0),v=(2s,s,0) and k be a scalar. Then u+v=(2(r+s),r+s,0) and ku=(2kr, kr, 0).The first component ofu+vis twice the second component, and the last component is zero; same for ku. Thus W is closed

under addition and scalar multiplication - it is a subspace. Line defined by vector (2, 1,0).

7.

1 2 1 0

1 3 1 0

3 7 1 0

1 2 1 0

0 1 2 0

0 1 2 0

1 0 5 0

0 1 2 0

0 0 0 0

. General solution (5r,-2r,r).


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26

Let W be the subset of vectors of the form (5r,-2r,r). Let u=(5r,-2r,r), v=(5s,-2s,s) and kbe

a scalar. Then u+v=(5(r+s),-2(r+s),r+s) and ku=(5kr, -2kr, kr).The first component ofu+v is five times the last component, and the second component is minus two times the last component; same for ku. Thus W is closed under addition and scalar multiplication - it

is a subspace. Line defined by vector (5, -2, 1).

8. General solution is (2r-2s, r -3s, r, s). Let u=(2r-2s, r -3s, r, s) and v=(2p-2q, p-3q, p, q). Thenu+v=(2r-2s+2p-2q, r-3s+p-3q, r+p, s+q) = (2(r+p)-2(s+q), (r+p)-3(s+q), r+p, s+q). The first component is twice the third minus twice the fourth. The second component is the third minus three times the fourth - the required form. Thus space is closed under addition. Let k be a scalar. Then ku= (2kr -2ks, kr -3ks, kr, ks) the required form. Thus the set of solutions is closed under addition and scalar multiplication; it is a subspace.

9. General solution is (3r + s, -r - 4s, r, s). (a) Two specific solutions: r=1, s=1,u(4, -5, 1, 1);

r=-1, s=2,v(-1, -7, -1, 2). (b) Other solutions: u+v=(3, -12, 0, 3) and say, -2u=(-8, 10, -2, -2), 4v=(-4, -28, -4, 8). (-2u)+(4v)=(-12, -18, -6, 6). (c) Are solutions for r=0,s=3; r=-2,s=-2; r=-4,s=8; r=-6,s=6 respectively.

10. General solution is (2r, 3r, r). (a) Two specific solutions: r=1 gives u(2, 3, 1); r=2 givesv(4, 6, 2). (b) Other solutions: u+v= (6, 9, 3); 4u = (8, 12, 4); -v= (-4, -6, - 2);uv= (-2, -3, -1). Are solutions for r = 3. r = 4, r = -2, r = -1 respectively.

11. General solution is (2r - s, -3r - 2s r, s). (a) Two specific solutions: r=1, s=1, u(1, -5, 1,1);

r=2, s=-1,v(5, -4, 2, -1). (b) Other solutions: u+v=(6, -9, 3, 0). and say 2u=(2, -10, 2, 2),

3v=(15, -12, 6, -3). (2u)+(3v)=(17, -22, 8, -1). (c) Are solutions for r=3,s=0; r=2,s=2; r=6,s=-3; r=8,s=-1 respectively.

Exercise Set 1.5

1. Standard basis forR2: {(1, 0), (0, 1)}. (a) Let (a, b) be an arbitrary vector inR

2. We can

write

(a, b) = a(1, 0) + b(0, 1). Thus vectors (1, 0) and (0, 1) spanR2. (b) Let us examine the

identity p(1, 0) + q(0, 1) = (0, 0). This gives (p, 0) + (0, q) = 0, (p, q) = (0, 0). Thus p=0 and q=0. The vectors are linearly independent.

2. Standard basis forR4: {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}.

(a) Let (a,b,c,d) be an arbitrary vector in R4. We can write

(a,b,c,d) = a(1,0,0,0) +b(0,1,0,0) +c(0,0,1,0) +d(0,0,0,1). Thus vectors in basis spanR4.

(b) Let us examine the identity p(1,0,0,0)+q(0,1,0,0)+r(0,0,1,0)+s(0,0,0,1) = (0,0,0,0). This gives (p,q,r,s) = 0. Thus p=0, q=0, r=0, s=0. The vectors are linearly independent.


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3. (a) (a, a, b) + (c, c, d) = (a+c, a+c, b+d). 1st components same. Closed under addition. k(a, a, b) = (ka, ka, kb). 1st components same. Closed under scalar multiplication. Subspace. (a, a, b) = a(1, 1, 0) + b(0, 0, 1). Vectors (1, 1, 0) and (0, 0, 1) span space and are linearly independent. {(1, 1, 0), (0, 0, 1)} is a basis. Dimension is 2.

(b) (a, 2a, b) + (c, 2c, d) = (a+c, 2(a+c), b+d). 2nd component is twice first. Closed under addition. k(a, 2a, b) = (ka, 2ka, kb). 2nd component is twice 1st. Closed under scalar multiplication. Subspace. (a, 2a, b) = a(1, 2, 0) + b(0, 0, 1). Vectors (1, 2, 0) and (0, 0, 1) span the space and are linearly independent. {(1, 2, 0, (0, 0, 1)} is a basis. Dimension is2.

(c) (a, 2a, 4a) + (b, 2b, 4b) = (a+b, 2a+2b, 4a+4b) =(a+b, 2(a+b), 4(a+b)). 2nd component is twice 1st, 3rd component four times 1st. Closed under addition. k(a, 2a, 4a) = (ka, 2ka,4ka). 2nd component is twice 1st, 3rd component four times 1st. Closed under scalar multiplication Subspace. (a, 2a, 4a) = a(1, 2, 4). {(1, 2, 4)} is a basis. Dimension is 1. Space is a line defined by the vector (1, 2, 4).

(d) (a, -a, 0) + (b, -b, 0) = (a+b, -a-b, 0) = (a+b, -(a+b), 0). 2nd component is negative of 1st. Closed under addition. k(a, -a, 0) = (ka, -ka, 0). 2nd component is negative first. Closed under scalar multiplication. Subspace. (a, -a, 0) = a(1, -1, 0). {(1, -1, 0)} is abasis. Dimension is 1. Space is line defined by the vector (1, -1, 0).

4. (a) (a, b, a) + (c, d, c) = (a+c, b+d, a+c). 3rd component same as 1st. Closed under addition. k(a, b, a) = (ka, kb, ka). 3rd component same as 1st. Closed under scalar multiplication. Subspace. (a, b, a) = a(1, 0, 1) + b(0, 1, 0). {(1, 0, 1), (0, 1, 0)} is a basis. Dimension is 2.

(b) (a, b, 0) + (c, d, 0) = (a+c, b+d, 0). Last component is zero. Closed under addition. k(a, b, 0) = (ka, kb, 0). Last component is zero. Closed under scalar multiplication. Subspace. (a, b, 0) = a(1, 0, 0) + b(0, 1, 0). {(1, 0, 0), (0, 1, 0)} is a basis. Dimension is 2. It is xy-plane.

(c) (a, b, 2) + (c, d, 2) = (a+c, b+d, 4). Last component not 2. Not closed under addition. Not a subspace.

(d) (a, a, a+3) + (b, b, b+3) = (a+b, a+b, a+b+6). Last component is not 1st plus 3. Not closed under addition. Not a subspace.

5. (a) a(1, 2, 3) + b(1, 2, 3) = (a+b)(1, 2, 3). Sum is a scalar multiple of (1, 2, 3). Closedunder addition. ka(1, 2, 3) = (ka)(1, 2, 3). It is a scalar multiple of (1, 2, 3). Closed under scalar

multiplication. Subspace ofR3. Basis {(1, 2, 3)}. Dimension is 1. Space is line defined by

vector (1, 2, 3).

(b) (a, 0, 0) + (b, 0, 0) = (a+b, 0, 0). Last two components zero. Closed under addition. k(a, 0, 0) = (ka, 0, 0). Last two components zero. Closed under scalar


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multiplication. Subspace ofR3. Basis {(1, 0, 0)}. Dimension is 1. Space is the x-axis.

(c) (a, 2a) + (b, 2b) = (a+b, 2a+2b) = (a+b, 2(a+b)). 2nd component is twice 1st. Closed under addition. k(a, 2a) = (ka, 2ka). 2nd component is twice 1st. Closed under scalar multiplication. Subspace. Basis {(1, 2)}. Dimension is 1. Space is line defined by vector

(1, 2) in R2.

(d) (a, b, c, 1) + (d, e, f, 1) = (a+d, b+e, c+f, 2). Last component is not 1. Not closedunder

addition. Not a subspace ofR4.

6. (a) True: Arbitrary vector can be expressed as a linear combination of (1, 0) and (0, 1). (a, b) = a(1, 0) + b(0, 1).

(b) True: (a, b) = a(1, 0) + b(0, 1) + 0(1, 1). (Some of the scalars can be zero).

(c) True: p(1, 0) + q(0, 1) = (0, 0) has the unique solution p=0, q=0.

(d) False: Consider the identity p(1, 0) + q(0, 1) + r(0, 2) = (0, 0). Does this have the unique solution p=0, q=0, r=0? No, can have 0(1, 0) + 2(0, 1) - 1(0, 2) = (0, 0).Vectors are not linearly independent - we say that they are linearly dependent.

(e) True: (x, y) = x(1, 0) - y(0, -1). Thus (1, 0) and (0, -1) spanR2. Further,

p(1, 0)+q(0, -1)=(0, 0) has the unique solution p=0, q=0. Vectors are linearlyindependent.

(f) True: (x, y) = x

2(2, 0) +

y

3(0, 3). Thus (2, 0) and (0, 3) span R

2. Further,

p(2, 0) + q(0, 3) = (0, 0)(2p, 3q) = (0, 0), has the unique solution p=0, q=0. Vectors linearly independent.

7. (a) True: (1, 0, 0) and (0, 1, 0) span the subset of vectors of the form (a, b, 0). Further, p(1, 0, 0) + q(0, 1, 0) = (0, 0, 0) has the unique solution p=0, q=0. Vectors are linearly independent. Subspace is 2D since 2 base vectors. The subspace is the xy plane.

(b) True: The vector (1, 0, 0) spans the subset of vectors of the form a(1, 0, 0). Further p(1, 0, 0)=(0,0,0) has unique solution p=0. Subspace is line defined by vector (1, 0,0). One vector in basis, thus 1D.

(c) True: Can write (a, 2a, b) in the form (a, 2a, b) = a(1, 2, 0) + b(0, 0, 1).

(d) True: Can write (a, b, 2a-b) in the form (a, b, 2a-b) = a(1, 0, 2) + b(0, 1, -1).


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(e) False: 1(1, 0, 0) + 1(0, 1, 0) -1(1, 1, 0) = (0, 0, 0). Thus vectors not linearly independent.

(f) False: R2is not a subset ofR

3. e.g., (1, 2) is an element ofR

2, but not ofR

3.

8. (a) Let (x, y) be an arbitrary vector inR2. Then (x, y) = x(1, 0) + y(0, 1). Thus (1, 0), (0, 1)

spanR2. Notice that both vectors are needed to span R2- we cannot just use one of them. Further, a(1, 0) + b(0, 1) = (0, 0) => (a, b) = (0, 0) => a=0, and b=0. Thus

(1, 0) and (0, 1) are linearly independent. They form a basis forR2.

(b) (x, y) can be expressed (x, y) = x(1, 0) + 3y(0, 1) - y(0, 2), or (x, y) = x(1, 0) + 5y(0, 1)

- 2y(0, 2); there are many ways. Thus (1, 0), (0, 1), (0, 2) spans R2. But it is not an

efficient spanning set. The vector (0, 2) is not really needed.

(c) 0(1, 0) + 2(0, 1) - (0, 2) = (0, 0). Thus {(1, 0), (0, 1), (0, 2)} is linearly dependent. It is not a basis.

9. (a) Let (x, y) be an arbitrary vector inR2. Then (x, y) = x(1, 0) + y(0, 1). Thus (1, 0), (0, 1)

spanR2. Notice that both vectors are needed to span R

2- we cannot just use one of

them. Further, a(1, 0) + b(0, 1) = (0, 0) => (a, b) = (0, 0) => a=0, and b=0. Thus

(1, 0) and (0, 1) are linearly independent. They form a basis forR2.

(b) (x, y) can be expressed (x, y) = x(1, 0) + y(0, 1) + 0(1, 1); or (x, y) = (x+y)(1, 0) + 2y(0, 1)

- y(1,1) - there are many ways. Thus (1, 0), (0, 1), (1, 1) span R2. But it is not an

efficient spanning set. The vector (1, 1) is not really needed.

(c) 1(1, 0) + 1(0, 1) - 1(1, 1) = (0, 0). Thus {(1, 0), (0, 1), (1, 1)} is linearly dependent. It is not a basis.

10. (a) Let (x, y, z) be an arbitrary vector inR3. Then (x, y, z) = x(1, 0, 0) + y(0, 1, 0) +

z(0, 0, 1). Thus (1, 0, 0), (0, 1, 0), (0, 0, 1) spanR3. Notice that all three vectors are

needed to spanR3- we cannot just use two of them. Further, a(1, 0, 0) + b(0, 1, 0) +

c(0, 0, 1) = (0, 0, 0) => (a, b, c) = (0, 0, 0) => a=0, b=0, and c=0. Thus

(1, 0, 0) and (0, 1, 0, (0, 0, 1) are linearly independent. They form a basis forR3.

(b) (x, y, z) can be expressed (x, y, z) = x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1) + 0(0, 1, 1); or

(x, y, z) = x(1, 0, 0) + (yz)2

(0, 1, 0) + (yz)2

(0, 0, 1) + (yz)2

(0, 1, 1).

- there are many ways. Thus (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1) spanR2. But it is not

an efficient spanning set. The vector (0, 1, 1) is not really needed.

(c) 1(1, 0, 0) + 1(0, 1, 0) + 1(0, 0, 1) - 0(0, 1, 1) = (0, 0, 0). Thus (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1) are linearly dependent. Do not form a basis.


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11. (a) Vectors (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1) spanR3but are not linearly

independent.

(b) Vectors (1, 0, 0), (0, 1, 0) are linearly independent but do not spanR3.

12. Separate the variables in the general solution, (2r - 2s, r - 3s, r, s)= r(2, 1, 1, 0) + s(-2, -3, 0,

1).

Vectors (2, 1, 1, 0) and (-2, 3, 0, 1) thus span W.

Also, identity p(2, 1, 1, 0) + q(-2, 3, 0, 1) = (0, 0, 0, 0) leads to p=0, q=0. The two vectors

are

thus linearly independent. Set {(2, 1, 1, 0), (-2, 3, 0, 1)} is therefore a basis for W.

Dimension of W is 2. Solutions form a plane through the origin inR4.

13. Separate the variables in the general solution, (3r + s, -r - 4s, r, s)= r(3, -1, 1, 0)+s(1, -4, 0,

1).

Vectors (3, -1, 1, 0) and (1, -4, 0, 1) thus span W.

Also, identity p(3, -1, 1, 0) + q(1, -4, 0, 1) = (0, 0, 0, 0) leads to p=0, q=0. The two vectors

are thus linearly independent. Set {(3, -1, 1, 0), (1, -4, 0, 1)} is therefore a basis for W.


14. (2r, 3r, r) = r(2, 3, 1). The set of solutions form the line through the origin in R3defined by

the vector (2, 3, 1). Set {(2, 3, 1)} is a basis. Dimension of W is 1.

15. Separate the variables in the solution, (2r - s, -3r - 2s, r, s)= r(2, -3, 1, 0) + s(-1, -2, 0, 1).

Vectors (2, -3, 1, 0) and (-1, -2, 0, 1) thus span W.

Also, identity p(2, -3, 1, 0) + q(-1, -2, 0, 1) = (0, 0, 0, 0) leads to p=0, q=0. The two vectors are thus linearly independent. The set {(2, -3, 1, 0), (-1, -2, 0, 1)} is therefore a basis for

W.


16. (2r, -r, 4r, r) = r(2, -1, 4, 1). The set of solutions form the line through the origin in R4


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defined by the vector (2,-1,4,1). Set {(2, -1, 4)} is a basis. Dimension of W is 1.

17. Separate the variables in the solution, (3r-s, r, s )= r(3, 1, 0) + s(-1, 0, 1).

Vectors (3, 1, 0) and (-1, 0, 1) thus span W.

Also, identity p(3, 1, 0) + q(-1, 0, 1) = (0, 0, 0) leads to p=0, q=0. The two vectors

are thus linearly independent. The set {(3, 1, 0), (-1, 0, 1)} is therefore a basis for W.


18. Separate the variables in the solution, (2r + t, 3r 2s, r, s, t) = r(2, 3, 1, 0, 0) +

s(0, -2, 0, 1, 0) + t(1, 0, 0, 0, 1). Vectors (2, 3, 1, 0, 0), (0, -2, 0, 1, 0), (1, 0, 0, 0, 1) thus

span W. The identity p(2, 3, 1, 0, 0) + q(0, -2, 0, 1, 0) + h(1, 0, 0, 0, 1) = (0, 0, 0, 0, 0)

leads to p = 0, q = 0, h = 0. The vectors are thus linearly independent. The set

{(2, 3, 1, 0, 0), (0, -2, 0, 1, 0), (1, 0, 0, 0, 1)} is a basis for W. The dimension of W is 3.

Solutions form a three-dimensional subspace of the five-dimensional spaceR5.

19. (a) a(1, 0, 2) + b(1, 1, 0) + c(5, 3, 6)=0gives a+b+5c=0, b+3c=0, 2a+6c=0. Unique solution a=0, b=0, c=0. Thus linearly independent.

(b) a(1,1,1) + b(2, -1, 1) + c(3, -3, 0)=0gives a + 2b + 3c = 0, a b -3c = 0, a + b = 0. Unique solution a=0, b=0, c=0. Thus linearly independent.

(c) a(1, -1, 1) +b(2, 1, 0)+c(4, -1, 2)=0gives a+2b+4c=0, -a+b-c=0, a+2c=0. Many solutions, a=-2r, b=-r, c=r, where r is a real number. Thus linearly dependent.

(d) a(1, 2, 1)+b(-2, 1, 3)+c(-1, 8, 9)=0gives a-2b-c=0, 2a+b+8c=0, a+3b+9c=0. Many solutions, a=-3r, b=-2r, c=r. Thus linearly dependent.

(e) a(-2,0,3) + b(5, 2, 1) + c(10,6,9)=0 gives 2a+5b+10c=0, 2b+6c=0, 3a+b+9c=0. Unique solution a=0, b=0, c=0. Thus linearly independent.

(f) a(3,4,1)+b(2,1,0)+c(9,7,1)=0gives 3a+2b+9c=0, 4a+b+7c=0, a+c=0. Many solutions, a=-r, b=-3r, c=r. Thus linearly dependent.

Exercise Set 1.6

1. (a) (2,1)(3,4) = 2x3 + 1x4 = 6 + 4 = 10


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(b) (1,- 4)(3,0) = 1x3 + - 4x0 = 3

(c) (2,0)(0,- 1) = 2x0 + 0x- 1= 0

(d) (5,- 2)(- 3,- 4) = 5x- 3 + - 2x- 4 = - 15 + 8 = - 7

2. (a) (1,2,3)(4,1,0) = 1x4 + 2x1 + 3x0 = 4 + 2 + 0 = 6

(b) (3,4,- 2)(5,1,- 1) = 3x5 + 4x1 + - 2x- 1 = 15 + 4 + 2 = 21

(c) (7,1,- 2)(3,- 5,8) = 7x3 + 1x- 5 + - 2x8 = 21 - 5 - 16 = 0

(d) (3, 2, 0).(5, -2, 8) = 3x5 + 2x-2 + 0x8 = 15 - 4 + 0 = 11

3. (a) (5,1)(2,- 3) = 5x2 + 1x- 3 = 10 - 3 = 7

(b) (- 3,1,5)(2,0,4) = - 3x2 + 1x0 + 5x4 = - 6 + 0 + 20 = 14

(c) (7,1,2,- 4)(3,0,- 1,5) = 7x3 + 1x0 + 2x- 1 + - 4x5 = 21 + 0 - 2 - 20 = - 1

(d) (2,3,- 4,1,6)(- 3,1,- 4,5,- 1) = 2x- 3 + 3x1 + - 4x- 4 + 1x5 + 6x- 1= - 6 + 3 + 16 + 5 - 6 = 12

(e) (1,2,3,0,0,0)(0,0,0,- 2,- 4,9) = 1x0 + 2x0 + 3x0 + 0x- 2 + 0x- 4 + 0x9 = 0

4. (a)

1

3 .

-2

5 = 1x-2 + 3x5 = -2 + 15 = 13

(b)

5

0 .

4

-6 = 5x4 + 0x-6 = 20 +0 = 20

(c)

2

0-5

.

3

6-4

= 2x3 + 0x6 + -5x-4 = 6 + 0 + 20 = 26

(d)

1

3-7

.

-2

8-3

= 1x-2 + 3x8 + -7x-3 = -2 + 24 + 21 = 43

5. (a) ||(1, 2)|| = 12+ 22 = 5 (b) ||(3, - 4)|| = 32+ (-4)2 = 25 = 5


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(c) ||(4, 0)|| = 42+ 02 = 16 = 4 (d) ||(- 3, 1)|| = (-3)2+ 12 = 10

(e) ||(0, 27)|| = 02+ 272 = 27

6. (a) ||(1,3,- 1)|| = 12+ 32 + (-1)2 = 11

(b) ||(3,0,4)|| = 32+ 02 + 42 = 25 = 5

(c) ||(5,1,1)|| = 52+ 12 + 12 = 27 = 3 3

(d) ||(0,5,0)|| = 02+ 52 + 02 = 25 = 5

(e) ||(7,- 2,- 3)|| = 72+ (-2)2 + (-3)2 = 62

7. (a) ||(5,2)|| = 52+ 22 = 29

(b) ||(- 4,2,3)|| = (-4)2+ 22 + 32 = 29

(c) ||(1,2,3,4)|| = 12+ 22 + 32 + 42 = 30

(d) ||(4,- 2,1,3)|| = 42+ (-2)2 + 12 + 32 = 30

(e) ||(- 3,0,1,4,2)|| = (-3)2+ 02 + 12 + 42 + 22 = 30

(f) ||(0,0,0,7,0,0)|| = 02+ 02 + 02 + 72 + 02 + 02 = 49 = 7

8. (a) ||

3

4 || = 32+ 4

2 = 25 = 5 (b) ||

2

-7 || = 22+ (-7)

2 = 53

(c) ||

1

23

|| = 12+ 22 + 32 = 14 (d) ||

-2

05

|| = (-2)2+ 02 + 52 =

29


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(e) ||

2

359

|| = 22+ 32 + 52 + 92 = 119

9. (a) (1,3)||(1,3)|| = ( 110 , 3

10 )

(b) (2,4)||(2,4) ||

= ( 22 5

, 4

2 5) = ( 1

5, 2

5)

(c)(1,2,3)

||(1,2,3)|| = ( 1

14,

2

14,

3

14)

(d)(- 2,4,0)

||(- 2,4,0)|| = ( 2

20,

4

20, 0) = ( 1

5,

2

5, 0)

(e)(0,5,0)

||(0,5,0)|| = (0,1,0)

10. (a)(4,2)

||(4,2)|| = ( 4

2 5,

2

2 5) = ( 2

5,

1

5)

(b) (4,1,1)

|| (4,1,1) ||= ( 4

3 2,

1

3 2,

1

3 2)

(c)(7,2,0,1)

||(7,2,0,1)|| = ( 7

3 6,

2

3 6, 0,

1

3 6)

(d)(3,- 1,1,2)

||(3,- 1,1,2)|| = ( 3

15,

1

15,

1

15,

2

15)

(e) (0,0,0,7,0,0)

|| (0,0,0,7,0,0) ||= (0,0,0,1,0,0)

11. (a)

4

3/||

4

3|| =

4/5

3/5 (b)

1

-3/||

1

-3 || =

1/ 10

-3/ 10


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(c)

3

40

/||

3

40

|| =

3/5

4/50

(d)

-1

2-5

/||

-1

2-5

|| =

-1/ 30

2/ 30

-5/ 30

(e)

3018

/ ||

3018

|| =

3/ 740

1/ 74

8/ 74

12. (a) cos=(- 1,1)(0,1)

||(- 1,1)|| ||(0,1)|| =1

2 . =

4 = 45

(b) cos=(2,0)(1, 3)

||(2,0)|| ||(1, 3)||

=24 =

12 . =

3 = 60

(c) cos=(2,3)(3,- 2)

||(2,3)|| ||(3,- 2)|| = 0. = 2 = 90

(d) cos=(5,2)(- 5,- 2)

||(5,2)|| ||(- 5,- 2)|| =- 2929 = - 1. = = 180

13. (a) cos=(4,- 1)(2,3)

||(4,- 1)|| ||(2,3)|| =

5

17 13 (=70.34620)

(b) cos=(3,- 1,2)(4,1,1)

||(3,- 1,2)|| ||(4,1,1)|| =13

14 18 =

13

6 7 (=35.02290)

(c) cos=(2,- 1,0)(5,3,1)

||(2,- 1,0)|| ||(5,3,1)|| =7

5 35 =

7

5 7 =

7

5(= 58.0519

0)

(d) cos=(7,1,0,0)(3,2,1,0)

||(7,1,0,0)|| ||(3,2,1,0)|| =23

50 14 =

23

10 7 (=29.62050)

(e) cos=(1,2,- 1,3,1)(2,0,1,0,4)

||(1,2,- 1,3,1)|| ||(2,0,1,0,4)|| =5

4 21 (=74.17070)


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36

14. (a) cos=

1

2 .

- 1

4

||

1

2 || ||

- 1

4 || =

7

5 17 (=40.60130)

(b) cos= 51 . 0- 3

||

5

1 || ||

0

- 3 || =

-3

26 9 =

-1

26 (=101.30990)

(c) cos=

1

- 3 0

.

2

5- 1

||

1

- 3 0

|| ||

2

5- 1

||

=-13

10 30 =

-13

10 3 (=138.63850)

(d) cos=

- 2

3- 4

.

2

5- 1

||

- 2

3- 4

|| ||

2

5- 1

||

=15

29 30 =

15

29 2 (=59.43290)

15. (a) (1,3)(3,- 1) = 1x3 + 3x- 1 = 0, thus the vectors are orthogonal.

(b) (- 2,4)(4,2) = - 2x4 + 4x2 = 0, thus the vectors are orthogonal.

(c) (3,0)(0,- 2) = 3x0 + 0x- 2= 0, thus the vectors are orthogonal.

(d) (7,- 1)(1,7) = 7x1 + - 1x7= 0, thus the vectors are orthogonal.

16. (a) (3,- 5)(5,3) = 3x5 + - 5x3 = 0, thus the vectors are orthogonal.

(b) (1,2,- 3)(4,1,2) = 1x4 + 2x1 + - 3x2 = 0, thus the vectors are orthogonal.

(c) (7,1,0)(2,- 14,3) = 7x2 + 1x- 14 + 0x3 = 0, thus the vectors are orthogonal.

(d) (5,1,0,2)(- 3,7,9,4) = 5x- 3 + 1x7 + 0x9 + 2x4 = 0, thus the vectors are orthogonal.

(e) (1,- 1,2,- 5,9)(4,7,4,1,0) = 1x4 + - 1x7 + 2x4 + - 5x1+ 9x0 = 0, thus the vectors areorthogonal.


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37

17. (a)

1

2 .

- 6

3 = 1x-6 + 2x3 = 0, thus the vectors are orthogonal.

(b) 5- 2 . 410 = 5x4 + -2x10 = 0, thus the vectors are orthogonal.

(c)

4

- 1 0

.

2

8- 1

= 4x2 + -1x8 + 0x-1 = 0, thus the vectors are orthogonal.

(d)

- 2

3 2

.

2

6- 7

= -2x2 + 3x6 + 2x-7 = 0, thus the vectors are orthogonal.

18. (a) If (a,b) is orthogonal to (1,3), then (a,b)(1,3) = a + 3b = 0, so a = - 3b. Thus anyvector of the form (- 3b,b) is orthogonal to (1,3).

(b) If (a,b) is orthogonal to (7,- 1), then (a,b)(7,- 1) = 7a - b = 0, so b = 7a. Thus anyvector of the form (a,7a) is orthogonal to (7,- 1).

(c) If (a,b) is orthogonal to (- 4,- 1), then (a,b)(- 4,- 1) = - 4a - b = 0, so b = - 4a. Thusany vector of the form (a,- 4a) is orthogonal to (- 4,- 1).

(d) If (a,b) is orthogonal to (- 3,0), then (a,b)(- 3,0) = - 3a = 0, so a = 0. Thus anyvector of the form (0,b) is orthogonal to (- 3,0).

19. (a) If (a,b) is orthogonal to (5,- 1), then (a,b)(5,- 1) = 5a - b = 0, so b = 5a. Thus anyvector of the form (a,5a) is orthogonal to (5,- 1).

(b) If (a,b,c) is orthogonal to (1,- 2,3), then (a,b,c)(1,- 2,3) = a - 2b +3c = 0, soa = 2b- 3c. Thus any vector of the form (2b- 3c,b,c) is orthogonal to (1,- 2,3).

(c) If (a,b,c) is orthogonal to (5,1,- 1), then (a,b,c)(5,1,- 1) = 5a + b - c = 0, soc = 5a+b. Thus any vector of the form (a,b,5a+b) is orthogonal to (5,1,- 1).

(d) If (a,b,c,d) is orthogonal to (5,0,1,1), then (a,b,c,d)(5,0,1,1) = 5a + c + d = 0, sod = - 5a- c. Thus any vector of the form (a,b,c,- 5a- c) is orthogonal to (5,0,1,1).

(e) If (a,b,c,d) is orthogonal to (6,- 1,2,3), then (a,b,c,d)(6,- 1,2,3) = 6a - b + 2c + 3d =0, so b = 6a+2c+3d. Thus any vector of the form (a,6a+2c+3d, c,d) is orthogonal to

(6,- 1,2,3).


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38

(f) If (a,b,c,d,e) is orthogonal to (0,- 2,3,1,5), then (a,b,c,d,e)(0,- 2,3,1,5)= - 2b +3c + d + 5e = 0, so d = 2b- 3c- 5e. Thus any vector of the form(a,b,c,2b- 3c- 5e,e) is orthogonal to (0,- 2,3,1,5).

20. If (a,b,c) is orthogonal to both (1,2,- 1) and (3,1,0), then (a,b,c)(1,2,- 1) = a + 2b - c =0 and (a,b,c)(3,1,0) = 3a + b = 0. These equations yield the solution b = - 3a and

c = - 5a, so any vector of the form (a,- 3a,- 5a) is orthogonal to both (1,2,- 1) and (3,1,0).

21. Let (a,b,c) be in W. Then (a,b,c) is orthogonal to (-1,1,1). (a,b,c)(-1,1,1)=0, -a+b+c=0, c=a-b. W consists of vectors of the form (a,b,a-b). Separate the variables. (a,b,a-b)=a(1,0,1)+b(0,1,-1). (1,0,1), (0,1,-1) span W. Vectors are also linearly independent. {(1,0,1),(0,1,-1)} is a basis for W. The dimension of W is 2.

It is a plane spanned by (1,0,1) and (0,1,-1).

22. Let (a,b,c) be in W. Then (a,b,c) is orthogonal to (-3,4,1). (a,b,c)(-3,4,1)=0, -3a+4b+c=0, c=3a-4b. W consists of vectors of the form (a,b,3a-4b). Separate the variables. (a,b,3a-4b)=a(1,0,3)+b(0,1,-4). {(1,0,3), ((0,1,-4)} is a basis for W. The dimension of W is 2. It is a plane spanned by (1,0,3) and (0,1,-4).

23. Let (a,b,c) be in W. Then (a,b,c) is orthogonal to (1,-2,5). (a,b,c)(1,-2,5)=0, a-2b+5c=0, a=2b-5c. W consists of vectors of the form (2b-5c,b,c). Separate the variables.

(2b-5c,b,c)=b(2,1,0)+c(-5,0,1). {(2,1,0), ((-5,0,1)} is a basis for W. The dimension of W is 2. It is a plane spanned by (2,1,0) and (-5,0,1).

24. Let (a,b,c,d) be in W. Then (a,b,c,d) is orthogonal to (1,-3,7,4).(a,b,c,d)(1,-3,7,4)=0, a-3b+7c+4d=0, a=3b-7c-4d. W consists of vectors of the form (3b-7c-d,b,c,d). Separate the variables. (3b-7c-4d,b,c,d)=b(3,1,0,0)+c(-7,0,1,0) +d(-4,0,0,1). {(3,1,0,0), (-7,0,1,0), (-4,0,0,1)} is a basis for W. The dimension of W is 3.

25. (a) d = (6- 2)2+(5- 2)2 = 5. (b) d = (3+4)2+(1- 0)2 = 50 = 5 2 .

(c) d = (7- 2)2+(- 3- 2)2 = 50 = 5 2 . (d) d = (1- 5)2+(- 3- 1)2 = 4 2 .

26. (a) d = (4- 2)2+(1+3)2 = 20 = 2 5 . (b) d = (1- 2)2+(2- 1)2+(3- 0)2 = 11 .

(c) d = (- 3- 4)2+(1+1)2+(2- 1)2 = 54 = 3 6 .


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39

(d) d2 = (5- 2)2+(1- 0)2+(0- 1)2+(0- 3)2 = 20, so d = 20 = 2 5 .

(e) d2 = (- 3- 2)2+(1- 1)2+(1- 4)2+(0- 1)2+(2+1)2 = 44, so d = 44 = 2 11 .

27. (a) (u+v)w= (u1+ v

1)w

1 + (u

2+ v

2)w

2 + . . . + (u

n+ v

n)w

n

= u1w

1 + v

1w

1 + u

2w

2 + v

2w

2 + . . . + u

nw

n+ v

nw

n

= u1w

1 + u

2w

2 + . . . + u

nw

n+ v

1w

1 + v

2w

2 + . . . + v

nw

n =u w+v w.

(b) cuv= cu1v

1 + cu

2v

2 + . . . + cu

nv

n = c(u

1v

1 + u

2v

2 + . . . + u

nv

n) = c(uv), and

cu1v

1 + cu

2v

2 + . . . + cu

nv

n = u

1cv

1 + u

2cv

2 + . . . + u

ncv

n =u cv.

28. uis a positive scalar multiple ofvso it has the same direction asv. The magnitude ofuis

||u|| =1

||v||(v

1)2+(v

2)2+...+(v

n)2 =

||v||||v||

= 1, souis a unit vector.

29. uand vare orthogonal if and only if the cosine of the anglebetween them is zero.

cos= uv

||u|| ||v|| = 0 if and only ifu v= 0.

30. If uv= uw then u(v- w) = 0 for all vectors uin U. Sincev- wis a vector in U thismeans that (v- w)(v- w) = 0. Therefore v- w=0, so v =w.

31. u(a1v

1+ a

2v

2+ . . . + a

nv

n) =u (a

1v

1) +u (a

2v

2+ . . . + a

nv

n)

=u (a1v

1) +u (a

2v

2) +u (a

3v

3+ . . . + a

nv

n) = . . . = u (a

1v

1) +u (a

2v

2) + . . .

+u (anv

n)

= a1(uv

1) + a

2(uv

2) + . . . + a

n(uv

n) = a

1u v

1+ a

2u v

2+ . . . + a

nu v

n.

32. (a) vector (b) not valid (c) not valid (d) scalar

(e) not valid (f) scalar (g) not valid

(h) not valid


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40

33. ||c(3,0,4)|| = 3cx3c+4cx4c = |c| 9+16 = 5|c| = 15, so |c| = 3 and c = 3.

34. ||u+v||2 = (u1+v

1)2 + (u

2+v

2)2 + . . . + (u

n+v

n)2

= u12 + 2u

1v

1+ v

12 + u

22 + 2u

2v

2 + v

22 + . . . + u

n2 + 2u

nv

n+ v

n2

= u12 + u

22 + . . . + u

n2 + 2u

1v

1 + 2u

2v

2 + . . . + 2u

nv

n + v

12 + v

22 + . . . +

vn2

= u12 + u

22 + . . . + u

n2 + 2(u

1v

1 + u

2v

2 + . . . + u

nv

n) + v

12 + v

22 + . . . + v

n2

= ||u||2 + 2(uv) + ||v||2 = ||u||2 + ||v||2 if and only ifu v= 0, i.e., if and only ifuand

vare orthogonal.

35. (a,b)(- b,a) = a - b + b a = 0, so (- b,a) is orthogonal to (a,b).

36. (u+v)(u - v) = (u1 + v

1)(u

1- v

1) + (u

2+ v

2)(u

2- v

2) + . . . + (u

n + v

n)(u

n- v

n)

= u12 - v

12 + u

22 - v

22 + . . . + u

n2 - v

n2

= u12 + u

22 + . . . + u

n2 - v

12 - v

22 - . . . - v

n2 = ||u|| - ||v||.

Thus ||u|| - ||v||=0 if and only if (u+v)(u-v) = 0. That is ||u|| = ||v|| if and only ifu +vandu -v are orthogonal.

37. (a) ||u||2 = u12+u

22+ . . . +u

n2 0, so ||u|| 0.

(b) ||u|| = 0 if and only if u12+u

22+ . . . +u

n2 = 0 if and only if u

1 = u

2 = . . . = u

n = 0.

(c) ||cu||2 = (cu1)2+(cu

2)2+ . . . +(cu

n)2 = c2(u

12+u

22+ . . . +u

n2) = c2 ||u||2,

so ||cu|| = |c| ||u||.

38. (a) ||u|| = |u1| + |u

2| + . . . + |u

n| 0 since each term is equal to or greater than zero.

|u1| + |u

2| + . . . + |u

n| = 0 if and only if each term is zero.

||cu|| = |cu1| + |cu

2| + . . . + |cu

n| = |c||u

1| + |c||u

2| + . . . + |c||u

n|


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41

= |c|(|u1| + |u

2| + . . . + |u

n|) = |c| ||u||.

||(1,2)|| = |1| + |2| = 3, ||(- 3,4)|| = |- 3| + |4| = 7, ||(1,2,- 5)|| = |1| + |2| + |- 5|= 8,

and ||(0,- 2,7)|| = |0| + |- 2| + |7| = 9.

(b) ||u|| = maxi1,...,n

| ui| 0 since the absolute value of any number is equal to or greater

than zero.

maxi1,...,n

| ui| = 0 if and only if all | u

i| = 0.

||cu|| = maxi1,...,n

| cui| = |c| max

i1,...,n| u

i| = |c| ||u||.

||(1,2)|| = |2| = 2, ||(- 3,4)|| = |4| = 4, ||(1,2,- 5)|| = |- 5| = 5, and ||(0,- 2,7)|| = |7| = 7.

39. (a) d(x,y) = ||x- y|| 0.

(b) d(x,y) = ||x- y|| = 0 if and only ifx- y=0if and only ifx =y.

(c) d(x,z) = ||x- y+y- z|| ||x- y|| + ||y- z|| = d(x,y) + d(y,z), from the triangle inequality.

Exercise Set 1.7

Exercises 1, 2, and 3 can be solved simultaneously since the coefficient matrices are thesame for all three.

a0

+ a1

+ a2

= b1

a0

+ 2a1

+ 4a2

= b2

a0

+ 3a1

+ 9a2

= b3

, where b1, b

2, b

3 are the y values 2, 2, 4 in Exercise 1;

14, 22, 32 in Exercise 2; and 5, 7, 9 in Exercise 3.

1 1 1 2 14 5

1 2 4 2 22 7

1 3 9 4 32 9

R2 (1)R1

R3 (1)R1

1 1 1 2 14 5

0 1 3 0 8 2

0 2 8 2 18 4

R1(1)R2

R3 (2)R2

1 0 2 2 6 3

0 1 3 0 8 2

0 0 2 2 2 0


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42

(1/2)R3

1 0 2 2 6 3

0 1 3 0 8 2

0 0 1 1 1 0

R1(2)R3

R2 (3)R3

1 0 0 4 8 3

0 1 0 3 5 2

0 0 1 1 1 0

, so the values of a0, a1, a2

are 4, - 3, 1 for Exercise 1; 8, 5, 1 for Exercise 2; and 3, 2, 0 for Exercise 3. Thus the equations of the polynomials are:

1. 4 - 3x + x2 = y 2. 8 + 5x + x2 = y 3. 3 + 2x = y

4. a0 + a1 + a2 = 8

a0

+ 3a1

+ 9a2

= 26

a0

+ 5a1

+ 25a2

= 60

1 1 1 8

1 3 9 26

1 5 25 60

R2 (1)R1

R3 (1)R1

1 1 1 8

0 2 8 18

0 4 24 52

(1/2)R2

1 1 1 80 1 4 9

0 4 24 52

R1 (1)R2

R3 (4)R2

1 0 3 10 1 4 9

0 0 8 16

(1/8)R3

1 0 3 10 1 4 9

0 0 1 2

R1 (3)R3

R2 (4)R3

1 0 0 50 1 0 1

0 0 1 2

,

so a0= 5, a1= 1, a2= 2, and the equation is 5 + x + 2x2 = y.

When x = 2, y = 5 + 2 + 8 = 15.

5. a0 - a1 + a2 = - 1

a0

= 1

a0 + a1 + a2 = - 3

1 1 1 1

1 0 0 1

1 1 1 3

R2 (1)R1

R3 (1)R1

1 1 1 1

0 1 1 2

0 2 0 2

R1R2

R3 (2)R2

1 0 0 1

0 1 1 2

0 0 2 6

(1/2)R3

1 0 0 1

0 1 1 2

0 0 1 3

R2 R3

1 0 0 1

0 1 0 1

0 0 1 3

so a0

= 1, a1 = - 1, a

2 = - 3, and the equation is 1 - x - 3x2 = y.

When x = 3, y = 1 - 3 - 27 = - 29.

6. a0 + a1 + a2 + a3 = - 3

a0

+ 2a1

+ 4a2

+ 8a3

= - 1

a0

+ 3a1

+ 9a2

+ 27a3

= 9

a0

+ 4a1

+ 16a2

+ 64a3

= 33

1 1 1 1 3

1 2 4 8 1

1 3 9 27 9

1 4 16 64 33

R2 (1)R1

R3(1)R1

R4 (1)R1

1 1 1 1 3

0 1 3 7 2

0 2 8 26 12

0 3 15 63 36


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43

R1(1)R2

R3(2)R2

R4 (3)R2

1 0 2 6 5

0 1 3 7 2

0 0 2 12 8

0 0 6 42 30

(1/2)R3

(1/6)R4

1 0 2 6 5

0 1 3 7 2

0 0 1 6 4

0 0 1 7 5

R1(2)R3

R2 (3)R3

R4 (1)R3

1 0 0 6 3

0 1 0 11 10

0 0 1 6 4

0 0 0 1 1

R1(6)R4

R2 (11)R4

R3 (6)R4

1 0 0 0 3

0 1 0 0 1

0 0 1 0 2

0 0 0 1 1

, so a0

= - 3, a1 = 1, a

2 = - 2, a

3 = 1 and the equation

is - 3 + x - 2x2 + x3 = y.

7. 1

+ 2

- 3

= 0

21 + 43 = 34

42

+ 43

= 28

8. 1

+ 2

- 3

= 0

1 + 23 = 9

32

+ 23

= 17

so that 1 = 5,

2 = 1,

3 = 6. so that

1 = 1,

2 = 3,

3 = 4.

9. 1

+ 2

- 3

= 0

33

= 9

42

+ 33

= 13

10. 1

+ 2

- 3

= 0

21

+ 23

= 4

42

+ 23

= 2

so that 1 = 2, 2 = 1,3 = 3. so that 1 = 1, 2 = 0,3 = 1.

11. 1

- 2

- 3

= 0

1

+ 32

= 31

1

+ 73

= 31

so that 1 = 10,

2 = 7,

3 = 3.

12.

1

2

3 0

3

4

5

0

2 4

4

4

5

4

13. 1

2

3

0

3

4

5

0

1 2 4

1

24

4

24

25

2

so that1 = 12,

2 = 4, so that

1 = 7/3,

2 = 5/3,

3 = 8,

4 = 4,

5 = 4.

3 = 2/3,

4 = 5/6,

5 =1/6.


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44

14. Assume3

flows from A to B.

1

- 2

- 3

= 0

1

+ 23

= 4

2

- 23

= 9

gives1 = 6,

2 = 7,

3 = - 1, so the current in AB is 1 amp flowing from B to A.

15. Let 1

be the current in the direction from the 16volt battery to A, let 2

be the current

from A to B, and let3

be the current in the direction from C to B.

1

- 2

- 3

= 0 : 0

51

+ 2

= 16 : 16

- 2

+ 53

= 9 : 23

We solve the two systems simultaneously:

1 1 1 0 0

5 1 0 16 16

0 1 5 9 23

R2 (5)R1

1 1 1 0 0

0 6 5 16 16

0 1 5 9 23

R2 R3

1 1 1 0 0

0 1 5 9 23

0 6 5 16 16

(1)R2

1 1 1 0 00 1 5 9 23

0 6 5 16 16

R1R2

R3 (6)R2

1 0 6 9 230 1 5 9 23

0 0 35 70 154

(1/35)R3

1 0 6 9 230 1 5 9 23

0 0 1 2 22 /5

R1(6)R3

R2 (5)R3

1 0 0 3 17 /5

0 1 0 1 1

0 0 1 2 22 /5

(a) 1 = 3,

2 = 1,

3 = 2. (b)

1 = 17/5,

2 = - 1,

3 = 22/5.

If 2 = 0, let the voltage at C be denoted by V.


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45

1

- 3

= 0

51

= 16

53

= V

Thus1 = 3 = 16/5, so V = 16.

16. A: x1 + x

4 = 300 B: x

1 + x

2 = 250

C: x2 + x

3 = 100 D: x

3 + x

4 = 150

1 0 0 1 300

1 1 0 0 250

0 1 1 0 100

0 0 1 1 150

...

1 0 0 1 300

0 1 0 -1 -50

0 0 1 1 150

0 0 0 0 0

.

x1 = - x

4 + 300, x

2 = x

4- 50, x

3 = - x

4 + 150.

Two solutions are x1 = 250, x

2 = 0, x

3 = 100, x

4 = 50

and x1 = 150, x

2 = 100, x

3 = 0, x

4 = 150.

The minimum value of x1 comes from taking the maximum value of x

4, which is 150.

Thus, the minimum value of x1 is 150.

17. A: x1

- x4 = 100 B: x

1- x

2 = 200

C: - x2 + x

3 = 150 D: x

3- x

4 = 50

1 0 0 -1 100

1 -1 0 0 200

0 -1 1 0 150

0 0 1 -1 50

...

1 0 0 -1 100

0 1 0 -1 -100

0 0 1 -1 50

0 0 0 0 0

.

x1 = x

4 + 100, x

2 = x

4- 100, x

3 = x

4 + 50.

x2 = 0 is min flow along BC; i.e., BC, closed to traffic.

In that case x4 = 100, x

1 = 200, x

3 = 150. Then alternative route would have to be

provided to get from B to towns in direction of C, and D.


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Sect i on 1. 7

46

18. x1= x2+100, x3=x2+90, x3=x4+130, x5=x4+110, x5=x6+80, x7=x6+75,

x7=x8+120, x1=x8+155. Since a flow cannot be negative these equs give:

x1100, x390, x3130, x5110, x580, x775, x7120, x1155.

Thus, must have x1155, x3130, x5110, x7120.

Is x1=155 possible, i.e., does it result in nonneg flows? Yes, gives

x2=55, x3=145, x4=15, x5=125, x6=45, x7=120, x8=0.

Minimum flow allowable along x1is 155. Note that this is attained by closing x8to

traffic. Alternative routes (clearly labeled diversions!) will then have to be provided for the x7traffic wanting to get to some of the towns that are accessed from other exits of

the roundabout.

19. A: x1

+x2

= 200 B: x1

- x3

- x4

= 0

E: x2

+x3

- x5 = 0 D: x

4+x

5 = 200

1 1 0 0 0 200

1 0 -1 -1 0 0

0 1 1 0 -1 0

0 0 0 1 1 200

...

1 0 -1 0 1 200

0 1 1 0 -1 0

0 0 0 1 1 200

0 0 0 0 0 0

.

x1= x3- x5+ 200, x2= -x3+ x5, x4= -x5+ 200.

Total time = k(x1+ 2x2+ x3+ 2x4+ x5) = 4(x3- x5+ 200 + 2(-x3+ x5) + x3+ 2(-x5+ 200) + x5) = 4(600) = 2400 minutes.

This averages to 12 minutes per car. It is interesting to note that the time is independent of the actual distribution of traffic for this model. This will not be true if k differs on different stretches of road due to different road conditions. Students can examine these situations.

20. Let y = a0+ a1x + a2x2. These polynomials must pass through (1, 2) and (3, 4). Thus

a0 + a1 + a2 = 2

a0 + 3a1 + 9a2 = 4 .

1 1 1 2

1 3 9 4

1 0 -3 1

0 1 4 1. a0= 3a2+1, a1= -4a2+1.

Let a2= r. The family of polynomials is y = (3r+1) + (-4r+1)x + rx2. r=0 gives the liney=1+x that passes through these points. When r >0 the polynomials open up and whenr


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Chapt er 1 Revi ew Exer ci ses

47

21. Let y = a0+ a1x + a2x2+ a3x3. Polynomials must pass through (1, 2), (3, 4) (4,8). Thus

a0 + a1 + a2 + a3 = 2

a0 + 3a1 + 9a2 + 27a3 = 4

a0 + 4a1 + 16a2 + 64a3 = 8 .

1 1 1 1 2

1 3 9 27 4

1 4 16 64 8

1 0 0 12 4

0 1 0 -19 -3

0 0 1 8 1.

a0= -12a3+4, a1= 19a3-3, a2= -8a3+1.

Let a2= r. The family of polynomials is y = (-12r+4) + (19r-3)x + (-8r+1)x2+ rx3.

When r=1, y = -8 + 16x - 7x2+ x3.

Chapter 1 Review Exercises

1. (a) 2 x 3 (b) 2 x 2 (c) 1 x 4 (d) 3 x 1 (e) 4 x 6

2. 0, 6, 5, 1, 9

3. I5=

1

0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

4. (a)

1 2

4 -3,

1 2 6

4 -3 -1 (b)

2 1 -4

1 -2 8

3 5 -7

,

2 1 -4 1

1 -2 8 0

3 5 -7 -3

(c)

-1 2 -7

3 -1 5

4 3 0 ,

-1 2 -7 -2

3 -1 5 3

4 3 0 5 (d)

1 0 0

0 1 0

0 0 1,

1 0 0 1

0 1 0 5

0 0 1 -3

(e)

-2 3 -8 5

1 5 0 -6

0 -1 2 3,

-2 3 -8 5 -2

1 5 0 -6 0

0 -1 2 3 5

5. (a) 4x1 + 2x2 = 0

-3x1

+ 7x2

= 8

(b) x1

+ 9x2

= -3

3x2

= 2

(c) x1

+ 2x2

+ 3x3

= 4

5x1

-3x3

= 6

(d) x1

= 5

x2

= -8

x3

= 2

(e) x1

+ 4x2

- x3

= 7

x2

+ 3x3

= 8

x3

= -5


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48

6. (a) Yes. (b) Yes.

(c) No. There is a 2 (a non zero element) above the leading 1 of row 2.

(d) Yes.

(e) No. The leading 1 in row 3 is not positioned to the right of the leading 1 in row 2.

7. (a)2 4 2

3 7 2

(1/2)R1

1 2 1

3 7 2

R2 (3)R1

1 2 1

0 1 1

R1(2)R2

1 0 3

0 1 1

,

so the solution is x1 = 3 and x

2 = -1.

(b)

1 2 6 172 6 16 46

1 2 1 5

R2 (2)R1

R3(1)R1

1 2 6 170 2 4 12

0 4 5 12

(1/2)R2

1 2 6 170 1 2 6

0 4 5 12

R1 (2)R2

R3(4)R2

1 0 2 5

0 1 2 6

0 0 3 12

(1/3)R3

1 0 2 5

0 1 2 6

0 0 1 4

R1(2)R3

R2 (2)R3

1 0 0 3

0 1 0 2

0 0 1 4

,

so that x1= 3, x2= -2, x3= 4.

(c)

0 1 2 6 21

1 1 1 5 12

1 1 1 4 9

3 2 0 6 4

R1R2

1 1 1 5 12

0 1 2 6 21

1 1 1 4 9

3 2 0 6 4

R3 (1)R1

R4 (3)R1

1 1 1 5 12

0 1 2 6 21

0 0 2 9 21

0 1 3 21 40

R1R2

R4 (1)R2

1 0 3 11 33

0 1 2 6 21

0 0 2 9 21

0 0 5 27 61

(1/2)R3

1 0 3 11 33

0 1 2 6 21

0 0 1 9 /2 21/2

0 0 5 27 61

R1(3)R3

R2 (2)R3

R4 (5)R3

1 0 0 5 /2 3/2

0 1 0 3 0

0 0 1 9 /2 21/2

0 0 0 9 /2 17/2

(2/9)R4

1 0 0 5 /2 3/2

0 1 0 3 0

0 0 1 9 /2 21/2

0 0 0 1 17 /9


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49

R1 (5/2)R4

R2 (3)R4

R3 (9/2)R4

1 0 0 0 56 /9

0 1 0 0 17 / 3

0 0 1 0 2

0 0 0 1 17 /9

, so that x1= 56/9, x2= 17/3, x3= 2, x4 = 17/9.

8. (a)

1 1 1 3

2 3 1 8

4 2 10 10

R2 (2)R1

R3 (4)R1

1 1 1 3

0 1 3 2

0 2 6 2

R1R2

R3 (2)R2

1 0 4 1

0 1 3 2

0 0 0 2

There is no need to continue. The last row gives 0 = 2 so there is no solution.

(b)

1 3 6 2 7

2 5 10 3 10

1 2 4 0 0

0 1 2 3 10

R2 (2)R1

R3 (1)R1

1 3 6 2 7

0 1 2 1 4

0 1 2 2 7

0 1 2 3 10

R1(3)R2

R3R2

R4 (1)R2

1 0 0 1 5

0 1 2 1 4

0 0 0 1 3

0 0 0 2 6

R1 (1)R3

R2 R3

R4 (2)R3

1 0 0 0 2

0 1 2 0 1

0 0 0 1 3

0 0 0 0 0

,

so there are many solutions, and the general solution is

x1 = 2, x

2 = - 1 - 2r, x

3 = r, x

4 = 3.

9. If a matrix A is in reduced echelon form, it is clear from the definition that the leading 1 in any row cannot be to the left of the diagonal element in that row. Therefore if A I

n,

there must be some row that has its leading 1 to the right of the diagonal element in that row. Suppose row j is such a row and the leading 1 is in position (j, k) where

j < k n. Then, if rows j + 1, j + 2, . . . , j + (n - k) < n all contain nonzero terms, the

leading 1 in these rows must be at least as far to the right as columns k + 1, k + 2, . . . ,k + (n - k) = n, respectively. The leading 1 in row j + (n - k) + 1 must then be to the rightof column n. But there is no column to the right of column n, so row j + (n - k) + 1 mustconsist of all zeros.

10. Let E be the reduced echelon form of A. Since B is row equivalent to A, B is also rowequivalent to E. But since E is in reduced echelon form, it must be the reduced echelonform of B.


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50

11. 12.

13. (a) u+w = (3,- 1,5) + (0,1,- 3) = (3,0,2).

(b) 3u+v= 3(3,- 1,5) + (2,3,7) = (11,0,22).

(c) u - 2w= (3,- 1,5) - 2(0,1,- 3) = (3,- 3,11).

(d) 4u - 2v+ 3w= 4(3,- 1,5) - 2(2,3,7) + 3(0,1,- 3) = (8,- 7,- 3).

(e) 2u - 5v - w = 2(3,- 1,5) - 5(2,3,7) - (0,1,- 3) = (- 4,- 18,- 22).

14. a(1, 2, 0) + b(0, 1, 4) + c(-3, 2, 7) = (-1, 3, -5). a-3c = -1, 2a+b+2c=3, 4b+7c=-5. Unique solution, a=2, b=-3, c=1. Is a linear combination.

15. (a) Let W be the subset of vectors of the form (a,b,a+2b). Letu=(a,b,a+2b),v=(c,d,c+2d), and k be a scalar. Then u+v=(a+c, b+d,(a+c)+2(b+d)) and ku=(ka,kb,ka+2kb).The third component ofu+vis the first plus two times the second; same for ku. Thus W is closed under addition and scalar multiplication - it is a subspace.

(b) Let W be the subset of vectors of the form (a,b,a+4). Letu=(a,b,a+4), v=(c,d,c+4) and k be a scalar. Then u+v=(a+c,b+d,(a+c)+8). Third component is not (a+c)+4. Thus

u+vis not in W. W is not a subspace. Let us check closure under scalar multiplication.ku=(ka,kb,ka+4k)). Unless k=1 the 4th component is not the first component plus 4. ku

is not in general in W. W Is not closed under scalar multiplication either.

16. Separate the variables in the general solution, (2r + s, 3s, r, s)= r(2, 0, 1, 0) + s(1, 3, 0,

1).

Vectors (2, 0, 1, 0) and (1, 3, 0, 1) thus span W.

Also, identity p(2, 0, 1, 0) + q(1, 3, 0, 1) = (0, 0, 0, 0) leads to p=0, q=0. The two vectors

are


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51

thus linearly independent. The set {(2, 0, 1, 0), (1, 3, 0, 1)} is a basis for W; dimension is 2.

17. a(1, -1, 2) + b(1, 0, 1) + c(6, -2, 10)=0gives a+b+6c=0, -a-2c=0, 2a+b+10c=0. Unique solution a=0, b=0, c=0. Thus linearly independent.

18. (a) (1,2)(3,- 4) = 1x3 + 2x- 4 = 3 - 8 = - 5.

(b) (1,- 2,3)(4,2,- 7) = 1x4 + - 2x2 + 3x- 7 = 4 - 4 - 21 = - 21.

(c) (2,2,- 5)(3,2,- 1) = 2x3 + 2x2 + - 5x- 1 = 6 + 4 + 5 = 15.

19. (a) ||(1,- 4)|| = (1,- 4)(1,- 4) = 1x1+- 4x- 4 = 17 .

(b) ||(- 2,1,3)|| = (- 2,1,3)(- 2,1,3) = - 2x- 2+1x1+3x3 = 14 .

(c) ||(1,- 2,3,4)|| = (1,- 2,3,4)(1,- 2,3,4) = 1x1+- 2x- 2+3x3+4x4 = 30 .

20. (a) cos=(- 1,1)(2,3)

||(- 1,1)|| ||(2,3)|| =1

2 13 =

1

26 .

(b) cos=(1,2,- 3)(4,1,2)

||(1,2,- 3)|| ||(4,1,2)|| = 0.

21. The vector (1,2,0) is orthogonal to (- 2,1,5).

22. (a) d = (1- 5)2+(- 2- 3)2 = 16+25 = 41 .

(b) d = (3- 7)2+(2- 1)2+(1- 2)2 = 18 = 3 2 .

(c) d2 = (3- 4)2+(1- 1)2+(- 1- 6)2+(2- 2)2 = 50, so d = 50 = 5 2 .

23. ||c(1,2,3)|| = |c| ||(1,2,3)|| = |c| (1,2,3)(1,2,3) = |c| 1x1+2x2+3x3 = |c| 14 , so if

||c(1,2,3)|| = 196 then c = 196

14 = 14 14 .

24. Let (a,b,c) be in W. Then (a,b,c) is orthogonal to (-3,5,1). (a,b,c)(-3,5,1)=0, -3a+5b+c=0, c=3a-5b. W consists of vectors of the form (a,b,3a-5b). Separate the variables.


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(a,b,3a-5b)=a(1,0,3)+b(0,1,-5). {(1,0,3), {(0,1,-5)} is a basis for W. The dimension of W is 2. It is the plane defined by (1,0,3) and (0,1,-5).

25. The equation is of the form a0 + a

1x + a

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