Jawaban Fismat Bab.1 - Bab.3

1 INFINITE SERIES,  POWER SERIES SECTION 1

1. Dike Diketa tahu huii : n = 10 2

2.

V= 3

3.

a = 1 yd

. !.

Ditanya : a. "10

#.

 $. S10

%.

&a'a$

:

(. ).

"10

= a* n+1

10.

= a*10+1 

11.

= a*)  

12.

 2        = 1 3  

13.

!12 = %!#

1.

= 0,#%%% d

)

1!. a (1 − r n ) 1#. S10 =

1 − r  1

  2  10  1 × 1 −       3    1− 1%.

=

1(.

= 1,# d

2 3

1). !

+

20. 3. 0.!!!!! 0.!!!!! = 10

!

+

100 100

! 1000

+ ....

21. . !

 s ∞

=

a 1 − r 

10 1

=

1−

22. !. 0.!( 0.!(33 3333 33

=

! )

10

= 0.!( 0.!( - 0.00 0.0033 3333 33 3

= 0.!( - 1000

#.

+

3 10000

3 !( 100

%.

+

=

1000 1 1− 10

1%! = 300

(.

%

= 12

). 10. %

%.

0.%%%%% = 10

+

% 100

+

% 1000

+ .... %

 s ∞ 11.

=

a 1 − r 

=

10 1

1−

% )

10

1(! 1(!

12. ).

=

0.1(!1(! = 1000

+

1(! 1(! 1000000

13.

2

+ ....

+ .....

 s ∞

=

a 1 − r 

=

1(!

1(!

1000 1

= 1000 =

1−

1.

))!

1000

! 2%

1000

1!. 1#. 1%. 23

0.2323 = 1000

1(. 11.

+

23

+ ....

1000000

1). 23 23

 s ∞

=

a 1 − r 

1000 1

=

1−

20.

=

23 ))!

=

3 3%

1000

21. 22. 22. 13. 13. 1 -

[ ( 0,06 ) ( 1,005 ) +( 0,06 ) ( 1,005) + …] 10 - [ ( 0,6 ) ( 1,005 ) + ( 0,6 ) ( 1,005 ) + … ] 2

2

23.

S 120=10 +

2.

a (r

−1 ) 120 r −1 120

a ( 1,005

−1 ) 120 1,005 − 1 120

2!. 2!.

= 10 -

2#. 2#.

= 10(, 10(,( (

2%. 2(. 2(. 1! a = ( 2). 2).

*=

30. 30.

S=

a 1 −r

 =

8 1 −3 / 4

 = 32

31. PROBLEM PROBLEM SECTION SECTION 2 32. ∞

1.

n ∑ = 2 n 1

n

1 2 3 + + +… 2 4 8

 =

33. ∞

3.

3.

∑= n+n 5 n 1

 =

1 2 3 + + +… 6 7 8

3

3!. ∞

2 n ( 2 n + 1) ∑ 3 n +5 =

3#. !.

6

 =

8

1

n

+

20

42

11

14

 +

+…

3%. 1 2 4 8 + +  + + … 3 5 7 9

3(. %. 3).

∞

1 1 1 1 n +1 − + − + …= (−1 ) 4 9 16 24 n=1

∑

0. ).

1

( n + 1 )2

1. ∞

1 1 1 − + − 1 + … = (−1 )n+ 1 4 8 16 32 n= 1

∑

2. 11.

1 n+1

2

3. 44. PROBLEM SECTION 4 45. 1

1.

2

n

 1

 1

1

1

2

4

8

16

=¿ + + +

+

1 32

+⋯

∞

¿ ∑ = 1

n

1 2

#.

S 1= a 1=

%.

S 2= a 1+ a2 =

(.

S 3= a1+ a2 + a3 =

).

S 4 =a1 + a 2+ a3+ a4 =

!0.

S 5= a1+ a2 + a3 + a 4 + a 5= 2

n

1 1 1 7 + + = 2 4 8 8 1 1 1 1 15 + + + = 2 4 8 16 16 1 1 1 1 1 31 + + + + = 2 4 8 16 32 32

−1

!1.

S n=

!2.

S = lim Sn

2

n

n→ ∞

n

!3.

1 1 2 + = 2 4 4

=

lim n →∞

2

−1

2

n



lim 1−

1

!.

=

!!.

=

56.

= 1 /adi ka*ena hainya 1 aka (konvergen)

n →∞

1−

2

n

 1

∞

57. ∞

3.

1  , ∑ = n ( n +1 )

1

 int :

n ( n+ 1 )

n 1

1 2

2 3

1

n

n +1

3 4

1 2

1 6

= + +

 4 5

!(.

S 1= , S 2= , S3=  , S 4 =  , S 5=

!).

S n=

1 1 1 + + +⋯ 12 20 30

5 6

n

#0.

n +1 S = lim Sn

#1.

¿ lim

n→∞

n →∞

n n+1 1

¿ lim

#2.

n →∞

1+

1

n

1

#3.

¿

#.

¿ 1 (konvergen )

1 +0

∞

!.

1

= −

1 ∑ = n2

n 1

n

1

 , int :

2n

< n

1 n

2

, n >1

1 2

#!.

S 1= a 1=

##.

S 2=  , S3 =

5 8

16 24

#%.

!.

n

(. ). 10. 11.

2 3

4 5

1

#.

n2

12.

n

5 8

16 24 131 192 661 960

13. 1. 1!.

%. 1%. 1(. 1). 20.

1 n

2 3 4 7

8

15 16 31 32

1

#(.

n2

%3. %.

!

<

#).&adi : 1

%0.

n2

n

(konvergen) %1. %2.

1#.

n

 1 n

2

%!. 76. PROBLEM SECTION 5

%%. %(. "e the 4*eiina*y tet t5 'hethe* the 655'in7 e*ie a*e di8e*7ent 5* *e9ui*e 6u*the* tetin7. a*e6u: d5 n5t ay that e*ie i ;5n8e*7ent< the  4*eiina*y tet ;ann5t de;ide thi. %). 1

1.

2

4

9

5

10

− +

(0.

−

16 17

+

25 26

−

36 37

∞

+ …  =

(− 1 ) ∑ = n 1

¿1

(3.

1 1

∞

1

+

∞

1

(.

¿1

(!. (#.

¿ ∞  Di8e*7ent>

0+0

∞

n+ 3 ∑ = n + 10 n

 = ....

2

n 1

.

¿ lim

!.

¿ lim

n →∞

n →∞

%. (. ). 10. 11. 12. 13.

2

+1 lim an

n →∞

2 n+ 1 n n +1 2 ( ) lim an= lim −1  di$a7i den7an (−1 ) n 2 n →∞ n→ ∞ n +1 1 ¿ lim 1 1 1 n →∞ + n+ 1 (−1 ) (−1 )n+ 1 n2

(2.

#.

n

2

"ntuk en7u/i kek5n8e*7enian aka ;a*iah

(1.

3.

n

n+ 1

1+

¿ ¿

n +3 2

+ 10 n 3 1+

n

 di$a7i den7an n

n

n + 10 3

∞

∞ + 10

1 +0

∞ ¿ 0  Tet 6u*the*>

#

∞

1!.

n! ∑ = n !+1 1

n

¿ lim

1#.

n →∞

n!  di$a7i den7an n !+ 1 1

¿ lim

1%.

n →∞

n!

1

1+

n!

1

1(.

¿

1). 20. 21. 22. 23. 2. 2!. 2#.

¿ 1  Di8e*7ent>

1 +0

∞

%.

(−1 )n n ∑ 3 n = 1 √ n + 1 ¿ lim

2%.

n →∞

¿ lim 2(.

n →∞

(−1 )2 n

di$a7i den7an

1

√ n3 + 1 (−1 )2 n

1

2).

¿

30. 31.

¿ 0  Tet 6u*the*>

∞

).

(− 1 ) n n √ n 3+ 1

∞

n

3

∑ 2 +3

n=1

n

n n

32.

¿ lim n →∞

33.

¿ lim n →∞

3 n

2

n

+3

n

2

n

3!.

¿

n

3

1 3

3.

 di$a7i den7an

+1

1

∞+ 1 ¿ 0  Tet 6u*the*>

3#. 11. "in7 .#> 7i8e a 4*556 56 the 4*eiina*y tet 65* ;5n8e*7en;e 56 an in6inite e*ie. int: 3%.

s n− s n−1 =an .

38. PROBLEM SECTION 6

%

3). 0. 1. h5' that n?@2n 65* a n@3. int : '*ite ;ut a 8ie' te*< than ;5nide*  uti4y $y t5 75 6*5, ay, !? T5 #? And 6*5 2 ! t5 2#. 1. 2. Buktikan $ah'a n?@2n, untuk n@3 3. 44.n

46.

45.

47.

∞

∞

∑ n!

∑2

n

1

1

48.1

49.1

50.<

51.2

52.2

53.2

54.<

55.4

56.3

57.6

58.<

59.8

60.4

61.24

62.>

63.16

64.5

65.120

66.>

67.32

68.6

69.720

70.>

71.64

%2. %3. Da4at diihat $ah'a: n?@2n, untuk n@3 %. ∞

%!. 3. 4*58e the k5n8e*7en;e 56

∑ n1

2

 $y 7*5u4in7 te* 5e 'hat a in

1

 4*5$e t'5. →

%#.

a

n=

 1

n

❑ deret yang diuji

2

→

%%.

m

n=

1

❑ deret konvergen

n+ 1

|a n|

78.n

79.

82.1

83.1

84.>

1 4 1 91. 9 1 95. 16 1 99. 25

88.<

86.2 90.3 94.4 98.5

80.

87.

81.mn 85. 89.

92.<

<

 

101.

1 6

(

1 3

1 4 1 97. 5 93.

96.< 100.

1 2

102.

|a n|< mn

Ca*ena

 aka

an konvergen mulai dari n > 1

103. ∞

10.

!.

∑ √ 1n

√ n

hint :'hi;h i a*7e*, n 5*

 

1

∞

10!.

n =¿

∑ √ 1n 1

 =

1+

a¿

1

√ 2

+

1

√ 3

+

1

√ 4

+

1

√ 5

+…

∞

10#.

n =¿

∑ n +1 1 = 12 + 13 + 14 + 15 + 16 +… 1

m¿ 107.

n

 

109.

108.

110.

|a n| 111.

1

112.

1

113.

>

 

114.

1 2 115.

2

 

117.

116.

>

 

1 3

1

√ 2 119.

3

 

121.

120.

>

 

√ 3 4

 

125.

124.

>

 

1 2 127.

5

 

129.

128.

|a n|> mn

an  Divergen

 aka

132. 133. 13. 13!. ∞

13#.

%.

>

 

130.

1 6

√ 5

Ca*ena

126.

1 5

1

131.

122.

1 4

1 123.

118.

1 ∑ = nlnn n 2

)

mn

a

13%. 13(. 13).

n=

 

1

nlnn

misal ; lnn =u 1

n

dn = du 1

u 10.

 du

¿ ∞

1

∫¿

 dn=¿ nlnn

2

∞

∫¿ 2

11.

¿ [ lnu ] ∞

12.

¿ ln ∞− ln2

13.

¿∞

2

an dn= ∞ ( Divergen) 1.

¿

∞

∫¿ 2

1!. 1#. 1%. 1(. 1). 1!0. 1!1. 1!2. 1!3. 1!. 1!!. 1!#. 1!%. 1!(. 1!). 1#0. 10

1#1. 1#2. 1#3. ∞

1#.

1 ∑ = n −4

).

2

n 3

1

n =¿

1#!.

2

n a¿

−4

n dn =¿ ∞

∞

3

3

∫ a ¿∫ n 1−4 dn

1##.

2

∞

1#%.

=

∫ ( n−2 1)( n +2 ) dn 3

1#(.

1

A

=

B

+

( n−2 ) ( n + 2) ( n− 2 ) ( n + 2) 1= A ( n + 2 ) + B ( n− 2 )

1#). →

→

1 4

1%0.

n =2 ❑ 1= 4 A + 0 ❑ A =

1%1.

n =−2 ❑ 1=0 + (−4 B )❑ B =

→

→

∞

1%2.

aka:

4

∞

1

∫ ( n−2 )( n+2 ) dn=∫ ( n A−2 ) + ( nB+2) dn 3

1%3.

−1

3

=

−1

1 4

∞

∫ ( n−2 ) + (n +4 2) dn 3

1%.

=

1 4

∞

∫ 3

1

1 dn − n −2 4

∞

∫ n 1+2 dn 3

1%!.

=

1 ∞ 1 ∞ ln ( n−2 ) ] −  [ ln ( n + 2 ) ]  [ 4 3 4 3

1%#.

=

1 ln ( ∞ −2 )− ln 1 4

1%%.

=

∞ − ∞ =∞

 [

1%(.

11

] − 14 [ ln ( ∞−2 )− ln 5 ]

an dn=∞ ( Divergen )

¿

1%).

∞

∫¿ 3

1 + lnn ¿

3 2

¿ n¿

1(0.

11.

1

¿ ∞

∑¿ 1

1 + lnn ¿

3 2

¿

n¿

1(1.

n=¿

 1

¿

a¿ 1(2. 1(3.

 

misal : 1+ lnn= u 1

dn = du

n

n dn =¿ 1 + lnn ¿

3 2

¿

dn

¿ n¿

1(.

1

¿ ¿ ¿

∞

∞

∫ a¿∫ ¿ 1

1

u¿

3 2

¿

du 1(!.

¿ ¿

1

¿ ¿ ¿ ∞

¿∫ ¿ 1

12

−3

u 1(#.

2

¿

du

∞

¿∫ ¿ 1

1(%.

[

¿ −2 u

1)0. 1)1.

2

]∞ 1

[− ( +

1((.

1().

−1

−1

2 1 lnn)

]

∞ 1

] − − ¿ −(  ) ¿

[

2

−2 ∞ √ 1 + lnn 1 2

2 1

∞

¿2 an dn=2 (C 0 nvergen )

¿

1)2.

∞

∫¿ 1

∞

1)3.

∑2

1).

2n

n= 0

1).

1)!.

3

an  =

n

3

2

n

2n

3

an +1  =

2

n+ 1

2 ( n+ 1 )

| || n+ 1

3

1)#.

| |

ρ n=

a n+ 1 an

=

2 ( n+ 1 )

2

3

2

1)%.

2

n +1

n→ ∞

|| 3 4

=3 4

1)(. 1)). 200. 201. 202.

13

2n

2

2( n + 1)

2n

ρ= lim | ρn| ¿ lim n→ ∞

=

n

 3

n

3

| || =

3 4

203. 20. 20!. 20#. 20%. ∞

20(.

∑ =

21.

( 2 n) !

n 0 2

n

210.

5

an +1  =

(( n + 1) ! )2 ( 2 ( n +1 )) !

| |

| |

ρn=

a n+ 1

=

an

(2 n) !

n+1

5

211.

5 (n ! )

an  =

20).

n+ 1

(( n + 1 ) ! )2 n+ 1 2 n ( 2 (n +1 ) ) !  3 2 = 2( n + 1) n 2 n 5 ( n! ) 2 3 (2 n)!

|

| | =

1

212.

lim | ρ n|¿ lim

ρ  =

2

n

5 (n !)

n →∞

n →∞

|

|

10 ( 2 n−1 ) n = lim n−2 1 n→ ∞

n

|

10 ( 2 n−1 )

n− 2

| | 20−

10

1−

2

n

=20

n

213. ∞

21.

n! ∑ = 100

23.

n

n 0

an  =

21!.

100

n +1 !

an +1  =

21#.

n!

100

n+ 1

21%.

| |

ρ n=

21(.

| || n+ 1!

n +1

100

=

n!

n +1 100

|

n

100

1

a n+ 1 an

=

n

ρ  =

21).

1

| |

¿ lim ¿ n →∞

n+ 1 n 100 1

=0

n lim n→ ∞

| ρn|¿

220. ∞

221.

2!.

n

e √ n !

∑ = n 0

222.

S5uti5n :

223.

e an = √ n !

22.

e an +1= √ n+ 1 !

22!.

¿ lim

n

n +1

n →∞

| | an + 1 an

| | n+ 1

e √ n + 1 !

22#.

¿ lim

22%.

e e ¿ lim × n →∞ √ n + 1 ! √ n !

n →∞

n

e √ n !

| |  √  |√ ( + ) √  | |√  + | |√  + | |√  | n +1

n

n

22(.

22).

¿ lim n →∞

¿ lim n →∞

230.

¿ lim n →∞

231.

¿ lim n →∞

232.

e e

n

e

1

 n!

×

n!

e

n

❑

n 1 e

❑

∞

 e

1

❑

∞

¿ 0 < 1 konvergen

233.

1!

∞

23.

n

100

∑ = n

2%.

200

n 0

23!. 23#.

S5uti5n : an =

100

n

200

n

n +1

23%.

an +1=

23(.

¿ lim n →∞

100

( n + 1 )200

| | an + 1 an

23).

| | n+ 1

100

20.

¿ lim n →∞

( n +1)

n

100

n

21.

22.

23.

¿ lim n →∞

¿ lim n →∞

¿ lim n →∞

|

200

| | |

200

 100

n +1

( n+ 1 )200

×

n

n

100

( n +1 )

200

n

|

200

n

100  100

200

×

n

n

100

|

|

100  100

( n +1 )200

100  ∞

100

|

2.

¿

2!.

¿ 100 > 1 divergen

∞

200

2#. 2%.

√ ( 2 n ) ! ∑ n! = ∞

2(.

2).

n 

2).

S5uti5n : √ ( 2 n ) !

2!0.

an =

2!1.

an +1=

n!

√ ( 2 ( n +1 ) ) ! ( n +1 ) !

1#

¿ lim

2!2.

n →∞

| | an + 1 an

| | √ ( 2 ( n + 1 ) ) !

¿ lim

2!3.

n →∞

( n +1) ! √ ( 2 n ) ! n!

| |

| |

2!.

( 2 ( n +1 ) ) ! × n ! ¿ lim √  ( n +1 ) ! n →∞ √ ( 2 n ) !

2!!.

( ( 2 n +2) ) ! × n ! ¿ lim √  n →∞ (n+ 1)! √ ( 2 n ) !

2!#.

( 2 n + 2 )  √ ( 2 n +1 ) √ ( 2 n ) ! n! ¿ lim √  × ( n + 1 )  n! n →∞ √ ( 2 n ) !

2!%.

| | |

lim

=

n →∞

2!(.

=

lim

n →∞

2!).

√ ( 2 n + 2 )  √ ( 2 n + 1 ) ( n +1)

√ 4 n + 6 n + 2 (n +1 ) 2

¿ lim

2#3. 264.

6  2

+ +

n n2

n

¿

 6

∞

1+

2#2.

:n

❑

❑

1+

4+

2#1.

|

2

| | √ | | √

4

n →∞

2#0.

:n

|

+

1

n

❑

2

∞

2

 1

∞

| | ¿|√  | ¿| | 4 +0+ 0 ¿ √  1+ 0

4

1

2 1

=2@1

divergen

265.

1%

|

2##.

31.

2#%.

2#(.

S5uti5n :

2#).

2%0.

2%1.

2%2. 2%3.

2%.

2%!.

2%#.

2%%. 2%(.

1(

2%).

33.

2(0.

S5uti5n :

2(1.

2(2.

2(3.

2(.

2(!.

2(#.

2(%.

2((. 2().

2)0.

3!.

1)

2)1.

2)2.

S5uti5n :

2)3.

2).

2)!.

2)#.

2)%.

2)(.

2)).

300.

20

301. 302. 303.

PROBLEM SECTION 7

21

30. 30!. 30#. 30%.

=−1> n

∞

%.

n

∑ 1+ n n =1

⇒ an =

2

n→∞

= i n →∞

n

1+ n

=

< a n +1

2

= −1>

an

n +1

=n + 1>

1 + =n + 1>

n +1

= −1> =−1>=n + 1> n

=1 + n

2

+ 2n + 1>

×

n

1+ n

= −1> n n

= ni →∞

1

1

n

n2

− =1 + +

2

2

= −1>

n +1

=n + 1>

1 + =n + 1>

2

×

1+ n

2

=−1> n n

− =n + 1 + n 3 + n 2 > = ni → ∞ = n + n 3 + 2n 2 + n

2

− =n + n + n + 1> = ni = ni → ∞ = n 3 + 2n 2 + 2n →∞ 3

308.

= −1>

=n + 1> = −1> n = ni : 2 → ∞ 1 + = n + 1> 2 1+ n

a n +1

 ρ  = i

= −1> n

1+

2

n

+

+

1

n3

>

= 1 ⇒ Gagal Uji, gunakan Uji lain

2

n2

PROBLEM SECTION 9

309.

∑ =n + n2>=−n1 + 3> = 0 + 201 + 302 + 23 + !# + %2! + )0# +  ∞

1.

n =1

310.

Den7an en77unakan u/i *ai5 4ada uku ke+# dan uku ke+%,

aka di4e*5eh:

 ρ n 311.

! !# 3! 3! =          =  ρ = n→∞it ρ n = 〈1  %2       3#  dan 3# , aka de*et te*e$ut i

e*u4akan de*et k5n8e*7en. 312. 313.

3. ∞

∑

1

n =1 n

n 3

= 1+

1 2n 3

+

1 3n 3

+

1 n 3

22

+

1 !n 3

+

1 #n 3

+

1 % n 3

+

31.

Den7an en77unakan u/i *ai5 4ada uku ke+% dan uku

ke+#, aka di4e*5eh:

1   # n 3     =  n 3      = 0,(213  %    1    dan

 ρ n 31!. it   ρ = ni →∞ ρ n

= 0,(213〈1 , aka de*et te*e$ut e*u4akan de*et k5n8e*7en.

31#. 31%.

!. ∞

∑ n

=1 n

1

−

3

=

1

−3

2

3



23

+ +

+

 #0

+

! 121

+

# 212

+

% 33)

+

( 132

)

+

%2!

+

31(. 31).

Den7an en77unakan u/i *ai5 4ada uku ke+) dan uku

ke+(, aka di4e*5eh:

 ρ n 320.

) 132 =          = 0,20( it   ρ = ni  %2!    (   →∞ ρ n = 0,20(〈1  ,  dan

aka de*et te*e$ut e*u4akan de*et k5n8e*7en. 321. ∞

322.

%.

= 2n >?

∑ 3 =n?> n =0

n

2

=  takada + 2 + 2 + 3

3

 2%

+ %0 + 2( + 30( +  (1

(1

%2)

=2n>? n

3 =n?>

2

323.

Den7an en77unakan u/i $andin7 de*et di8e*7en $e*ikut: 1+

32.

1 2

1

1

1

1

1

1

3



!

#

%

n , /ika kita $andin7kan uku

+ + + + + +

=2n>? n

eetaknya, uai ke+!, e;a*a uu $e*aku hu$un7an 3 =n?>

 〉

2

1

n

, aka

de*et itu di8e*7en. ∞

32!.

).

nn

∑ n? n =1

2

= 1 + 2 + 3 + 10 +  3

e$ea* a4ai

23

, ka*ena de*etnya eakin

- ∞ , aka de*etnya di8e*7en.

32#. 32%. 32(.

11. ∞

2n

= n

2

∑ n

−)

32).

=

(

−1

+

10 1#

+

12 2!

+

1 0

+

1# !!

+

1( %2

+

20 )1

+

22 112

+

Den7an en77unakan u/i *ai5 4ada uku ke+( dan uku

ke+%, aka di4e*5eh:

 ρ n 330.

22 )1 =          = 0,()3%! it   ρ = ni ρ n = 0,()3%!〈1  112   20   → ∞  dan  ,

aka de*et te*e$ut e*u4akan de*et k5n8e*7en. 331. 332.

13.

333.

Jawab:

33.

P*eiina*y tet

33!.

33#.

=

33%.

=

33(.

= 0 TF>

33).

S4e;ia ;54a*i5n tet

30.

$n =

31.

2

32.

33.

.

3.

=

3!.

=1

3#.

&adi, de*et te*e$ut k5n8e*7en.

3%. 3(.

1!.

3).

Jawab:

3!0.

P*eiina*y tet

3!1.

3!2.

=

3!3.

=

3!.

&adi, de*et te*e$ut di8e*7en.

3!!. 3!#.

1%.

3!%.

Jawab:

3!(.

Rai5 tet

3!).

3#0.

=

2!

3#1.

=

3#2.

=

3#3.

=

3#.

=

3#!. 3##.

= Ca*ena

aka de*et te*e$ut k5n8e*7en.

3#%. 3#(.

1).

369.

Jawab:

3%0. 3%1.

P*eiina*y tet

3%2.

=

TF> 3%3.

Rai5 tet

3%.

3%!.

=

2#

3%#.

=

3%%.

=

3%(.

=

3%).

=

3(0.

=

3(1.

=

3(2.

Ca*ena

aka de*et te*e$ut k5n8e*7en.

3(3. 3(.

21.

/ika

385.

Jawab:

3(#. 3(%.

=

3((.

Rai5 tet

3().

3)0.

=

3)1.

=

3)2.

=

3)3.

Ca*ena

aka de*et te*e$ut k5n8e*7en.

3). 2%

3)!. 396.

PROBLEM SECTION 10

397.

1.

3)(.

Rai5 tet

3)).

00.

=

01.

=

02. 03. 0. 0!. 0#.

"/i:

0%.

"ntuk

, de*et en/adi:

0(. 0).

di8e*7en> "ntuk

, de*et en/adi:

10. 11.

di8e*7en> &adi, de*et ini k5n8e*7en daa dae*ah

12. 13. 1. 1!.

Find the inte*8a 56 ;5n8e*7e 56 ea;h the 655'in7 45'e* e*ie<

 $e u*e in8eti7ate the end45int 56 the inte*8a in ea;h ;ae.

2(

1#.

( − 1) n x n n =1 n( n + 1) ∞

∑

1%.

3.

1(.

S5uti5n:

 x n 1).

 x n

+1

 x n

n( n + 1) < a n +1 = ( n + 1)( n + 1 + 1) = ( n + 1)( n + 2)

an =  ρ n

=

i it  n→ ∞

 ρ n

=

i it  n→∞

 x n

+1

 x n

:

( n + 1)( n + 2) n( n + 1)

20.

21.

 x n . x

n( n + 1)

( n + 1)( n + 2)

 x n

1  x.n n n+2 1 n

i it  n →∞

22.

 ρ n  ρ n

=

 x

it  = i n →∞

1+

23.

2

n

 x

1+

2.

 ρ n

2!.

Tet :

2#. 2%.

=

2

 x

∞

=1

( x ) 2 = 1  <

n

=

a

= x <1

 ;5n8e*7ent>

 x n n( n + 1)

 x ± 1

2(. 2).

he;k 65* x=1 ∞

30.

11

32.

12

13

1

∑ = 1(1 + 1) + 2( 2 + 1) + 3( 3 + 1) + (  + 1) + ... n =1

1

31.

+1

=2

1

1

#

12

+ +

+

1 20

+ ...

 ;5n8e*7ent>

he;k 65*  x = −1

2)

∞

33.

∑ = − 12 + 1# − 121 + 201 − ... n =1

 ;5n8e*7ent>

3. 3!.

S5, the e*ie ;5n8e*7e 65*

 x

≤1

3#. ∞

 x n

∑ ( n)?

2

3%.

!. n =1

3(.

S5uti5n: n

3).

=

a

 x n

( n?) 2

n +1

=

 < a  x n +1

0.

 ρ n

=

i it  n →∞

 x n

( n + 1)?2  x n

:

( n + 1)?2 ( n)?2  x n . x

1.

 ρ n

=

i it  n →∞

+1

( n )?2

( n + 1)?2 ( n )?2  x n 1  x

n ( n + 1)? 1 n 2

2.

 ρ n

=

i it  n →∞

 x n 3.

 ρ n

i it 

= n →∞

1   ( n + 1)?2     

 n  

 x

∞

.

1   ( ∞ + 1)?       ρ n ∞     =

!.

Ca*ena

=0

2

 ρ n

untuk eua ha*7a  A >. #. %. (.

30

= 0, aka de*et dikatakan k5n8e*7en

∞

∑ n =1

 x

3n

n

).

%.

!0.

S5uti5n : an

!1.

=

 x 3 n n  <

+1

 x 3n  ρ n

!2.

=

n +1

i it  n →∞

=

a n +1

:

 x 3n +1 n +1

 x 3n n

 x 3n . x n  ρ n

!3.

=

n + 1  x 3 n

i it  n →∞

1

 ρ n

!.

 x.n n n +1 1

i it 

n

= n →∞  x i it  n →∞

1+

1

!!.

 ρ n

=

!#.

 ρ n

= n→∞ =  G 1 ;5n8e*7ent>

Tet :  = 1  <  = ±1

!(.

#0. #1.

i it 

2

!%.

!).

n

an

=

 x 3 n n

he;k 65* x=1 ∞

13.1

n =1

1

∑=

+

13.2 2

+

13.3 3

+

13. 

+ ...  di8e*7ent>

he;k 65*  = +1 ∞

#2.

∑= = − 1 + 12 − 13 + 1 + ...

#3.

S5, the e*ie ;5n8e*7e 65*

n 1

#. #!. ##.

31

 ;5n8e*7ent>

−1 ≤ x < 1

∞

∑ ( − 1)n  x 3

n

n →∞

#%.

).

#(.

S5uti5n :

#).

an

= n 3 x n  <

an+1

3 n +1 1 + ( ) n  x =

( n + 1) 3 x n +1 %0.

 ρ n

=

n 3 x n

i it  n →∞

( n + 1) 3 . x n . x %1.

 ρ n

=

n 3 . x n

i it  n →∞

( n + 1) 3 . x %2.

 ρ n

=

n3

i it  n →∞

+ 3n 2 + 3 xn +  x>. x

=n 3 %3.

 ρ n

=

i it  n →∞

n

3

1  xn

+ 3 xn + 3 xn +  x

3

2

n %.

%!.

 ρ n

 ρ n

=

=

3

i it  n →∞

i it  n →∞

3

n 1 n

1

 x + 1

3 xn

+

1

+

+

1 3 xn

2

+

3

 x n3

 x

%#.

 ρ n

=-

%%.

 ρ n

=  - 0 - 0 - 0 =  G 1 ;5n8e*7ent>

∞ ∞ ∞

%).

an = n 3 x n 1 = Tet :   <  = ±1

(0.

he;k 65*  = 1

%(.

2

∞

(1.

∑= = 1 .1 + 2 .1 3

1

3

2

+ 33.13 + 3.1 + ...

n 1

(2.

= 1 - ( - ) - # - H di8e*7ent> ∞

(3.

∑ = −1 + ( − ) + # − ... n =1

32

 ;5n8e*7ent>

(.

S5, the e*ie ;5n8e*7e 65* ∞

n

(!. (#.

S5uti5n:

n =1

n

an (%.

1   x  

 ρ n

n +1

  x      n + 1  !  

=

i it  n →∞

 ρ n

:

  

n  !  

n

i it 

= n →∞

  x  .n     !   n +1 )0.

n

1   x  

  x     x   n       n + 1  !    !    x  n     !   1

 ρ n

n +1

  x   =    an+1 =    n  !   < n + 1  !   1

1

().

<1

1   x  

   ∑ n  !   11.

((.

 x

i it 

1

n 1 n

= n →∞

 x

! )1.

 ρ n

=

1+

i it  n →∞

1 n

 x )2.

 ρ n

!

= 1+ 0

=  x < 1 !

 ;5n8e*7ent>

)3. 2

).

  x   = 1    Tet :  !   < 2



)#.

=

)%.

he;k 65*  = !

)(.

an

=    n  !  

= 2!

)!.

∞

n

1   x  

1

±! 2

1 !   1  !  

3

1  !  



1  !  

∑ = 1 !    + 2  !    + 3  !    +   !    + ... n

=1

33

1

1

1

3



+ + + ...

)).

=1- 2

!00.

he;k 65*  = +!

 di8e*7ent>

∞

!01.

∑= = −1 + 12 − 13 + 1 + ...

!02.

S5, the e*ie ;5n8e*7e 65* +! ≤ x

n 1

;5n8e*7ent>


!03.

n( −  x )

∞

∑n

!0.

13. n =1

!0!.

S5uti5n:

2

+1

n( −  x ) 2 an= n

!0#.

!0%.

!0(.

n

n

+1

<

( n + 1)( −  x ) n+1 an +1 = ( n + 1) 2 + 1

 ρ n

( n + 1)( −  x ) 2 ( n 2 + 1)  x = ( n + 1) 2 + 1 n( −  x ) n

 ρ n

( n + 1)( −  x ) n ( −  x ) 1 .( n 2 + 1) = [( n + 1) 2 + 1]n( −  x ) n ( −  x ) ( n 2 + 1)( n + 1)

 ρ n

= n( n 2 + 2n + 2 + 1)

 ρ n

−  x( n 3 + n + n 2 + 1) = n( n 2 + 2 n + 3 )

 ρ n

−  x( n 3 + n + n 2 + 1) = n 3 + 2n 2 + 3n

!12.

 ρ n

1 1 1 −  x 1 + 2 + + 3      n n n   1 3 1+ + 2 = n n

!13.

 ρ n

= + G 1 ;5n8e*7ent 65*  ≥ 1 >

!0).

!10.

!11.

!1. !1!.

1!.

3

!1#.

!1%.

,

!1(.

!1).

,

!20.

G1

!21.

G3

!22.

Sehin77a

!23.

+3G +2G3

!2.

+1G 

G!

!2!. !2#.

"ntuk  = !

!2%.

aka

!2(.

"ntuk  = +1

!2).

aka

!30.

Sehin77a de*et

ehin77a 1 - 1 - 1 - H e*u4akan de*et di8e*7en

ehin77a +1 - 1 +1 - H e*u4akan de*et di8e*7en

k5n8e*7en 4ada +1 G  G !

!31. !32.

!33.

1%.

,

!3.

,

3!

!3!.

!3#.

,

!3%.

G1

!3(.

Sehin77a

!3).

+1 G -1G1

!0.

+2 G  G0

!1. !2.

"ntuk  = 5

!3.

aka

ehin77a 1 -

H

e*u4akan de*et di8e*7en !.

"ntuk  = +1

!!.

aka

ehin77a +1+

e*u4akan de*et di8e*7en !#.

Sehin77a de*et k5n8e*7en 4ada

k5n8e*7en 4ada +2 G  G0 !%. !(.

1).

ia  = y

!).

,

!!0.

!!1.

,

3#

!!2.

,

!!3.

,

!!.

,

!!!.

, ,

!!#.

,

!!%. !!(.

, "ntuk  = 3

!!).

!#0.

=1-1-1-H

"ntuk  = + 3

!#1.

!#2.

di8e*7en

=1-1-1-1-H

di8e*7en

Sehin77a de*et

 4ada

k5n8e*7en .

!#3. !#.

21.

!#!.

!##.

,

3%

!#%.

!#(.

, aka

!#).

G1

!%0.

G1

!%1.

+1 G  G 1

!%2.

"ntuk  = 1

!%3.

=

!%.

"ntuk  = +1

!%!.

=

!%#.

-

-

-

 

+

 

-H

Sehin77a de*et

di8e*7en

k5n8e*7en 4ada

!%%. !%(.

k5n8e*7en

-H

23.

!%).

!(0.

,

3(

!(1.

=

!(2.

, aka

!(3.

G1

!(.

G3

!(!.

 - 1 G +3 dan  - 1 @ 3

!(#.



!(%.

"ntuk  = +

G + dan

!((.

@2

=

k5n8e*7en !().

"ntuk  = 2

!)0.

=

 

!)1.

Sehin77a de*et k5n8e*7en 4ada

H

k5n8e*7en 4ada  G + dan  @ 2 !)2. !)3.

2!.

!). !)!.

3)

di8e*7en

!)#.

=

!)%.

G 1 aka

!)(.

in  G

!)). #00. #01. #02.

"ntuk  =

#03.

di8e*7en #0.

"ntuk  =

#0!. #0#.

 

#0%.

Sehin77a

#0(.

k5n8e*7en 4ada

#0). #10. 611.

di8e*7en

PROBLEM SECTION 13

612.

0

#13.

1

#1.

3

#1!.

!

#1#.

%

#1%.

)

#1(. 2 x 2 x

11.

e −1

=

#1).

#20.

#21.

Penyeeaian :  x De*et daa* e  =

2 x Sehin77a e  =

1+

1+

 x

+

1?

2 x

+

1?

 x 2

+

2?

=2 x >

 x 3

+

3?

2

+ .....

?

=2 x >

+

2?

x

3

=2 x >

+

3?



?

2 x

2 x #22.

e 2 x − 1

=

1 + 2 x +

=2 x >

2

2?

+

=2 x >

3

+

3?

=2 x >



?

+ ..... − 1

2 x

#23.

=

2 x +

=2 x>

2

2?

=2 x >

+

3

3?

+

=2 x > ?



+ .....

1 1+ #2.

 2 x> 2?

=

 2 x>

+

3?

2

+

 2 x> ?

#2!. .

2

in  x = #2#.

4enyeeaian:  x −

 x 3 3?

+

 x ! !?

−

 x % %?

........

#2%.

de*et daa* in  =

#2(.

2 keudian 7anti  den7an  x  ehin77a:

#2).

#30.

2 in  x =

 x

2

−

 x 2 = 3>

3?

+

 x 2 = ! >

!?

−

x 2= % >

#31.

1

%?

........

3

+ ....

#32.

2 in  x =

 x

2

−

 x # 3?

+

 x10 !?

−

x1 %?

........

#33. e in x  =

#3.

1!.

#3!.

Penyeeaian :

#3#.

#3%.

 x de*et daa* e  =

in  =

 x −

 x 3 3?

+

1+

 x ! !?

 x

1?

−

+

 x % %?

 x 2

2?

+

 x 3

3?

+

 x 

?

........

in  x in 2  x in 3  x in   x 1+ + + + in x 1? 2? 3? ? aka e  =

#3(. #3).

  x3  x !  x %   x − 3? + !? − %? ..... in x e  = 1- 

#0.

  x 3  x ! x %   x − 3? + !? − %1 .....  

2

.

1 2

+ .....

#1.  x

∫ ;5 t  = 2

#2. #3.

#.

#!.

1%.

0

Penyeeaian:

de*et daa*: ;5 t = 2 untuk ;5 t  =

 x

 x

t 

t 

2? t (

2?

+

t  ?

−

t (

t 12

?

#?

+ −

0

t 12

0

 x

#%.

=

 t ! t ) t 13  t − !.2? + ).? − 13.#? + ......  0

2

t # #?

......

......

∫ ;5 t  = ∫ 1 − 2? + ? − #? ....... 2

##.

1−

1−

t 2

#(.

=

#).

  x!  x) x13  − + − +  x ......  !.2? ).? 13.#?  − [ 0]  

=

 x −

 x ! !.2?

+

 x )

−

x13

).? 13.#?

+ ....

#!0.

#!1.

1 +  x n 1 −  x

1).

x

= ∫  0

dt 

1 − t 2 1

#!2.

1 +  x  2 1 − x 

n 

"ntuk

#!3.

= 1 n 1 +  x 2 1 −  x = 1 [ n (1 +  x) − n (1 −  x ) ] 2

=

1    x 2  x − 2? 2  

 x       x 2 − ...  −  −  x − ?   2?    

 x 3 + 3?

 x 3 − 3?

 x    − ...   ?    

3     = 1  2 x + 2  x ...  2   3?    

 x 3 =  x + ... 3?  x

dt 

∫ 1− t 

2

#!.

"ntuk  x

0

dt 

∫ 1− t 

x

2

#!!.

0

#!#.

 =

arc tan t 

= a*; tan t

Sehin77a kita da4at enui a*; tan t

− ( ) t  1 + den7an 2

1

(1 + t  ) − 2

#!%.

0

1

2  # = 1 − t  + t  − t  + .... <

3

 x

dt 

∫ 1− t 

t  −

2

#!(.

 =

0

#!).

=

 x −

##0.

 x 3 3?

+

 x ! !?

t 3

3?

−

t !

t %

!?

%?

 x

+ − + .... 0 x% %?

+ .....

aka 4e*nyataan diata $ena*, $ah'a

##1.

1 +  x n 1 −  x

##2.

x

= ∫  0

dt 

1 − t 2

 =

 x −

 x 3 3?

+

 x ! !?

−

x% %?

+ .....

##3.

664. 2 665. COMPLEX NUMBER  666.

PROBLEM SECTION 4

##%.

1. 1- i

##(.

S5uti5n :

669.

!1

670.

"!1

#%1.

# =

=

#%2.

#%3. #%.

3. 1 + i

#%!.

S5uti5n :

676.

!1



677.

"!$

678.

#=

=2

#%).

680.

681.

#(2.

!. 2i

#(3.

S5uti5n :

684.

!0

685.

# =

"!0

=2

#(#.

#(%.

#((.

%. +1

#().

S5uti5n :

690.

 ! $1

691.

"!0

#)2.

# =

=2

#)3.

#). #)!.

). 2i  2

#)#.

S5uti5n :

697.

 ! $2

698.

"!2

!

699.

# =

=2

%00.

%01. %02.

= 3J K 

%03.

11. 2  ;5

%0.

S5uti5n :

%0!.

#=

- i in

>

!2

=

706.

%0%.

 = tan

708.

!1

709.

"!

=

=

%10. %11.

1!. ;5

%12.

S5uti5n :

%13.

=

%1.

 = tan

715.

#=

=

=0 =1

%1#.

=1

%1%.

1-

%1(.

1-0=

=

#

%1). 720.

 ! $1

721.

"!0

%22. %23.

1%.

%2.

S5uti5n :

%2!.

L = *.

%2#.

# =

=

= %2%.

=2

%2(.

1-

%2).

1-1=

 = tan

=

%30. 731.

!1

732.

" ! $1 %&a#'(a ) &*(+,-a'/

733. 734.

PROBLEM SECTION 5

735.

%3#.

1. 

%3%. %3(. %3).

=

!.  i -

=

=1 > 2 = +3 =

%

=

=1

%0.

aka e4unyai Re+M = 1,# dan +M

= +2,% %1.

aka e4unyai Re+M = 1% dan +M

= 12 %2.

13. !  ;5

%3.

> = ! e i2JK!

- i in =

=

%. %!.

=

= ,) i 1). L +1 =

%#.

=

%%.

=

=

=

%(. %).

"e e9uati5n !.1> t5 6ind the a$5ute 8aue 56: 2 + 3i

%!0.

2%.

1− i

+ 3i 1− i

2

%!1.

An'e*:

=

=

+) 1+1

=

13 2

%!2.

%!3.

− 3i 1 + 3i 2

%!. %!!. %!#. %!%.

2). 1-2i>3

%!(.

An'e*:

1-2i>3 =  1-2i> 2 1-2i>

(

%!).

= 1-i+> 1-2i>

%#0.

= i+3> 1-2i>

%#1.

= i+(+3+#i

%#2.

= +11+12i

%#3.

=−11>2 + =−2> 2

=

%#.

=

121 + 

%#!.

=

12!

%##.

=! ! ! − 2i

%#%.

31. ! + 2i

%#(.

An'e*:

! − 2i

! + 2i

! + 2i

! − 2i

=

2! +  2! + 

=

2) 2)

=

1 = 1

%#). 2!

%%0.

33. 3 + i

%%1.

An'e*:

2!

2!

=

3 + i 3 − i %%2.

#2! ) + 1#

=

#2! 2!

=

2!

=!

S58e 65* a 45i$e 8aue 56 the *ea nu$e*  and y in the

655'in7 e9uati5n. %%3. %%.

3!.

 - iy = 3i  

An'e*:

 = +  and y = 3

%%!. %%#. %%%.

3%.

 - iy = 0

An'e*:

 = 0 and y = 0

%%(. %%). %(0. %(1. %(2.

3).  - iy = y - i An'e*:

 = y and y = 

1. 2  3y+!>-i-2y-1>=0 An'e*:

2 + 3y +! =0 .... kai 2

)

%(3.

-2y-1 =0 .....kai 3

%(.

-

%(!.

 + #y +10=0

%(#.

3 + #y -3 =0

%(%.

 -

%((.

%

=%

%().

Q=1

%)0.

1-2y-1=0

%)1.

2y=+2

%)2.

y =+1

%)3. %).

3.  - iy> 2 = 2i

%)!.

 - iy>2 = 2i

An'e*:

%)#.

2 - 2iy + y 2 = 2i

%)%.

2  y2 -2iy = 2i •

2  y2 = 0

•

2y = 2

%)(.

y=1

%)).

2  y2 = 0

(00.

2  1> 2 = 0

(01.

2

(02. (03.

=1 

= 1 = 1

5  = 1 5*  = +1 and y = 1

(0. (0!.

!.   - iy> 2 =    iy > 2

(0#.

An'e*: 2 - 2iy + y 2 = 2  2iy - y 2 2+ y2 - 2iy =  2 - y2  2iy

(0%. (0(.

2+ y2

= 2 - y2

(0).

(10.

). 1+-iy> =-iy

(11.

!0

(12. (13.

(1.

(1!. (1#. (1%. (1(. (1).

and y = 0

(20. (21. 822. 823.

PROBLEM SECTION 6

824.

(2!.

Tet ea;h O6 the 655'in7 e*ie 65* ;5n8e*7en;e

(2#. (2%.

3.

(2(.

S58e :

(2).

= i

(30.

n

(31.

= i

(32.

n





!1

(33.

=

=

G 1 CONVEREN>

(3. (3!.

!.

(3#.

S58e :

(3%.

= i

(3(.

n

(3).

= i

(0.

n

- i

n





- i

n





(1.

=

-

(2.

= 2 @ 1 DIVEREN>

(3. (.

%.

(!.

S58e :

(#.

= i

(%.

n

((.

= i

().

n

(!0.





= i  1>

!2

(!1.

=

=

DIVEREN>

(!2. (!3.

).

(!.

S58e :

(!!.

= i

(!#.

n

(!%.

= i

(!(.

n

(!).

=i

(#0.

=

(#1.

DIFFERENT TEST>





=

(#2. (#3. 

INTERA

(#. (#!.

(##.

i= 1 /adi in = 1

(#%. (#(. 

ehin77a:

(#).

(%0.

DIVEREN>

871. 872. 873. 874.

PROBLEM SECTION 7

!3

e

 z 2  z  + − ..... 1− 2? !? n 2 ∞  z  =−1> n + 2 =2n?> n =0

∞

∑

∑

 ρ  =  ρ  =

n

n→∞

875.

 ρ  = 0

(%#.

1.

 z  =−1> ( 2n + 2?) =2n + 1> 2

1

 ρ 

(%(. (%). ((0. ((1. ((2. ((3.

ka*ena G1,

((.

aka k5n8e*7en 4ada eua niai M

((!. ((#.

3.

((%. (((. ((). ()0. ()1. ()2. ()3.

Ca*ena G1,

().

aka k5n8e*7en 4ada eua niai M

!

2?

+

3

 z 

+ ......

3?

n

 z 

= =

i n

n

it 

n

 z 

+1

→ ∞ ( n + 1?)

i

=0  ρ  < 1  ρ 

(%%.

2

 z 

=0 n?

 ρ  n+3 2 n+ 2  z 2 n =−1> n +3 i it   z  =−1> : n → ∞ =2n + 2>? =2n>?

i it 

= 1 +  z  +

 x

it 

→∞

 z  n

+1

n

:

 z 

n?

n

∞

  z       ∑ =   2   0

n

n

  z     ρ  =    n → ∞   2   i it 

()!.

  z       n → ∞   2     z     ρ  =      2  

()#.

!.

 ρ 

=

+1

n

  z    :     2  

i it 

()%. ()(. ()). )00. )01. )02.

C5n8e*7en /ika  ρ  < 1  z 

2

)03.

<1

− 2 <  z  < 2

)0. )0!. n

n

)0#.

n

   z    =   − 2         ∑   2   =   2   n = 0 →1 n = 1 → −1 n = 2 →1 n = 3 → −1 ∞

0

a. Pada M = +2,

)0%. )0(. )0). )10. )11. )12.

de*et : 1 + 1 - 1 + 1 - ..

)13.

De*et di8e*7en

)1.

$. Pada M =2

!!

∞

    =   2         ∑   2   =   2   = 0 →1 =1→1 = 2 →1 = 3 →1 n

n

n

 z 

0

n n n

)1!.

n

)1#. )1%. )1(. )1). )20.

De*et : 1 - 1 - 1 - 1 - ....

)21.

De*et k5n8e*7en

)22.

− 2 < z  ≤ 2 &adi de*et te*e$ut k5n8e*7en

 4ada inte*8a ∞

=−1> n +1 z n + 2

n =0

=2n>?

∑

 ρ  =  ρ  =

i it  =−1> n +1 z 2 n + 2

→∞

=2n + 2>?

i it 

=−1> z 2 n

n n

→∞

:

=−1> n  z 2 n =2n >?

=2n + 2>=2n + 1>

 ρ  = 0

)23.

 ρ  < 1

)2.

%. )2!. )2#. )2%. )2(. )2).

)30.

Ca*e a*ena G G1, ak akaa k5 k5n8e*7en 4a 4ada e eu ua ni niai M

!#

∞

∑ n

n

 z 

n

=1

 ρ  =  ρ  =

 z  +1 n

i it 

→∞

n

n +1

i it 

 ρ  =  z 

)31.

 ρ  = z 

)32.

)

n

n +1

i it 

n

:

 z  n

→∞

n

n

 z 

→ ∞ 1+

1 1

n

)33. )3. )3!. )3#. )3%. )3(. )3). )0.

<1  z  < 1 − 1 <  z  < 1 C5n8e*7en /ika

 ρ 

)1. )2. )3. ).

a. 4ada M = +1 ∞

∑ n

=1

 z 

n

n

−1

n

= n

= 1 → −1

n

=2→

n

n

)!.

n

1

=3→−

2 1 3

=→ 1

2

!%

)#. )%. )(. )). )!0. )!1. )!2. )!3.

de*et te te*e$ut e e*u4akan de de*et di di8e*7en ∞

∑ n

 z 

n

n

=1

n

=

n

n

=1→1

n

=2→

n

n

)!.

1

1 2 1

=3→ =→

3 1 2  $. 4ada M = 1

)!!. )!#. )!%. )!(. )!). )#0. )#1. )#2. )#3.

De*et te te*e$ut ad adaah de de*et k5 k5n8e*7en. ∞

)#.

∑ − 1 < z  ≤ 1 n=1

)#!.

&adi de*et

 z n n k5n8e*7en 4ada

)##.

!(

∞

∑ n

( n?) 3 z 

n

=3n>?

=0

( n?)   z  +1 ( n?) 3  z   ρ  = : =3n >? n → ∞ =3n + 1>? i it  ( n?)  z   ρ  = n → ∞ 3n i

 ρ 

=  z 

i n

n

it 

it 

→∞

n

1 3n n?

=

)#%.

 ρ 

)#(.

11.

 z 

)#). )%0. )%1. )%2. )%3. )%. )%!. )%#.

C5n8e*7en /ika

<1  z  < 1 − 1 <  z  < 1

 ρ 

)%%. )%(. )%). )(0.

a. 4ada M = +1 ∞

∑ n

=1

 z 

n

n

−1

n

= n

= 1 → −1

n

=2→

n

n

)(1.

n

1

=3→− =→

2 1 3

1 2

)(2. !)

)(3. )(. )(!. )(#. )(%. )((. )().

de*et te*e$ut e*u4akan de*et di8e*7en

))0.

$. 4ada M = 1 ∞

∑ n

n

n

=1

=

1n n

n

=1→1

n

=2→

n

n

))1.

 z 

1 2 1

=3→ =→

3 1 2

))2. ))3. )). ))!. ))#. ))%. ))(. ))).

De*et te*e$ut adaah de*et k5n8e*7en. ∞

1000.

∑ − 1 < z  ≤ 1 n=1

1001.

&adi de*et

 z n n k5n8e*7en 4ada

1002. 1003. 100.

#0

= z  − i > n n n =1 ∞

∑

 ρ  =

i it  = z  − i > = n +1> = z  − i> n : n → ∞ = n + 1> n

 ρ  =

i it  = z  − i > n n → ∞ = n + 1>

100!.

i it  = z  − i > 1 n→∞ 1+ n  ρ  = 0

100#.

13.

 ρ  =

100%. 100(. 100). 1010. 1011. 1012. 1013. 101. 101!.

Ca*ena G1, aka k5n8e*7en 4ada eua niai M

101#. ∞

= z  + 2 − i > n

=0

2n

∑ n

101%.

 ρ 

=

 ρ 

=

i n

it 

= z  + 2 − i > = n +1> 2 = n+1>

→∞

i

it 

:

2n

= z  + 2 − i >

→∞ = z  + 2 − i >  ρ  = n

2

2

1!.

101(. 101). 1020. 1021.

= z  + 2 − i > n

 z  + 2 − i > < 2

#1

1022. 1023.

&adi de*et te*e$ut k5n8e*7en 4ada

102. 1025. −

1. e

PROBLEM SECTION 9

i µ 



−

102#.

e

i µ 



 = 

1 102%. = 2 1

102(. =



2

− i. 1

2

−

2

 + i

in

µ 



>

2

2

i

2

2

e  =  - iy 1

= 2

2

1

1031.

 µ 

iθ 

102). 1030.

;5

2

y=+2

3 µ i

1032.

3.

2

)e

1033. 3 µ i 2

;5

103.

)e

 = ) 

103!.

= 0 - i +1>

103#.

= +i

103%.

e

iθ 

 =0

103).

y = +1 !.

e

2

+ i in

3µ 

=  - iy

103(.

100.

3 µ 

! µ i

101. ! µ i

 =  ;5 3 µ  + i in ! µ   >

102.

e

103.

=  ;5 µ  + i in  µ   >

10.

= +1 - 0

10!.

= +1

#2

2  >

10#.

iθ 

e

10%.

 =  - iy

 = +1

10(.

y =0

10). 10!0.

%.

3e

2 =1+ iµ >

10!1. 10!2.

3e 2

2 =1+ iµ >

 = 3e

2 + i 2 µ 

i 2 µ 

10!3.

= 3e . e

10!.

2 = 3e =;5 2 µ  + i in 2 µ >

10!!.

= 3e  1 - 0>

10!#.

= 3e

2

2

10!%.

e

10!(.

 = 3e

10!).

y=0

iθ 

 =  - iy

2

10#0.

10#1.

).

2e

−i

 µ  2

10#2.

10#3.

2e

−i

 µ  2  = 2 

;5

 µ 

2

10#.

= 2 0  i.1>

10#!.

= +2i

10##.

e

iθ 

− i in

µ 

2>

 =  - iy

10#%.

 =0

10#(.

y = +2

10#). 10%0. ! i µ 

10%1.

2e

11. ! i µ 

10%2.

2e





2 =;5

=

! µ  

+ i in !µ > 

#3

10%3.

=

2 =;5 22! + i in 22!>

10%.

=

2 =− ;5 ! − i in !> 2 =−

10%!.

=

1!76.

= +1  i

10%%.

e

iθ 

1 2

1

2 −i.

2>

2

=  - iy

10%(.

 = +1

10%).

y = +1

10(0.

=i − 3> 10(1.

10(2.

1− i

13.

=i − 3> 1− i

3

1+ i  1+ i

=

=i 2 10(3.

3

=i − 3 > 3 =1 + i > 1 − = −1>

−2

=

=2 − 2 3i >=i + i

2

−

3i + 3>=i − 3 − i 3 > 2 =

3 − i 3>

2 =2 − 2 3i >=−1 − 3 + i − i 3>

10(.

2

=

=2 − 2 3i >=−1i + 3> + i =1 − 3>

10(!.

=

10(#.

=

− 2=1 +

2

3 > + 2i =1 − 3 > + 2 3i =1 + 3 > − 2 3 =i >=1 − 3> 2

2

− 2=1 + 10(%.

2

=

−2− 2 10((.

=

3 + (i + 2 3 − # 2

=

− ( + (i 10().

3> + 2iV=1 − 3 > + = 3 + 3>U + 2 3 − #

2

= − − i #

10)0.

 = +

y=

10)1.

=1 + i> 2

10)2.

1!.

10)3.

=1 + i>

10).

2

+ =1 + i> 

+ =1 + i>   =

=1 + i> V1 + =1 + i> U 2

2

= 1 + 2i − 1V1 + =1 + 2i − 1>U

10)!.

= 2iV1 + 2iU

10)#.

= 2i + =−1>

10)%.

=

10)(.

 = +

10)).

y=2

−  + 2i

1

1100.

3 + ( ) i 1 1%.

1101.

&a'a$ : 1

1102.

−3 (1 + i ) 3  = (1 + i ) =

   

2e

iπ 

−3



1103. %

110.

=

20

110!.

(1 − i ) (

110#.

1).

110%.

&a'a$ :

110(.

(1 − i ) (  =

   

2e

−i π  

(

    

− i 2π  = 1#e

110). 1110. 1111.

0

 1 − i        2  

1112.

21.

1113.

&a'a$ :

#!

e

−iπ  

    

0

 1 − i        2  

111.

=

     

2e

0

       

−i π  

2

1 x10 # e −i10π  111!.

=

111#.

=

1 x10 # e −i10π 

(1 + i ) (

( 3 − i)

111%.

23.

111(.

&a'a$ :

(1 + i )

2

(

( 3 − i)

111).

2!

2

2!

(

!0

 =

1120.

   

=

1121.

1122.

2

( e (  ) π 

  e −!0π  #       

(−!0

  (π  !0#π          e +         ((π 

−2

=

2

=

1 e 2

e

2

11π  3

1123. 12

 i 2        1  i +    

112.

2!.

112!.

&a'a$ : 12

=

  − 1 2         π  i  e 2    

=

  − 1    π     e i       

12

112#.

 i 2        1 +  i    

12

112%. 112(. 2%.

Sh5' that 65* any *ea 2(. y,

e i"

 = 1

##

2). &a'a$ : 30. 31.

e i"

 z . z 

2

= * = *   =

e i"

−i"

i"

 = e . e

32.

= 1 e

iπ 

2

33. 2). 3. &a'a$ : e

iπ 

2

3!.

 = e

iπ 

2

 . e

− iπ  2

3#. 3%.

= 1

3(. !e

2π i

3

3). 31. 0. &a'a$ : !e

2π i

!e

3

1.

i 2π 

3

 = i 2π 

2.

= !e

3.

= !.1

.

=

!. 33.

3

. !e

− i 2π  3

!

3 2e +iπ 

#. &a'a$ : %.

2e

3+ iπ 

 =

2e 3

.

(.

= 2 (e e

).

= 2.1.1

!0.

= 2

3

e iπ  −3

. (e e iπ 

− iπ 

!1. !2. !3.

#%

3e !i . %e −2 i

!. 3!.

!!. &a'a$ : !#. !%.

3e !i . %e −2 i

! ( e 3

=

i

e

−! i

)

(e −

2i

. %

e 2i )

!(. = 3 (1)  . % (1)  = 21

!). #0. #1.

62. PROBLEM SECTION 10

#3. 3

#. 1.

1

#!. &a'a$ : 0

##. 1 eiiki  = 1 < y = 0 < * = 1 dan θ   = #%.

3

1K 3

#(.

3

1  e4unyai * =

z 

1K 3

 =

0

1

iθ  K 3

r  e

π  + 2nπ 

#). θ   =

3

 den7an n = 0,1, 2, 3, H.

0

0 + 0 3#0 n

3

%0. θ   = 0

%1. θ   =

= 120

0

0

0

n

0 ,120 , 20 , 3#0

0

%2. %3.

3

0

1 = 1  ;5

0

0

- i in

0

> =1

0

%.

= 1  ;5120 - i in120 > = +1K2 - 1 K 2 3  i 0

%!.

0

0

= 1  ;5 20 - i in 20 > = +1K2 + 1 K 2 3  i 0

%#.

0

= 1  ;5 3#0 - i in 3#0 > = 1

%%. %(. #(



%). 3.

1

(0. &a'a$ : 0

(1. 1 eiiki  = 1 < y =0 < * = 1 dan θ   = (2.



1K 

(3.



1  eiiki * =

z 

0

1  = 1

iθ  K 

1K 

r  e

 =

π  + 2nπ 

(. θ   =



 den7an n = 0,1, 2, 3, H..

0

0 + 0 3#0 n



(!. θ   = 0

(#. θ   =

0

= )0

n

0

0

0 , )0 ,1(0 , 2%0

0

(%. ((.



0

1  = 1  ;5

0

0

 - i in 0 > = 1 0

= 1  ;5 )0

(). )0.

= 1  ;5

)1.

= 1  ;5

0

- i in )0 > = i

1(0

0

2%0

0

0

- i in1(0 > = +1 0

- i in 2%0 > = + i

)2. )3. #

). !.

1

)!. &a'a$ : 0

)#. 1 eiiki  = 1 < y =0 < * = 1 dan θ   = )%.

#

1  eiiki * = 1K #

)(.

z 

1  = 1

iθ  K #

1K #

 =

#

r  e

π  + 2nπ 

)). θ   =

#

 den7an n = 0,1, 2, 3, H.. 0

0 + 3#0 100.

θ   =

#

0

n

0

= #0

n

#)

0

0

θ   =

101.

0

0

0

0

0 , #0 ,120 ,1(0 , 20 , 300

0

102. #

103.

0

0

0

1  = 1  ;5

 - i in 0 0

10.

= 1  ;5

>= 1 0

#0

- i in #0

> = 1K2 - 1 K 2 3  i

10!. 0

0

= 1  ;5120 - i in120 > = +1K2 - 1 K 2 3  i

10#. 10%.

0

10(.

= 1  ;5 1(0

10).

= 1  ;5 20

0

 - i in1(0

0

> = +1

 - i in 20

0

> = +1K2 + 1 K 2 3  i

110. 111.

300

= 1  ;5

0

0

- i in 300 > = 1K2 + 1 K 2 3  i

112. 113.%. .

(

1#

11.&a'a$ : 0

11!.1# eiiki  = 1# < y =0 < * = 1# dan θ   = ( 11#. 1#  eiiki * = 1K (

(

1#  =

0

2

iθ  K (

1K (

11%. z   = r  e π  + 2nπ 

11(. θ   =

(

 den7an n = 0,1, 2, 3, H..

0

0

0 + 3#0

n

(

11). θ   =

= 0

120.

θ   =

0

! n

0

0

0

0

0

0

0 , ! , )0 ,13! ,1(0 , 22! , 2%0 , 31!

121. 122.

(

1# =

0

2   ;5

0

0

 - i in 0 0

123.

0

=

2   ;5

>=

2 0

! - i in ! > = %0

1 -i

0

12.

2   ;5

=

)0

0

- i in )0 > = 0

12!.

2   ;5

=

12#.

0

13!

 - i in13! > = +1 - i

0

12%. 12(.

22!

2   ;5

=

2   ;5

=

2%0

0

31!

2   ;5

=

0

0

- i in1(0 > = + 2 0

- i in 22! > = +1 + i 0

- i in 2%0 > = + 2  i

0

12).

1(0

2   ;5

=

2  i

0

- i in 31! > = 1  i

130. !

1

131.

).

132.

&a'a$ :

133.

1 eiiki  = 1 < y =0 < * = 1 dan θ   =

0

13.

!

1K !

13!.

!

1  eiiki * =

0

1  = 1

iθ  K !

1K !

z   = r  e π  + 2nπ 

13#.

θ   =

!

 den7an n = 0,1, 2, 3, H..

0

0 + 3#0 13%.

!

θ   = 0

13(.

θ   =

0

n

0

%2

= 0

n 0

0

0

0 , %2 ,1 , 21# , 2((

13). 10.

!

0

1   =  ;5

0

0

 - i in 0 0

11.

=

 ;5 %2

0

- i in %2

0

12. 13.

=

>= 1

0

 ;51 - i in1

=  ;5

21#

0

> = +0,(0 - 0,!( i 0

- i in 21# > = +0,(0  0,!( i 0

1.

> = 0,30 - 0,)! i

0

=  ;5 2((  - i in 2(( > = 0,30  0,)! i

1!.

%1

1#. 3

1%.

11. .

1(.

&a'a$ :

1). 1!0.

+( eiiki  = +( < y =0 < * = ( dan θ   = 1(0 3

− (  eiiki * = 1K 3

1!1.

−(

3

0

(  = 2

iθ  K 3

1K 3

z   = r  e π  + 2nπ 

1!2.

θ   =

3

 den7an n = 0,1, 2, 3, H.. 0

1(0 + 3#0 1!3.

θ   =

n

3

θ   =

0

0 + = #0 120 n

0

0

1!.

0

#0 ,1(0 , 300

0

,

1!!. 1!#.

3

0

− ( = 2 ;5 #0

1!%. 1!(.

0

 - i in #0 0

=

2 ;5 1(0

=

2 ;5 300

0

>= 1-

 - i in1(0

3  i

0

 - i in 300

>= +2 0

>=1+

3  i

1!). 1#0. 

−

1#1.

13.

1#2.

&a'a$ :

1#3.

+ eiiki  = + < y =0 < * =  dan θ   =

1#.



−   eiiki * = 1K 

1#!.

3

  =

2

iθ  K 

1K 

z   = r  e π  + 2nπ 

1##.

θ   =



 den7an n = 0,1, 2, 3, H.. 0

0 + 1(0 3#0 n

1#%.

θ   =



0

0

+ = ! )0

%2

n

1(0

0

0

0

1#(.

0

0

! ,13! , 22! , 31!

θ   =

1#). 1%0.



0

−   =

0

!  - i in ! > =

2   ;5

0

1%1.

0

13!

- i in13!

=

2   ;5

=

2   ;5

22!

=

2   ;5

31!

0

1%2.

0

1%3.

1-i > = +1 - i 0

 - i in 22!

> = +1 + i

0

- i in 31!

>= 1i

1%. 1%!. #

− #

1%#.

1!.

1%%.

&a'a$ :

1%(. 1%).

+# eiiki  = +# < y =0 < * = # dan θ   = 1(0 #

− #  eiiki * = 1K #

1(0.

#

# = 2

iθ  K #

1K #

z   = r  e π  + 2nπ 

1(1.

θ   =

#

 den7an n = 0,1, 2, 3, H.. 0

1(0 + 3#0 1(2.

#

θ   = 0

1(3.

θ   =

0

n =

0

30

0

0

+ #0

0

0

0

n 0

30 , )0 ,1!0 , 210 , 2%0 , 330

0

1(. 1(!.

#

− #  =

0

0

2  ;5 30 - i in 30

0

0

1(#.

=

2  ;5 )0 - i in )0 > =

2  ;51!0

1(%.

=

1((.

= 2  ;5 210

1().

=

3  - i

>=

0

0

- i in1!0

0

>= +

 - i in 210

2  ;5 2%0

0

%3

2i

3  - i

0

3  + i

>= +

- i in 2%0

0

> = +2 i

0

1)0.

330

= 2  ;5

0

0

 - i in 330 > =

3  + i

1)1. !

−1

1)2.

1%.

1)3.

&a'a$ :

1).

+1 eiiki  = +1 < y =0 < * = 1 dan θ   = !

1)!.

− 1 eiiki * = 1K !

!

1(0

0

1  = 1

iθ  K !

1K !

z   = r  e

1)#.

π  + 2nπ  θ   =

1)%.

!

 den7an n = 0,1, 2, 3, H.. 0

1(0 + 3#0 0

θ   =

1)).

n

!

θ   =

1)(.

0

0

0

0 + = 3# %2 n 0

0

3# , )( ,1(0 , 2!2 , 32

0

200. !

201.

0

− 1  = 1  ;5 3#

0

- i in 3# > = 0

0

= 1  ;5 )( - i in )(

202.

0

203.

= 1  ;51(0

20.

=1  ;5 2!2

- i in1(0

0

20!.

0

> = + 0,13 - 0,)) i

0

- i in 2!2

= 1  ;5 32

0,(0 - 0,!( i

> = +1 0

> = + 0,30  0,)! i

- i in 32

0

> = 0,(0  0,!( i

20#. 3

i

20%.

1).

20(.

&a'a$ : 0

i eiiki  = 0 < y =1 < * = 1 dan θ   =

20).

3

210. 1K 3

211. z   =

i  eiiki * = 1K 3

#

1  = 1

iθ  K 3

r  e

%

)0

π  + 2nπ 

212.

θ   =

3

 den7an n = 0,1, 2, 3, H..

0

0 + )0 3#0 n

213.

θ   =

21.

θ   =

3

30 +120 0

= 0

0

30 ,1!0 , 2%0

0

n

0

21!. 21#.

3

0

30  - i in 30

= 1  ;5 1!0

21%. 21(.

0

i = 1  ;5 ;5

= 1  ;5 2%0

0

0

> = 1 K 2 3  - 1K2

- i in1!0

- i in 2%0

0

> = + 1 K 2 3  - 1K2

0

>=+i

21). 220.

2 + 2i 3

221.

21.

222.

&a'a$ :

223. 22. 22!.

2 + 2i 3  eiiki * = 1K 2

22#.

0

2 + 2i 3 eiiki  = 2 < y =2 3

z 

#0

  = 2

iθ  K 2

1K 2

 =

< * =  dan θ   =

r  e

π  + 2nπ 

22%.

θ   =

2

 den7an n = 0,1, 2, 3, H..

0

#0 + 3#0 22(.

2

θ   = 0

22).

θ   =

0

n =

30 , 210

30 +1(0 0

0

n

0

230. 231. 232.

2 + 2i 3  = 2  ;5

0

0

30  - i in 30 > =

= 2  ;5 210

233.

%!

0

- i in 210

3  - i 0

> = + 3  + i



(i 3 − (

23.

23.

23!.

&a'a$ :

23#.

(i 3 − ( eiiki  = +( < y =(

23%.



(i 3 − ( eiiki * = 1K 

23(.

#

3 < * = 1# dan θ   = 120

1# = 2

iθ  K 

1K 

z   = r  e π  + 2nπ 

23).

θ   =



 den7an n = 0,1, 2, 3, H.. 0

120 + 3#0 20.

n



θ   =

30

= 0

0

21.

0

0

+)0

0

θ   = 30 ,120 , 210 , 300

0

0

n

0

22. 

23.

(i 3 − ( = 2  ;5

2.

=

0

30

0

- i in 30

2  ;5 120

2  ;5 210

2!.

=

2#.

= 2  ;5 300

0

0

0

- i in120

- i in 210 - i in 300

3  - i

>=

0

0

>= +

0

>= 1+

> = + 1-

3  i

3  + i 3  i

2%. !

−1− i

2(.

2!.

2).

&a'a$ : 0

2!0. 2!1.

!

− 1 − i  eiiki * = 1K !

2!2.

2  dan θ   =

+ 1  i eiiki  = +1 < y = + 1< * = !

2=

1 K 10

2

iθ  K !

1K !

z   = r  e π  + 2nπ 

2!3.

θ   =

!

 den7an n = 0,1, 2, 3, H.. 0

22! + 3#0 2!.

θ   =

!

0

n =

!

%#

0

+ %2

0

0

n

22!

0

2!!. 2!#.

θ   = !

0

0

0

0

0

! ,11% ,1() , 2#1 , 3! 1 K 10

−1− i

2

=

0

  ;5 ! - i in ! > = 2 1 K 10

2!%.

=

−) K 10

0

2

  ;5 11%

0

- i in11%

0

−) K 10

 - 2

 i

1 K 10

>=

2

+0,! -

0.() i> 1 K 10

2!(.

2   ;51()

=

0

- i in1()

1 K 10

2

>= 0

0

1 K 10

0

+0.)(  0,1! i > 1 K 10

= 2   ;5 2#1 - i in 2#1 > = 2

2!).

0

1 K 10

= 2   ;5 3!

2#0.

0

- i in 3!

 +0,1!  0,)(i>

1 K 10

2

>=

0.)#  0,2!i>

2#1.

(e ) − iθ ) ( ) i ;5 θ  in θ  − ( e θ  θ  ;5 2  + i in 2 > =  = 2%. ;5 2 θ   - i in 2 θ  > =

2#2.

( ;5 θ +i in θ )

2

iθ 

 =

n

2

2#3.

n

2#.

#  z  + #  z  + #  z  # +# +# 1

2#!. 2##.

2). M = aa

1

1

3

2

3

3

# = # =# =# 2

1

2

#= z  + z  + z  >  z 1 + z 2 + z 3 3# 3  = 2

1

2#%.

2

2

aka M =

3

 te*$ukti>

2#(. 

1#

2#).

31.

2%0.

&a'a$ :

2%1.

1# eiiki  = 1# < y =0 < * = 1# dan θ   =

0

2%2.



1# eiiki * = 1K 

2%3.

z 

1K 

 =



1# = 2

iθ  K 

r  e

π  + 2nπ 

2%.

θ   =



 den7an n = 0,1, 2, 3, H..

%%

0

0

0 + 0 3#0 n

2%!.

θ   =

2%#.

θ   =

 0

2%%.



0

= )0

0

n

0

0 , )0 ,1(0 , 2%0

1# = 2  ;5

0

0

0

0

 - i in 0

>= 2 0

0

2%(. 2%). 2(0.

= 2  ;5 )0 - i in )0 = 2  ;5 1(0 = 2  ;5 2%0

0

0

- i in1(0

- i in 2%0

>=2i

0

> = +2

0

> = + 2i

2(1.

Pen/uahan da*i hainya = 0

2(2.

2 - 2 i +2 +2 i = 0  te*$ukti>

2(3. 284.

PROBLEM SECTION 11

2(!. 1. De6ine in M and ;5 M $y thei* 45'e* e*ie. W*ite the 45'e* e*ie 65* eiM. By ;54a*in7 thee e*ie 5$tain the de6initi5n II.> 56 in M and ;5 M. 2(#.

2(%.

2((.

2(). 2)0.

2)1.

2)2. 2)3. 2). 2)!.

-

2)#.

%(

2)%. 2)(.

2)). 300. 301. 302.

+

303. 30. 30!.

30#. 30%. 30(.

Find ea;h 56 the 655'in7 in *e;tan7ua* 6*5 -iy

3.

30). 310. 311.

312. .

313. 31.

%)

31!. 31#. 31%. 31(. .

31). 320.

321. 322. .

323.

32.

32!.

32#.

32%.

32(.

32).

330. 331.

(0

332. 333.

33. 33!.

In the 655'in7 inte7*a e4*e the ine and ;5ine in

e45nentia 65* and then inte7*ate t5 h5' that: .

33#.

33%.

=

33(.

=

33).

30.

=

=

(1

31.

=

32.

=

33.

=

3.

=

3!.

=

3#.

=

3%. . 3(.

3). 3!0.

=

3!2.

=

3!1.

3!3.

(2

3!.

3!!.

=

=

3!#.

=

3!%.

=

3!(.

=

3!).

=

3#0.

=

. 3#1.

3#2.

=

(3

3#%.

3#3.

=

3#.

=

3#!.

=

3##.

=

=

3#(.

=

3#).

=

3%0.

=

3%1.

=

3%2.

=

(

3%3.

E8auate

3%. 3%!.

3%#.

3(0.

3%%.

=

3%(.

=

3%).

=

=

3(1.

=

3(2.

=

(!

3(3.

3(.

3(!.

=

=

=

3(#.

=

3(%.

=

3((.

=

3().

390.

P#*b' S'*( 12

3)1. 1.

in M = in  - iy> = in  ;5h y - I ;5  inh y 3)2.

An'e* :

3)3.

(#

e iz  − e −iz  3).

2i

1.

3)!.

 e ix − e −ix   e " + e − "    e ix + e −ix   e " − e − "   =  2i     2   + i 2    2                  =

e ix + "

+ e ix− " − e −ix+ " − e −ix − " +   e ix + " − e ix− " + e −ix + " − e − ix− "    i   i     

3)#.

= e ix + "

+ e ix− " − e − ix+ " − e −ix− " − e ix+ " + e ix− " − e −ix+ " + e −ix− " i 2e

3)%.

ix − "

i

= 2=e

3)(.

ix − "

3)).

− e −ix + " 2i

= e

iz 

− e −iz  2i

− e −ix+ " > i

=

e ix − "

400.

− 2e − ix+ "

e iz  − e −iz  2i

=

%T'#b,&/

401.

3.

inh M = inh  ;5 y - i ;5h  in y 02.

An'e* : e z  − e − z 

03.

0.

2

 e x − e − x   e i" + e −i"    e x + e − x   e i" − e − i"         + i       2    2     2    2i   =   =

  e x +i" + e x−i" − e − x+i" − e − x−i"           -

 e x+i" − e x −i" + e − x+i" − e − x             x + i"

2=e

0!.



=

e x +i" 0#.

=

− e − x−i" >

− e − x −i" 2

(%

e z  − e − z 

e z  − e − z 

2

407.

2

=

%T'#b,&/

408. 409. 410. 411. 412.

13.

!.

in 2M = 2 in M ;5 M

1.

An'e* :

1!. e

2 iz 

− e −2

 e − e −  2i = 2   iz 

iz 

2i

1#.

e

1%.

2 iz 

iz 

  e + e −       2 iz 

iz 

      

+ e 0 − e 0 − e −2iz  2i

=

e 2iz  − e −2iz  2i

=

418.

%T'#b,&/

1). %.

inh 2M = 2 inh M ;5h M 20.

An'e* : e 2 z  − e −2 z  2

21.

 e − e −  2  = 2    z 

e 2 z  + e 0 22.

 z 

  e + e −       2  z 

 z 

      

− e 0 − e −2 z  2

=

e 2 z  − e −2 z  2

=

423.

%T'#b,&/

2. d 

;5 z 

2!.

). dz 

2#.

An'e* : d 

2%.

dz 

 = + in M

e

;5 z 

 =

iz 

− e − iz  2

((

1

= 2

2(.

( ie + ie − ) iz 

iz 

− e z  + e − z  2).

2

=

 e z  − e − z      −    2    

30.

=

31.

= + in M

32.

11. ;5h2 M  inh2 M

33.

An'e* :

=

1

3. 2

 e z  + e − z         2    

3!.

3#.

 e z  − e − z      −    2    

=1

 e 2 z  + e −2 z  + 2 − e 2 z  − e −2 z  + 2      = 1       

3%. 3(.

=1 1=1

439.

%T'#b,&/

440.

;5 3M =  ;5 3M  3 ;5 M

1.

13.

2.

An'e* : 3

 e 3iz  + e −3iz     e iz  + e −iz     e iz  + e −iz        =  2     − 3 2     2            

3. .

=

 e 3iz  + e −3iz  + 3e iz  + 3e −iz     3e iz  + 3e −iz      −      ( 2         e 3iz  + e −3iz  =

445.

#. 1!. inh iM = i in M

()

2

%T'#b,&/

%.

An'e* :

(.

e iz  − e

− iz 

2

). e

iz 

− e − iz  2

450.

 e iz  − e −iz       = i 2 i     e

iz 

− e − iz  2

=

%T'#b,&/

451.

!2.

1%. tanh iM = i tan M

!3.

An'e* : inh iz  ;5h iz 

!.

=i

in  z  ;5 z 

 eiz  − e −iz     e − e − 2      x iz  −iz    i 2 e + e     2i   = iz 

!!.

− e −iz  e iz  − e − iz  e iz  + e −iz  = e iz  + e −iz  e

456.

!).

1).

tanh z  =

tanh z  = i tan z  #0.

= i tan  x $ i">

#1.

=

#2.

=

#3.

=

#.

=

#!. 466. 467.

2!.

in  x % i") in  x % i") = in x ;5 i" % in x ;5 i" 46&.

= in x ;5h " % i ;5 x inh "

)0

 x

2 e

iz 

+ e−

iz 

      

iz 

457.

!(.

iz 

%T'#b,&/

%1.

46'.

R = in x ;5h "

%0.

I = + ;5 x inh "

*=

=

%2.

=

%3.

=

%.

=

%!.

=

%#.

=

%%. 47&.

2%.

%).

in  - 3i> = in  ;5 3i - ;5  in 3i = in  ;5h 3 - i ;5  inh 3

(0.

R = in  ;5h 3

(1.

I = + ;5  inh 3

(2.

=

(3.

=

(.

=

)1

(!.

=

(#.

=

(%. ((.

31.

inh n z  -

>

().

= inh n z > ;5h

- ;5h n z > inh

)0.

= inh n z > ;5h

- i ;5h n z > in

)1.

= inh n z > ;5h

> - i 

)2. )3.

)#.

499.

33.

tan i =

).

=

4'5.

= i tan 1

3!.

;5h i - 2>

)%.

= ;5h i ;5h 2  inh i inh 2

)(.

= ;5 1 ;5h 2  i inh 1 inh 2

PROBLEM SECTION 14

500.

(

n i + 3

)=

 z  = i + 3 , r  = 2,θ  = π  K #

(

n i + 3 !01.

3.

) = n 2 + i  π  + 2nπ     #

 

!02.

)2

;5h n z >>

n

(−

2 −i 2

 z  = − 2 − i 2 , r  = 2,θ  =

n !03.

(−

2 −i 2

  π 

 

!0. i

= e n ( −1) = e [ n 1+ ( ± 2 i

i

505.

9.

i

nπ  

)]

2i

=e !0#.

11.

=e

i

n i

i

  n 

 π  ± 2 nπ        2  

2 +i

!0%. i

2i K 2i

=e

n i

π  

2i 

!0(.

13.

=e

π  

 π      n 1+ i   2 ± 2 nπ      

!0).

( − 1) i n

i

= e i n n ( −1) = e i n [ n 1+ ( ± 2 i

!10.

15.

i

i

π 

) = n 2 + i + 2nπ   

5.

( − 1)

3

)]

n π  

)3

 

511.

( i − 1) i +1

= e ( i + i ) n ( i −1) !12.

=e

17.

 

 π  ± 2 nπ        2  

( i + i )  n

2 +i

!13. 514.

PROBLEM SECTION 15

!1!. !1#.

Find ea;h 56 the 655'in7 in the  - iy 65*.

1. a*; in 2 !1%.

 z  = a*; in 2 in  z  =

51&.

5*

e iz  − e − iz  2i

in z  = 2

=2

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

!1). u −u

−1

2i

52!.

=2

!21.

u2 + 1 = iu

!22.

u2 + iu +1 = 0

!23.

i ± 1#i 2 2

u

=

+ = ± 2i

( − 1) + 1 = 2i ±

−3

= u = 2i ± − 3

!2.

e iz 

!2!.

iz  = n 2i ±

!2#.

a*;in 2 =  z  = 2nπ  − i n 2i ±

(

−3

(

−3

!2%. !2(.

It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i 2. F5* 

(

iz  = n 2i ± !2).

−3 e iz 

= e n ( 2i ±

−3 )

= (2i ± − 3 )

!30.

e

531.

−iz 

=

Then

1 e iz 

=

1 e

(

n 2 i ±

in  z  =

=

−3 )

1

( 2i ± − 3 )

e iz  − e − iz 

=

2i

2i ±

=

( 2i  − 3 = (2i  − 3 = −( 2i  − 3 ) i 2

−+3

+3

− 3 + 2i  − 3 2i

=2

!32. 2. a*; tan 2i !33.

 z  = a*; tan 2i 5*

tan z  = 2i

e iz  − e

tan  z  =

in  z  ;5 z 

=

2i − iz  e + e iz 

u − u −1

i(u + u

−1

537.

−1

= −2(u + u −1 )

53&.

u −u

−1

= − ( u 2 + u −1 2 )

53'.

u2

!2.

u=

−2±

e iz  = u

 − ( 2 − 1) 2

=

−2±

=

−2±

#− 2

2

2

 − 2 ± iz  = n   2  

2  

 2i 2 ±    = n  2    

2  

π 

=  z  = + nπ  + 2i n 3 2

544. 3. ;5h+1 > !!.

−iz 

)

= 2i

+ u2 + 2 −1 = 0

a*;tan ( 2i )

!3.

i ( e iz  + e

= 2i

)

u −u

!1.

−iz 

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

!3!.

!0.

=

e iz  − e

2

534.

536.

− iz 

 z  = a*; ;5  5*

;5 z  = 

     

=

−2± 2

2

e iz  + e −iz 

;5  z  =

546.

=1

2

2

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

!%. u + u −1 2

54&.

1

=

2

54'.

u2 - 1 = u

!!0.

u2 + u - 1 = 0 u

!!1.

=

1± 1−  2

 1 ± i  2  

1± i 3

e iz 

2

5* 

=u=

1± i 3 2

3  

iz  = n 

!!2.

=      

 1  =  z  = i ± π  + 2nπ          2     3  

a*;;5

!!3. !!. !!!.

It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i . F5* 

 1 ± i  2  

iz  = n 

e !!#.

iz 

3  

      = n  1 ± − 3        2  

1± n

=e

−3

=

2

1±

−3 2

−3

1

e −iz 

=

1

e iz 

e

!!%.

55&.

Then

;5 z  =

e

1

=

iz 

1± n

−3 2

=

1 1±

−3

=

2

1 − ( − 3)

+e 2

1±

=

2

− 3 + 1 − 3 2

2

!!). 4. inh+1 iK2> !#0.

 z  = a*; in iK2>

in  z  = 561.

e iz  − e

2i

− iz 

5*

=   i     2  

2

=2=1

in z  = iK2>



−3

1



2 −iz 

=

2

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

!#2.

u−u

−1

i =       2  

2i

563.

564.

u2 + 1 = i2u

!#!.

u2 - u +1 = 0

!##.

!#%.

u=

1± 1+  2

e iz  = u

1± ! 2

1± !

=

2

 1 ±

!#(.

=

iz  = n 

!  

     

2

 

 i  =  z  = i 2nπ  + π   , i 2nπ  + !π            #     #    2    

a*;in 

!#). !%0.

It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i iK2. F5* 

 1 ±  2  

iz  = n 

e

!  

     

iz 

!%1.

=e

n

1± ! 2

= 1±

!

2 1 !

e −iz 

=

1

e

iz 

e

!%2.

573.

Then

in  z  =

1

=

e

iz 

1± ! n 2

−e

=

1 1± !

2i

(

=

2

1− !

= −1





2

−iz 

=

1± ! + 1 ! 2 2i

) =

1 2i

!%. 5. a*; ;5 i ( > !%!.

576.

 z  = a*; ;5 i ( > ;5 z  =

e iz  + e − iz  2

5*

=i

(

;5 z  = i ( >

2

!

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

!%%. u + u −1

=i

2

57&.

(

57'.

u2 - 1 = 2i ( u

5&!.

u2 + 2i ( u - 1 = 0

!(1.

u

=

2i ( ± 32i 2 2 iz 

−  = 2i

=u=i

(±

!(2.

e

!(3.

iz  = n i ( ± 3i

(±

− 32 −  = 2i

(±

2

−) = i

2

− 3# = i

( ± 3i

( ± 3i

(

(

a*;;5 i (

!(.

) =  z  = ± π  + 2nπ  − i n (3 + ( ) 2



!(!. It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i i ( >. F5* 

!(#.

(

iz  = n i ( ± 3i e iz 

!(%.

) = n ( − ( ± 3i ) ( − ± ) = ( − ( ± 3i )

= e n

( 3i

!((.

e −iz 

5&'.

=

Then

1 e

iz 

=

1 e

n

( −( ± 3i )

;5 z  =

=

1

( − ( ± 3i )

e iz  + e − iz  2

=

=

− ( 3i − ( 3i = = − ( 3i 2 −(+) − ( − )i

− ( ± 3i + − ( 3i 2

= −( = i

6. tanh+1 +i> 5'!. 7. a*; tan i 2 > !)1.

 z  = a*; tan i 2 >

5*

tan z  = i 2 >

(

e

tan  z  =

in  z  ;5 z 

iz 

2i − iz  e + e iz 

=

i( e iz  + e

−iz 

)

=i

2

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

!)3.

u−u

−1

i(u + u

−1

=i

)

2 2 ( u + u −1 )

5'5.

u − u −1

=−

5'6.

u − u −1

= −( u

5'7.

u 2 + u 2 + 2 − 1 = 0

u

=

2 +u

−1

2

(

−

)=−

2 ± 2 −  2 −1 2

!)(.

e iz  =  u

=

−

 − iz  = n    

(

a*;tan i 2

2 ± #− 2 2

2 ± #− 2 2

!)).

#00.

=

e iz  − e −iz 

2

5'2.

5'4.

− e − iz 

2 ± #−  2 2

   −  = n       

2±

2  

2

 i 2    = n     

2 ± 2   2

) =  z  = i  π  + nπ     2

#01.

 

#02. &. a*; in !K3> #03.

6!4.

 z  = a*; in !K3> in  z  =

2i

=

in z  = !K3>

! 3

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

#0!. u − u −1 6!6.

e iz  − e − iz 

5*

2i

=

! 3

#0%.

3u2  1> = 10iu

#0(.

3u2 + 10iu +3 = 0

     

#0).

10i ± 100i 2 + 3# u= # e iz  #10.

=u=

= 10i ± − 100 + 3# = 10i ± − # = 10i ± (i = !i ± i #

#

#

3

!i ± i 3

 !i ± i       3  

iz  = n  #11.

#12.

a*;in 2 =  z  =

π  2

+ 2nπ  ± i n 3

#13. #1. #1.

It i int int*u *u;t ;ti8 i8ee n5' n5' t5 6in 6ind ;5 ;5 M and and ee ee th that it i !K3 !K3. F5 F5* 

 !i ± i       3  

iz  = n 

e

iz 

=e

 ! i ±  i      3  

n 

#1!.

=   !i ± i      3  

#1#.

617.

 !i i    !i i    !i i            1 1 1 3 3 3             = − !i i e −iz  = iz  =  !i ± i   = = 2 = = 2 − 2! + 1# n e −1 3    !i ± i   2!i − 1#i   e   3    ) )   3   !i ± i !i i + −iz  iz  e −e 10i ! 3 3 = = = in in  z  = 2i

Then

2i

#i

3

#1(. '. tanh+1 i  i 3 > #1).

 z  =  = a*; tan i i 3 >

tan  z  == 62!.

#21.

e iz  − e

− iz 

i( e iz  + e

−iz 

tan z  =  = i i 3 >

5*

)

=i

3

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

u − u −1

622.

i( u + u −

1

623.

u − u−

624.

u − u −1

625.

u2

#2#.

#2%.

#2(.

1

+u

u=

)

=i

3

= − 3 (u + u −1 ) = − (u

3 + u −1 3

3 + 3 −1 = 0

(

−

) = − 3 ± %−

3 ± 3 −  3 −1 2

−

e iz  = u =

3

2

3± %− 3 2

 − iz  = n    

(

3± %− 3 2

a*;tan i 3

   −  = n        

3 ± 1  

 i 2    = n     

3 ± 1  

2

2

     

) =  z  = i  π  + nπ     3

#2).

 

#30. 1!. a*; ;5 !K> 631. 11. 11. inh+1 i  iK 2 > #32.

 z  =  = a*; in i iK 2 >

in in  z  =

e iz  − e

2i

633.

i

=

2

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

#3.

u − u −1

2i

635.

636.

u − u −1

637.

u2

#3(.

−iz 

in z  =  = i iK 2 >

5*

+u

u=

−

i

=

2

= i2

2

2 −1 = 0

2± 2

2+ 

=

−

2± # 2

e iz  = u

#3).

−

=

2

 −  

2 ± #  

iz  = n 

#0.

π 

=  z  = + nπ  + i 2

2

#1.

     

2

i

a*;in

2 ± #

2 n #

#2. It i int int*u *u;t ;ti8 i8ee n5' t5 6in 6ind ;5 ;5 M and and ee ee that that it i iK 2 >. F5* 

#3. #3.

 −  

2 ± #  

iz  = n 

     

2

e

iz 

#.

=e

−

n

2± # 2

=−

2± # 2

#!.

− e −iz 

=

1

e

##.

iz 

1

= e

n

−

2± #

1

=

−

2

2± #

2 # 2

=

2−#

=

−

2 #

−

=−

−

2 # 

2

2

Then

in in  z  =

e

iz 

−e

−iz 

2i

−

2

±

( −− −

#

2

=

2i

2 2



#

=−

2

±

#

−

2



i

12. ;5h+1  3 K2> #%.

64&.

 z  =  = a*; ;5  3 K2> ;5 z  =

u

651.

2

=

;5 z  =  =  3 K2>

3 2

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

#). 65!.

e iz  + e − iz 

5*

+ u −1 2

u2 - 1 =

=

3 2

3u

#

=−

2 2i

=

i

2

#!2.

3 u - 1 = 0

u2 + u

#!3.

iz 

#!.

±

3

=

2

  = n        

#!!.

     e

e

iz 

5* 

=u =

3±i 2

3   2

π           =  z  = i 2nπ   ± #      

3 ± i  

     

2

iz 

=e

   

n 

3 ± i  

   

2

#!%.

Then

3±i 2

It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i  3 K2>. F5* 

iz  = n 

65&.

=

3 ± i     2    

a*;;5

#!#.

3− 

;5 z  =

e

iz 

  =    −iz 

3 ± i  

+e = 2

    

2

3±i + 3 i 2 2 2

=2

3 

=

3 2

65'. 66!.

##1. 13. ;5h+1 +1> ##2.

663.

 z  = a*; ;5 +1> ;5 z  =

− iz 

2

u + u −1 2

= +1

= −1

666.

u2 - 1 = +2u

##%.

u2 - 2u - 1 = 0 u=

##(. ##).

;5 z  = +1>

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

##.

665.

e iz  + e

5*

iz  = n ( − 1)

−2±

− 2

= −1

5* 

e iz  = u = −1

#%0. #%1.

a*;;5( − 1)

=  z  = i + 2nπ 

It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i +1. F5* 

iz  = n ( − 1) #%2.

e iz 

= e n( −1) = −1

e −iz  #%3.

674.

Then

=

1

e

;5 z  =

= iz 

1

e

=

n ( −1)

e iz  + e −iz 

2

1

−1

= −1

1 ( 1) = − + − = −1 2

#%!. 14. a*; in 3iK> #%#.

 z  = a*; in 3iK>

e iz  − e

in  z  =

− iz 

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

#%(.

u −u

−1

=   i         3

2i

67'.

6&!.

2 ( u − u −1 ) = −3

6&1.

2u − 2u −1

#(2.

2u2 - 3u + 2 = 0 u

#(3.

#(.

in z  = 3iK>

=   3i        

2i

677.

5*

=

= −3

3 ± ) + 1# 2

e iz  = u

=

=

3±

2! 2

=

3± ! 2

3± ! 2

 3 ± !       2  

iz  = n 

#(!.

 3i  =  z  = 2nπ  + i n 2, ( 2n + 1)π  − i n 2       

a*;in 

#(#. #(%. 15. a*; tan 2 - i> #((.

 z  = a*; tan 2 - i>

5*

tan z  = 2 - i>

tan  z  = 6&'.

e iz  − e − iz  i ( e iz  + e −iz  )

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

#)0.

u − u −1

6'1.

= 2+i

i ( u + u −1 )

= 2+i

= i ( 2 + i ) ( u + u −1 )

6'2.

u − u −1

6'3.

u −u

6'4.

u − u −1

6'5.

( 2 − 2i ) u − 2iu −1 = 0

6'6.

( 2 − 2i ) u 2 − 2i = 0

#)%.

u=

−1

= ( 2i + i 2 )( u + u −1 ) = 2iu + 2iu −1 − u − u −1

0 ± 0 − ( 2 − 2i )( − 2i ) 2

#)(.

e iz  = u = i + 

#)).

iz  = n i − 

%00.

a*;tan ( 2 + i )

1#i − 1#i 2 2

=

π 

i

(



=  z  = nπ  + 3 +

=

i + 

n 2

7!1. 16. tanh+1 1  2i> 7!2. 17. Sh5' that tan z  ne8e* take the 8aue  i. *int : t*y t5 58e the e9uati5n tan z = i and 6ind that it ead t5 a ;5nt*adi;ti5n. %03.

 z  = a*; tan i

5*

e iz  − e

in  z 

tan  z  =

;5 z 

7!4.

%0!.

− iz 

2i − iz  e + e iz  2

=

e iz  − e

−iz 

i( e iz  + e

−iz 

)

=i

et u  eiz . Then eiz  = u+1, and the e9uati5n $e;5e

u − u −1 7!6.

=

tan z  = i

i( u + u −

1

)

=i

= − ( u + u −1 )

−1

7!7.

u −u

7!&.

2u − 2u −1

7!'.

2u 2

%10.

u=

=0

−2=0 0 ± 0 + 1# 2

%11.

e iz  = u

%12.

iz  = n ( ± 2)

=

± 2

= ±2

= ±2    

a*;tan i =  z  = i ±

%13.

+ 2nπ      2  

π 

714. It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i i. F5* iz  = n ± 2

%1!. %1#.

e iz  = e n( ±2 ) e −iz 

%1%.

=

1

e

tan z  = 71&.

Then

iz 

= ±2

=

1

e

n ( ± 2)

−iz 

=

1

±2

± 2−

1

−e ±2 =i −iz  = iz  1   i( e + e )   i ± 2 +   ± 2     e

iz 

719. 3 720. LINER EUTION ECTORS MTRICES N ETERMINNTS 721. 722.

PROBLEM SECTION 2

723.

S*' ;' <**w(- ='= *< '>,a*(= b" #'?,(- ;' a# *

#*w ';'*( <*#.

%2. %2!. %2#. %2%. %2(. %2).

1.

Penyeeaian :

2 x +  " =  % x − 2 " = 3  R2 - 2R1

%30. R2 - 2R1

%31.

 2 1   % − 2

  

 2   11

  

  3    

1

  11   

0

 2   11

1 0

11   

 2   1

1

  

2 x +  " =  2(1) +  " = 

2 x +  " = 

%33.

 " = 2

 x = 1

%32.

R3 + 2R1  2 0 1 %    2 0     0 1 !      0 1   # − ! 1    0 #

aka  , y> = 1 , 2>

%3.

1 !

−%

       − 13   3. %

%3!. %3#.

%3%.

R3 + #R2  2  0 0  

0

1

1

!

0

− 3%

0

1

1

!

0

1

   2        0  − 3%      0 %

0

1

1

!

#

−%

       − 13    %

%3(. %3).

%0. %1. %2.

 2  0 0  

%   2 R3 : +3%    0  1     0

0

1

1

!

0

− 3%

       − 3%     %

0

  

   

1    

%3. u$titui

%. 2 - M = % %!.y - !M =  M=1 %#.

2 - 1 = % 2 = # =3

y-!= y = +1

%%. ,y,M> = 3,+1,1> %(. %). +  x  "

 0  1 1   0

%!0.

!.

%!1.

 0  1 1   0

 0  0 0   1

 z 

3R1  R

1

−2

0

0

2 !

0

1

0

2

3

0

−!

1

0



0

−#

! 11

0

0

2

!

0

1

0

2

3

0

−!

1

0

−#

3

0

0

1

−! −2

0

0

2

  0   1  1 0 − 3        !   0

! 11

             

  0     0   0    1   

0

0

3

0

0

1

0

0

− % − %     − ! 1   − 2 − 3     2 !    

0

−#

! 11

0

0

2

!

0

1

3

3

0

−2 −!

1

  0    1  1     0   

0

−#

! 11

0

0

2

!

0

1

0

2

3

0

−!

1

             

%!2. %!3.

R2  R

%!. %!!. %!#.

R3  R2

%!%. %!(. %!). %#0. %#1. %#2.

R1-#R3

R1  R

%#3. %#.

%#!.

R : +%

 1  0 0   0

0

0

3

0

0

1

0

0

2

!

−! 1 −2 −3 −% −%

   1       0    0      0    

%##. %#%. %#(.

+ + 2 z  = ! u$titui M

3 x

+ + 2 = !  " +=3

%#).

− ! z  = 1 − 2 z  = − 3  z  = 1

%%0. 3 x

%%1.

− ! =1  x = 2u$titui M

%%2.  " %%3.

− 2 = −3  " = − 1

%%.

u$titui M

%%!. %%#. = 3,2,+1,1> ',,y,M> %%%.

%%(.

 1 − 2 0  ! 0 1 1 2 − 1 %.  

%%).

 1 −R32 - R20  ! 0 1 # 2 0  

%(0.

  

    3    

%

    %   10     

0

0

3

0

0

1

0

0

2

!

−! 1 −2 −3 1

1

             

%(1.

 1 − 2  ! 0 # 2  

%(2.

 1 −R32 - R10 ! 0 1 1 0 0  

  1 − 2   %   ! 0  10     % 0

0



1 0

  

 

0 1 0

    %   1     

R3 : %

%  

2    

%(3. 2 −  "

=   x −  " =   " = − 2 2 +  z  = %  x +  z  = %  z  = !  x = 2

%(. %(!. %(#.

%(%. ,y,M> = 2+2,!> 788. 789. 790.

PROBLEM SECTION 3

%)1.

E8auate the dete*inant in P*5$e 1 t5 # $y *edu;in7 the a

h5'n in Ea4e  and uin7 the a4a;e de8e54ent.

1.

−2

3



3



!

#

−2 −3

%)2.

%)3.

D=

An'e* :

−2

3

3



!

#

−2 %).

D=

D=

−2 −3 →

3

11 0

1

0

1

0

+  ,1 2 ,3 − 3 ,2 U 2 ,2

*5' *edu;ti5n :





 11 %)!.



→

0

= −11

a4a;e de8e54entU

%)#.

2.

1 1

1

1

1 2

3



1 3

#

10

1  10

20

%)%.

An'e* :

%)(.

D=

%)).

D=

(00.

D=

(01.

D=

1 1

1

1

1 2

3



1 3

#

10

1  10

20

1

1

1

1

0

1

2

3

0

1

3

#

1

 10

1

0 1 2

3

0 1 3

#

3

1 3

#

3 ) 1)

D=

0 1

 a4a;e de8e54entUU

−  ,1  , − 3 ,2  *5' *edu;ti5n : 3 U  ,2

 a4a;e de8e54entU

=1

(0.

3.

U

− ,1  U

→

→

0 0 1 1 3

(03.

  *5' *edu;ti5n :  ,

2 3

0 1 3 D=

→

 ,3 −  ,2

→

0 3 ) 1) 1 2

 ,2 −  ,1  *5' *edu;ti5n :

20

1 1 1

1

(02.

→

%

0

1

−3

!

2

0

1



%

−1 −3

2



(

#

1

3

−2 −!

−1 −% %

!



(0!.

An'e* :

(0#.

(0%.

(0(.

(0).

D=

D=

%

0

1

−3

!

2

0

1



%

−1 −3

2



(

#

1

3

−2 −!

−1 −% %

!

%

0

1

−3

!

−!

2

−2

0

0

0

−3

1

2

1

(

#

1

3

#

−3

#

− 10

0

−! −(

2

0

0

−# #

−1 −%

0

( 1

3

−2 −2 −2 −!

%

!

D=

−( (

(11.

(12.

(13.

D=

D=

D=

D=

1

−2 −% −! %

%

0

1

0

0

(

#

1

3

− −2 −!

%

0

1

0 0

−

# 0

1

1

 ,3 −  ,1

  *5' *edu;ti5n :

 ,2

− 2 ,2 U

→  ,1 −  ,!

 !

 *5' *edu;ti5n :

 ,3

  a4a;e de8e54entU

3

%

1

#

1

1

−!

−3 −( → −% %

 *5' *edu;ti5n :

 ,3

− 2 , U

−3 −( → − 21

−!

%

 a4a;e de8e54entU

−3 − 21 → %

− , U

→

− 3 # − 10 − # − 2 −1 →  ,1 +  , # −2 −%  , +  ,3 3 −! %  *5' *edu;ti5n : 2 U

#

(10.

→

+ 3 ,3 % ,1 −  ,2 U

 ,2  *5' *edu;ti5n :

(1.

D=

3

#

0

3

−2 −!

0

1 3

(1!.

D=

1

%

→  a4a;e de8e54entU

−2 = −1! + 2 = −13 −!

(1#. 817.

).

PROBLEM SECTION 4

A (1(. (1).

(20. (21.

A B

B3

(22.

(23.

B-2A

A-B>

(2.

(2!. +A = +2i  3/ (2#. 3B = 3i + !/> (2%. A  B = 2+>i - 3++!>/ (2(. = +2i - (/ (2). B - 2A = i  !/ - 22i - 3/> (30. = (i - / (31.  A-B> =  2 ->i - 3+!>/> (32. = 3i  1/ (33. 11. We ha8e: (3. A = (3!. (3#. (3%. (3(.

S5,

(3).

(0. (1. (2. (3. (. (!. (#. (%. ((. (). (!0.

0 4*58ed>

&51. 13. We ha8e (!2. (!3. (!.

Xueti5n: 6ind ;aa* 465/e;ti5n 56 A 5n B 

and 4*5/e;ti5n

56 B 5n ABA> (!!. An'e*: (!#. (!%. (!(. (!).

tie 4*5/e;ti5n 56 B 5n A> (#0.

2

(#1. (#2.

tie 4*5/e;ti5n 56 A 5n B> (#3.

(#. (#!.

(##.

1!.

a> unit 8e;t5* in ae di*e;ti5n a A=2i+/-2k 

(#%. (#(. (#)."nit 8e;t5* =

(%0.

$> Ve;t5* a7intude 12, ae di*e;ti5n a A.

(%1.

(%2.

;> Ve;t5* 4e*4endi;ua* t5 AVA>.

(%3.T5 6ind 8e;t5* 4e*4endi;ua* t5 A, 'e need t5 ;*5 A t5 an5the* 8e;t5*, he*e 'e ue - = (%.

(%!.

d> Vekt5* unit VA =

1).

We ha8e:

(%(.

Vekt5* 4*e4endi;ua* t5 $5th the i A  B

(%#. (%%.

(%). ((0.

((1. &&2. ((3.

23.

a> I6

ea4e, B =

and A Y B = 0, B Z 0. F5*

((. ((!. ((#.

4*58ed>

((%. (((.

$> A = 2i  3/ - k. A  B = 0, B Z 0, 65* 4*58eent, 'e ue B =

(().

()0. ()1. 892.

4*58ed>

PROBLEM SECTION 5

()3. 1.

E9uati5n 56 the t*ai7ht ine th*5u7h 2, +3>, 54e 3K , * = * [  - At ().

5uti5n :

()!.

A = i - 3/ , ine th*5u7h 2, +3>, 5 :

()#.

*[ = 2i  3/

()%.

5 the e9uati5n 4a*aet*i; i

()(.

* = 2i  3/> - i - 3/> t

()). 3. W*ite in 4a*aet*i; 65*  a in 4*5$e 1>. The e9uati5n 56 the t*ai7ht ine that /5in 1, +2> and 3, 0 > )00.

5uti5n :

)01.

A = 3  1> i - 0  + 0 > > /

)02.

=2i-2/

)03.

ine th*5u7h 1, +2>, 5

)0.

*[ = i + 2 /

)0!.

5 the e9uati5n 4a*aet*i; i

)0#.

* = *[  - At

)0%.

* = = i  2/> - 2i - 2/> t

)0(. !.

S5uti5n : )0).

the 4a*aet*i; i : y = y[ - $t

)10. %.

ine th*5u7h 2, 3, > and !, 1, +2> )11.

5uti5n :

)12.

the e9uati5n yet*i; i :

)13.

2

)1.

2

=

y3

M +

=

+3

+#

)1!. )1#.

).

ine th*5u7h +1, 3, %> and +1, +2, %>

)1%.

S5uti5n :

)1(.

A = +1  +1> > i - +2 + 3> / - %  %> k  

)1).

= +!/

)20.

*[  = + i - 3 / - % k 

)21.

5 the e9uati5n 4a*aet*i; i

)22.

* = *[  - At

)23.

* = + i - 3 / - % k> - + ! /> t

)2. )2!. 11.

ine th*5u7h , +1, 3> and 4a*ae t5 i  2 k  )2#.

5uti5n :

)2%.



1 )2(.

13.

=

0

y  +1>

=

M +3

+2

ine th*5u7h 3, 0, +!> and 4a*ae t5 the ine * = 2, 1, +!> - 0+,

+3, 1> t )2).

S5uti5n :

)30.

N =3i!k

)31.

*[ =  2 i - /  ! k

* + * [ = + 3 / - k 

)32.

5 the e9uati5n 56 4ane :

)33.

N . * + *[ >  = 0 \ 3 i  ! k> . + 3 / - k> = 0 a. ! k = 0  $. ! M = 0

)3. 1!.

Pane th*5u7h the 5*i7in and the 45int in 4*5$e (. ine th*5u7h 0,+ 2, > and 3, + 2, + 1> )3!. )3#.

5uti5n :

)3%.

A =3  0> i - +2  +2> > / - +1  > k 

)3(.

=3i!k  

)3).

*[ = + 2 / - 

)0.

5 the e9uati5n i :

)1.

* = + 2 / -  k> - 3 i  ! k> t

)2. 1%.

Pane th*5u7h the 45int and the 4e*4endi;ua* t5 the ine in 4*5$e 13 )3. ).

5uti5n :

)!.

* = 2 i - / + !> -  +3 / - k > t

, 5 :

)#.

ine th*5u7h i 2, 1, + !>

)%.

the 4e*4endi;ua* i 8e;t5* + 3 / - k >

)(. )). )!0.

1).

Pane ;5ntainin7 the t'5 4a*ae ine in 4*5$e 13

)!1.

S5uti5n :

)!2.

* = 2 i - / + !> -  +3 / - k > t

)!3.

5 the e9uati5n i :

)!#.

)!.

2

)!!.

0

=

y 1

=

+3

21.

An'e* :

)!%.

A = 2i - #/ +3k  

)!(.

B = !i - 2y  k  

M -! 1

)!). )#0.

 2i - #/ +3k > !i - 2y  k > =

)#1. )#2. )#3.

)#. )#!.

23.

An'e*:

)##.

A = 2i - /  2k  

)#%.

B = 3i  #/ +2k  

)#(. )#).

2i - /  2k > 3i  #/ +2k> =

)%0. )%1. )%2.

2!.

)%3.

An'e*:

The e9uati5n 56 the ine 56 inte*e;ti5n 56 the 4ane: ;*5

 4*5du;t 56 thee n5*a 8e;t5*. )%.

=  2i  /  k>  3i  2/  #k>

)%!.

= +k - 12/ - 3k - #i  3/  2i

)%#.

= +i - )/  k  

)%%. )%(. )%). )(0.

2). •

)(1. •

An'e*:

One 45int in the 4ane 1!  ! , +2 , +1 > ;a thi 45int B, then  %, 1, +2 > 1!A AB =  2, 3, +1 > F*5 the e9uti5n 56 the 4ane 'e 7et n5*a 8e;t5* V = 2i  /  k 

•

We 7et $y n di8idin7 N $y

)(2. )(3.

=

)(.

)(!. )(#.

31.

An'e*  

•

One 45int in the 4ane 1!  2 , 1 , 0 > ;a thi 45int B, then  +2, , ! >

• •

1!A )(%. AB =  +, 3, ! > F*5 the e9uti5n 56 the 4ane 'e 7et n5*a 8e;t5* N = 2i - # / 3 k  We 7et $y n di8idin7 N $y

)((. )().

=

))0.

991.

1.

PROBLEMS SECTION 6

A ;5e7e kee4 a *e;5*d in the 65* 56 a at*i  that i, a ta$e > 56 the nu$e* 56 tudent *e;ei8in7 8a*i5u 7*ade in 8a*i5u ;5u*e ea;h yea*. ))2. *ade 5u*e 1)(1 1)(2

1000.

))3. Nu$e* )). A ))!. B ))#.  ))%. D ))(. F ))). What d5e the u 56 thee at*i;e *e4*eent What d5e thei*

di66e*en;e *e4*eent 1001. In 4*5$e 3, 6ind -/ -/  $ -/ 5/ 3-/ 53-. 5$e*8e that 0 -. Sh5' that det  -)  det  -)  (det ) (det -)/ $ut that det  $-)

Z det  $det -/ det 505 det/ and det 3-03det -. in 4*5$e 3, det 3B = 2% det B. 1002. 1003. 3.  100. 100!. 100#.

100%. 100(.

.-

An'e*:

S5 AB Z BA

100). • Det AB> = det BA> = det A> det B> 1010. 11# = 11# = 11# • DetA-B> Z det A - det B • Det!A Z ! det A • Det 3B> Z 3 det B • Det 3B = 2% det B 1011. 1012.

!.

54ute the 4*5du;t 56 ea;h 56 the at*i;e in 4*5$e  'ith it

t*an45e  in $5th 5*de*, that i, 1013. 101.

An'e*:

and

, et;.>

101!.

101#.

101%. 101(.

%.

I6 'e uti4y a ;54e nu$e*

$y

, 'e 7et

, that i a ;54e

nu$e* 'ith the ae r , $ut 'ith it an7e in;*eaed $y that yhe 8e;t5* the an7e

6*5 the 5*i7in t5 the 45int

. We ;an ay

ha $een *5tated $y

. In 6i7u*e #.1, 'e *5tated the ae th*5u7h an7e

i e9ui8aent t5 *5tatin7 the 8e;t5* # th*5u7h the an7e

ae un;han7ed. Thu i6

'hi;h

, ea8in7 the

, 'e h5ud 6ind

. Take *ea and ia7ina*y 4a*t 56 thi e9uati5n t5 5$tain e9uati5n #.3>. 101).

1020.

).

"e *ae*] *ue t5 58e the *5tati5n e9uati5n #.3> 65* x and  "

in te* 56  x and " . h5' that y5u* *eut ;5**e45nd t5 a *5tati5n th*5u7h the an7e 1021. 1022.

.

An'e*:

1023. 102.

=

102!.

102#.

102%. 102(. 11. 102).

1030.

The e9uati5n

Re4*eent *5tati5n 56 ae in th*ee dieni5n. Find the at*i 56

the *eutant *5tati5n and de;*i$e 7e5et*i;ay the net *eut 56 the t'5 *5tati5n. 1031.

1032 1032..

13. 13.

a.

Find Find the the at*i at*i  4*5d 4*5du;t u;t

. 1033. 103.

An'e*:

 $. Sh5', $y uti4yin7 5ut the the at*i;e, that the 655'in7 e9uati5n *e4*eent an ei4e. 103!. 103#.

= 30

An'e*:

103%.

= 30

103(.

103). 100. 101. 101.

1!.

The 655'in7 655'in7 at*i 4*5du;t 4*5du;t i ued in di;uin7 di;uin7 t'5 thin ene

in ai*  102.

=

103. 103. Whe*e Whe*e

and

a*e the 65;a 65;a en7th en7th 56 the ene ene and d  i  i the

ditan;e $et'een the. A in P*5$e 1, eeent

i

'he*e 6 i the 65;a en7th 56 the ;5$inati5n. Find , det , and 1K6. 10. 10!.

1%.

= " and 10#.

F5* the at*i;e in #.1%> and #.22>, 8e*i6y that A A = ". Penyeeaian :

10%.

10(. 10).

10!0.

10!1.

10!2.

10!3.

10!.

Den7an et5de k56akt5* :

10!!.

10!#.

10!%.

10!(.

10!).

10#0. det

10#1.

10#2.

A

=

10#3.

=

10#.

10#!.

10##. &adi 10#%.

In 4*5$e 1( t5 21, 6ind the in8e*e 56 the 7i8en at*i a> $y *5'

*edu;ti5n,  10#(.

=

$> $y the 65*ua #.2>.

10#).

1) 10%0. Penyeeaian : a> By *5' *edu;ti5n 10%1.

10%2.

S5

 $> By the 65*ua #.2> 10%3. A= 10%. 10%!. 10%#. 10%%. 10%(. 10%). 10(0.

10(1. 10(2.

21.

10(3. Penyeeaian : a. By *5' *edu;ti5n 10(.

=

10(!.

10(#.

10(%.

10((.

10().

10)0.

10)1.

10)2.

10)3.

10).

=

10)!.

10)#.

10)%.

10)(. 10)).23.

P*5$e 22$> i a 4e;ia ;ae 56 the 7ene*a the5*e that

the in8e*e 56 a 4*5du;t i the 56 at*i;e i the 4*5du;t 56 the in8e*e in *e8e*e 5*de*, that i,

. P*58e thi

the5*e . int  Thi i eay? 1100. 1101.

In 4*5$e 2 t5 2%, 58e ea;h et 56 e9uati5n $y the eth5d 56

6indin7 the in8e*e 56 the ;5e66i;ient at*i. 1102.

2!.

1103. Penyeeaian: 110.

110!.

110#.

110%.

110(. 110). 1110. 1111. 1112.

1113. 111.

111!.

111#. &adi P = , +3> 111%. 111(.

2%.

111). P'("''=aa( : 1120.

1121.

1122.

1123.

112.

112!.

112#.

112%.

112(.

112).

1130.

1131.

1132.

1133.

113.

113!.

113#.

113%. &adi 4 =  113(. 113).

2).

"e #.2> t5 6ind

110. A =

111. Penyeeaian : 1142.

1143.

, 7i8en

11.

11!. 11#.

11%. 11(.

11). 11!0.

11!1. 11!2.

11!3. 11!.

11!!. 11!#.

11!%.

=

11!(.

11!). 11#0.

11#1.

11#2. 11#3.

11#. 11#!.

=1

11##.

11#%.

=

1168. 1169.

PROBLEMS SECTION 7

11%0.

A*e the 655'in7 inea* 6un;ti5n P*58e y5u* ;5n;ui5n $y

h5'in7 that  *> ati6ie $5th 56 e9uati5n %.1> 5* that d5e n5t ati6y at eat 5ne 56 the. 1.    *> = A . * - 3, 'he*e A i a 7i8en 8e;t5*. 3.   *> = * . *  11%1. a*e the 655'in7 inea* 8e;t5* 6un;ti5n P*58e y5u* ;5n;ui5n uin7 %.2>. !. F*> = A  *, 'he*e A i a 7i8en 8e;t5*. 11%2.

A*e the 655'in7 54e*at5* inea*

%. De6inite inte7*a 'ith *e4e;t t5 x 6*5 0 t5 1< the 5$/e;t $ein7 54e*ated 5n a*e 6un;ti5n 56 x. ). Find the 9ua*e< 54e*ate 5n nu$e* 5* 5n 6un;ti5n. 11. Find the a$5ute 8aue< 54e*ate 5n ;54e nu$e*. 13. A in 4*5$e 12, i D2 - 2D - 1 inea* 1!. Find the t*an45e< 54e*ate 5n 9ua*e at*i;e. 1%. Find the dete*inant < 54e*ate 5n 9ua*e at*i;e. 11%3. 1174.

PROBLEMS SECTION 8

11%!.

S58e the 655'in7 et 56 iutane5u e9uati5n $y *edu;in7 the

at*i t5 *5' e;he5n 65*. 1.

11%#. at*iknya : 11%%.

=

11%(. 11%). S5uti5n : 11(0.

11(1. O* in the 8e;t5* 65* : 11(2. * = , y, M>

5*

 = 1 and y = M

11(3.

= 1, M, M>

11(.

= 1,0,0> - 0,1,1>M

11(!. 3.

11(#. at*iknya : 11(%.

=

11((. S5uti5n : 11().

5*  = + and y = 3

11)0. O* in the 8e;t5* 65* : 11)1. * = ,y> = ,+3> 11)2. !.

11)3. at*iknya : 11).

=

11)!. S5uti5n : 11)#.

11)%. Atau 11)(.  - M = 11)).

 =

1200.

yM =

-! +M

1201.

y ==

1202.

M

-M

=0

1203. O* in the 8e;t5* 65* : 120. * = , y, M> =

120!.

=

-

M

120#. %.

120%. at*iknya : 120(.

120).

1210. S5uti5n :

=

=

1211.

1212. O* M

=

1213.

y

=

121.

 = 0

121!. * = , y, M> = 0,

,

>

121#. ).

121%. at*iknya : 121(.

121). Sehin77a en/adi : 1220. 1221.

1222. Sehin77a en/adi :

=

1223.

=

122. S5uti5n : 122!. ! = +10 122#.  = +2 122%. +2y - M = +! 122(. 2y = M - ! 122). y = 2M 1230. O* in the 8e;t5* 65* : 1231. * = , y, M> = +2, 2M -

, M> = +2,

, 0> - 0, 2,1>M

1232. 11.

1233. at*iknya : 123.

123!.

=

a. The e9uati5n ;5**e45ndin7 t5 the *edu;ed at*i a*e 123#.

123%.  $. By $a;k u$tituti5n 'e 6ind

=

123(.

 - y  +y - M> = 0 123). -y = +y - M 120.  = +M 121. y = M ;. In the 8e;t5* 65* 122. * = , y, M> = +M, M, M> = +1, 1>M 123. 13.

12. 12!.

at*iknya :

12#.

=

=

a. The e9uati5n ;5**e45ndin7 t5 the *edu;ed at*i a*e 12%.

 $. By $a;k u$tituti5n 'e 6ind 12(.  = +! 12). y = +2 12!0. M = +3 ;. In the 8e;t5* 65* 12!1. * = , y, M> = +!, 2, +3> 12!2. 1!.

12!3.

at*iknya :

12!.

12!!.

12!#.

=

=

=

The e9uati5n ;5**e45ndin7 t5 the *edu;ed at*i a*e

12!%.

12!(.

Atau

12!).

y = M-

12#0.

 = +M + #

12#1.

O* in the 8e;t5* 65* :

12#2.

* = , y, M> = +M + #

,M-

> = + #

,

1,1,1>M 1%.

12#3.

at*iknya :

12#.

=

:

12#!. 12##.

aka en/adi :

, 0> - +

12#%.

12#(.

=

The e9uati5n ;5**e45ndin7 t5 the *edu;ed at*i a*e

12#).

12%0.

O*  

12%1.

2y = M  3

12%2.

y =

12%3.

 =+

12%.

O* in the 8e;t5* 65* :

12%!.

* = , y, M> = +

M+ M+

12%#.

= +

12%%. 1).

12%(. 12%).

M+

at*iknya :

, ,+

M+

, M>

, 0> - +

,

, 1>M

12(0.

=

:2

12(1.

en/adi :

12(2.

12(3.

The e9uati5n ;5**e45ndin7 t5 the *edu;ed at*i a*e

12(.

12(!.

O*  

12(#.

u = 2M  1

12(%.

8 = !M  #

12((.

=0

12().

y=0

12)0.

O* in the 8e;t5* 65* :

12)1.

* = , y, M, u, 8> = 0, 0, M, 2M  1, !M  #>

12)2.

= 0, 0, 0, +1, +#> - 0, 0,1, 2, !>M

12)3. 12).

Find the *ank 56 ea;h 56 the 655'in7 at*i;e

12)!.

12)#. 12)%.

=

:2

en/adi : =

12)(. 12)). 1300. 23.

The *ank 56 at*i i 1 =

=

1301. 1302. 1303. 2!.

The *ank 56 at*i i 2

: 2 dan

130.

=

:

130!.

130#. 130%. 130(. 2%.

=

The *ank 56 at*i i 1 =

:2 130). 1310.

aka en/adi : =

1311.

The*e a*e 5ny t'5 inde4endent 8e;t5*

1312. 1313.

PROBLEMSECTION 9

1. Find - and - 7i8en 131.

131!.

,

O$e*8e that - i the nu at*i< i6 'e ;a it 0, then  - =

0, $ut neithe*  n5* - i 0. Sh5' that  i in7ua*. 131#.

An'e*: AB =

131%.

&adi te*$ukti $ah'a AB = 0

131(.

3. i8en the at*i

131).

1320.

Find the t*an45e, the ad/5int, the in8e*e, the ;54ete

;5n/u7ate, and the t*an45e ;5n/u7ate 56 . Ve*i6y that +1 = +1  = the unit at*i. 1321.

An'e* : AT

1322.

A =

1323.

Ad/ A =

=

132. 132!.

Det A = 1 0> - 0 0> - !i  +2i  2i  2> = 10 A+1 = ad/ A 132#. =

132). 1330.

132%.

54ete ;5n/u7ate A =

132(.

T*an45e ;i5n/u7ate 56 A =

!. a>

Sh5' that (->T = -T T, that i, the t*an45e 56 a 4*5du;t

56 t'5 at*i;e i e9ua t5 the 4*5du;t 56 the t*an45e in *e8e*e 5*de*. a> "e a> t5 h5' that the at*i T i yet*i;. ^See ).%>_ 1331. An'e*: a. AB>T = BTAT 1332. ia: A = dan

1333.

AB> =

133.

AB> T =

133!.

BT =

=

dan AT =

133#.

BTAT =

=

133%. &adi, te*$ukti $ah'a AB> T = BTAT  $. AAT =

=

133(. 133).

%. Sh5' that the at*i 56 a *5tati5n #.> i 5*th575na< h5' that a unit

at*i i 5*th575na. 130. 131.

). Sh5' that the dete*inant 56 an 5*th575na at*i i -1 5* +1. int:

What i the dete*inant 56 T 'hen  i 5*th575na and h5' d5 det  and det  T  ;54a*e 132. 133.

11. Sh5' that a *ea e*itian at*i i yet*i;. Sh5' that a *ea

unita*y at*i i 5*th575na. 13.

13. Sh5' that the 655'in7 at*i i unita*y at*i.

13!.

1!. Sh5' that t'5 eanin7 56 ad/5int in #.#> a*e e*itian. 13#. 1347.

PROBLEM SECTION 10

13(. 13).

In 4*5$e 1 t5 (, the 7i8en et 56 e9uati5n $y *edu;in7 the

at*i t5 e;he5n 65*. Say 7e5et*i;ay 'hat the 5uti5n i 5ne 45int 5n a ine a 4ane,5* n5 5uti5n>.I6 the 5uti5n i a ine, '*ite it 8e;t5* e9ueti5n.

13!0.

1

3 x + 2 " −  z  = 1  x −  " + 2 z  = !  ! x + ( " −  z  = −3 .

 3 2 − 1 1       1 −  2 !    ! ( −  − 3     S5uti5n:  

13!1.

 % 0 0 %     %      , − , 2 , − ,       → 1 −  2 !        → − 3  ! ( −  − 3   !       1

3

2

3

0

0

0

(

(

−

    13   − 3    %

13!2.

  !   , ↔ ,       → − 3 %   1

3

−  − 3     !    , := % >     → − 3 ( 13     1 0 %      

( 0

3

0

 0   , −! ,       → 0 1   1

13!3.

3

( 0 0

−  − 3    !    , +3 ,     → 0 ( 13     1 0 1      

( 0

2

0

( 0

3

0

−  − 3     ( 1#   0 1    

−  − (     ( 1#   0 1     ( " −  z  = −( ( " − =2> = −(

( " −  z  = −(

( " − ( = −(

( z  = 1# →  z  = 2

=0  " = 0 ( "

13!.

 x = 1

13!!.

The 5uti5n i ,y,M>=1,0,2> and * = /- 2k 

$y $a;k u$tituti5n 'e 6ind

2 x + ! " + 2 z  =   x + 2 " + 2 z  = %  3 x +  " = 1 

13!#.

3.

13!%.

S5uti5n:

13!(.

 2  1 3  

!

2   

2

2

 0

 2 !   2 , − ,     → 0 − 1 %     3  1       2

1

2 2 0

  

 2 #    , − ,     → 0 − 1 10     3  1       1

2

0 2 0

− #    1 3   , := 2>      → 0 − 1 10     3  1       1

0 2 0

− 3     10   1    

13!).

 0 1       → 0 − 1 3   

0

3 ,1 − ,3  ,1:= ! >

2 0

− 2    3     , ↔ ,     → 0 − 1 10     0 −1 1      

   3  0 1      , + ,       → 0 0 2 (   2 10      0 1 0 − 2   0 − 2         3 x +  " = 1 3 x + =−2> = 1 3 x − ( = 1 3 x = ) $y $a;k u$tituti5n 'e 6ind  x = 3 1

3 x +  "

0

1

3

2

13#0.

=1 2 z  = ( →  z  =   " = −2

13#1.

The 5uti5n ,y,M> = 3,+2,> and * = 3i+2/-k 

13#2.

!.

 x +  " − 2 z  = ! 2 x +  " −  z  = 2   x −  " − 2 z  = 1 

%

2 x +  " −  z  = !  x + 3 " − 2 z  = 0 ! x + 2 " − 10 z  = 13  2 x − 2 " −  z  = (

3

13#3.

13#.

13#!.

S5uti5n:

 2 1  1 3 ! 2   2 − 2

 2 1 −   1 3 −2  , − ,       →  ! 2 − 10  0 − 3 0 

1

  2  1 3 , − 2 ,          → 13   0 3

13##.

2

− −2 − 10 −

!  

 

0  

13 

  (    

   2 1 −     0    , := 3>  1 3 −2         →  ! 2 − 10 13      3      0 − 1 0 0 −       3 − 2 0   0 − 2 3)     − 1 0 1    !



   2 0 −     0    , + ,  1 3 −2         →  ! 2 − 10 13      1      0 − 1 0 !

1



  

    13     1     0

13#%.

 , − 2 ,      → 3

1

 2 1 −  !    2 0 −  #    2 0 −  #            0 ! 0 ! 0 ! 0 ! 0 ! 0 ! − − −                2 , − ,  , + ,    →    →  1 0 − 2 3    1 0 − 2 3    0 0 0 0                0 − 1 0 1    0 − 1 0 1    0 − 1 0 1    2 0 −  #    1 0 − 2 3       , := 2>    0 ! 0 ! 0 1 0 1 − −           , ↔ , , := ! >       →   →  →   0 −1 0 1 0 −1 0 1           0 0 0 0 0 0 0 0         13#(. 1

3



3

1

1 2



13#). 13%0.

). "e *ae*] *ue t5 58e P*5$e 3.

13%1.

S5uti5n:

2 x + ! " + 2 z  =   2   x + 2 " + 2 z  = %  1 3 x +  " = 1   →  3

13%2. 13%3.

Find 

13%.

   % 1    x =  2  1 3  

13%!.

Find y

13%#.

 2  1 3    " =  2  1 3  

!

2  

2

2  

!

2  

! 2 

2    x  

           2   "  =  %         0      z    1  

 

 0    

 

2   = −(> + !=2> + 2= 2#> − 32 + 10 + !2  0    = 2=−(> + !=#> + 2=−2> = − 1# + 30 −  2



2  

%

2  

1

 

0    

!

2  

2

2   2= −2> + =#> + 2= −20>



 

0    

=

2=−(> + !=#> + 2=−2>

−  + 2 − 0 − 20 = = −2 − 1# + 30 −  10

=

=

30 10

=3

13%%.

Find M

13%(.

 2  1 3    z  =  2  1 3  

13%).

S5, *eut 6*5 thi 4*5$e i

13(0.

11. Sh5' that i6 ea;h eeent 56 5ne *5' 5* ;5un> 56 a

!

  

 

2 %    1     !

2  

 

2   2=−2#> + !=20> + =−2>  0    = 2=−(> + !=#> + 2=−2> 2

=

− !2 + 200 − ( 0 = = − 2# + 30 −  10

{3, −2,}  5, * =3i+2/-k 

dete*inant i the u te*, 13(1.

The dete*inant ;an $e '*itten a a u 56 t'5 dete*inant <65*

ea4e,

a11 a21 13(2. 13(3. 13(. 13(!. 13(#. 13(%. 13((. 13(). 13)0. 13)1. 13)2. 13)3. 13). 13)!. 13)#. 13)%. 13)(. 13)).

a31

+ 512 a22 + 522 a32 + 532 a12

a13

a11

a12

a13

a11

a12

a13

a23

a21

a22

a23

a21

a22

a23

a31

a23

a33

a31

a23

a33

a33

=

 -

"e thi *eut t5 8e*ity Fa;t $ 56 Se;ti5n 3.