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%. 23
0.2323 = 1000
1(. 11.
+
23
+ ....
1000000
1). 23 23
s ∞
=
a 1 − r
1000 1
=
1−
20.
=
23 ))!
=
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 hainya 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*eiina*y tet t5 'hethe* the 655'in7 e*ie a*e di8e*7ent 5* *e9ui*e 6u*the* tetin7. a*e6u: d5 n5t ay that e*ie i ;5n8e*7ent< the 4*eiina*y tet ;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$a7i den7an (−1 ) n 2 n →∞ n→ ∞ n +1 1 ¿ lim 1 1 1 n →∞ + n+ 1 (−1 ) (−1 )n+ 1 n2
(2.
#.
n
2
"ntuk en7u/i kek5n8e*7enian aka ;a*iah
(1.
3.
n
n+ 1
1+
¿ ¿
n +3 2
+ 10 n 3 1+
n
di$a7i den7an n
n
n + 10 3
∞
∞ + 10
1 +0
∞ ¿ 0 Tet 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 Tet 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 Tet 6u*the*>
3#. 11. "in7 .#> 7i8e a 4*556 56 the 4*eiina*y tet 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 ;5nide* uti4y $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 diihat $ah'a: n?@2n, untuk n@3 %. ∞
%!. 3. 4*58e the k5n8e*7en;e 56
∑ n1
2
$y 7*5u4in7 te* 5e 'hat a in
1
4*5$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 10.
du
¿ ∞
1
∫¿
dn=¿ nlnn
2
∞
∫¿ 2
11.
¿ [ lnu ] ∞
12.
¿ ln ∞− ln2
13.
¿∞
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.
S5uti5n :
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#.
S5uti5n : an =
100
n
200
n
n +1
23%.
an +1=
23(.
¿ lim n →∞
100
( n + 1 )200
| | an + 1 an
23).
| | n+ 1
100
20.
¿ lim n →∞
( n +1)
n
100
n
21.
22.
23.
¿ 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).
S5uti5n : √ ( 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#(.
S5uti5n :
2#).
2%0.
2%1.
2%2. 2%3.
2%.
2%!.
2%#.
2%%. 2%(.
1(
2%).
33.
2(0.
S5uti5n :
2(1.
2(2.
2(3.
2(.
2(!.
2(#.
2(%.
2((. 2().
2)0.
3!.
1)
2)1.
2)2.
S5uti5n :
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
= ni →∞
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 > = ni → ∞ = n + n 3 + 2n 2 + n
2
− =n + n + n + 1> = ni = ni → ∞ = n 3 + 2n 2 + 2n →∞ 3
308.
= −1>
=n + 1> = −1> n = ni : 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 *ai5 4ada uku ke+# dan uku ke+%,
aka di4e*5eh:
ρ n 311.
! !# 3! 3! = = ρ = n→∞it ρ n = 〈1 %2 3# dan 3# , aka de*et te*e$ut i
e*u4akan de*et k5n8e*7en. 312. 313.
3. ∞
∑
1
n =1 n
n 3
= 1+
1 2n 3
+
1 3n 3
+
1 n 3
22
+
1 !n 3
+
1 #n 3
+
1 % n 3
+
31.
Den7an en77unakan u/i *ai5 4ada uku ke+% dan uku
ke+#, aka di4e*5eh:
1 # n 3 = n 3 = 0,(213 % 1 dan
ρ n 31!. it ρ = ni →∞ ρ n
= 0,(213〈1 , aka de*et te*e$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 *ai5 4ada uku ke+) dan uku
ke+(, aka di4e*5eh:
ρ n 320.
) 132 = = 0,20( it ρ = ni %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
eetaknya, uai ke+!, e;a*a uu $e*aku hu$un7an 3 =n?>
〉
2
1
n
, aka
de*et itu di8e*7en. ∞
32!.
).
nn
∑ n? n =1
2
= 1 + 2 + 3 + 10 + 3
e$ea* a4ai
23
, ka*ena de*etnya eakin
- ∞ , aka de*etnya di8e*7en.
32#. 32%. 32(.
11. ∞
2n
= n
2
∑ n
−)
32).
=
(
−1
+
10 1#
+
12 2!
+
1 0
+
1# !!
+
1( %2
+
20 )1
+
22 112
+
Den7an en77unakan u/i *ai5 4ada uku ke+( dan uku
ke+%, aka di4e*5eh:
ρ n 330.
22 )1 = = 0,()3%! it ρ = ni ρ n = 0,()3%!〈1 112 20 → ∞ dan ,
aka de*et te*e$ut e*u4akan de*et k5n8e*7en. 331. 332.
13.
333.
Jawab:
33.
P*eiina*y tet
33!.
33#.
=
33%.
=
33(.
= 0 TF>
33).
S4e;ia ;54a*i5n tet
30.
$n =
31.
2
32.
33.
.
3.
=
3!.
=1
3#.
&adi, de*et te*e$ut k5n8e*7en.
3%. 3(.
1!.
3).
Jawab:
3!0.
P*eiina*y tet
3!1.
3!2.
=
3!3.
=
3!.
&adi, de*et te*e$ut di8e*7en.
3!!. 3!#.
1%.
3!%.
Jawab:
3!(.
Rai5 tet
3!).
3#0.
=
2!
3#1.
=
3#2.
=
3#3.
=
3#.
=
3#!. 3##.
= Ca*ena
aka de*et te*e$ut k5n8e*7en.
3#%. 3#(.
1).
369.
Jawab:
3%0. 3%1.
P*eiina*y tet
3%2.
=
TF> 3%3.
Rai5 tet
3%.
3%!.
=
2#
3%#.
=
3%%.
=
3%(.
=
3%).
=
3(0.
=
3(1.
=
3(2.
Ca*ena
aka de*et te*e$ut k5n8e*7en.
3(3. 3(.
21.
/ika
385.
Jawab:
3(#. 3(%.
=
3((.
Rai5 tet
3().
3)0.
=
3)1.
=
3)2.
=
3)3.
Ca*ena
aka de*et te*e$ut k5n8e*7en.
3). 2%
3)!. 396.
PROBLEM SECTION 10
397.
1.
3)(.
Rai5 tet
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 daa dae*ah
12. 13. 1. 1!.
Find the inte*8a 56 ;5n8e*7e 56 ea;h the 655'in7 45'e* e*ie<
$e u*e in8eti7ate the end45int 56 the inte*8a in ea;h ;ae.
2(
1#.
( − 1) n x n n =1 n( n + 1) ∞
∑
1%.
3.
1(.
S5uti5n:
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!.
Tet :
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(.
S5uti5n: 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 eua ha*7a A >. #. %. (.
30
= 0, aka de*et dikatakan k5n8e*7en
∞
∑ n =1
x
3n
n
).
%.
!0.
S5uti5n : 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>
Tet : = 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 →∞
#%.
).
#(.
S5uti5n :
#).
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 = Tet : < = ±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
(!. (#.
S5uti5n:
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 Tet : ! < 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!.
S5uti5n:
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).
ia = 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.
Penyeeaian : x De*et daa* 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#.
4enyeeaian: x −
x 3 3?
+
x ! !?
−
x % %?
........
#2%.
de*et daa* in =
#2(.
2 keudian 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!.
Penyeeaian :
#3#.
#3%.
x de*et daa* 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
Penyeeaian:
de*et daa*: ;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 enui 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*, $ah'a
##1.
1 + x n 1 − x
##2.
x
= ∫ 0
dt
1 − t 2
=
x −
x 3 3?
+
x ! !?
−
x% %?
+ .....
##3.
664. 2 665. COMPLEX NUMBER 666.
PROBLEM SECTION 4
##%.
1. 1- i
##(.
S5uti5n :
669.
!1
670.
"!1
#%1.
# =
=
#%2.
#%3. #%.
3. 1 + i
#%!.
S5uti5n :
676.
!1
677.
"!$
678.
#=
=2
#%).
680.
681.
#(2.
!. 2i
#(3.
S5uti5n :
684.
!0
685.
# =
"!0
=2
#(#.
#(%.
#((.
%. +1
#().
S5uti5n :
690.
! $1
691.
"!0
#)2.
# =
=2
#)3.
#). #)!.
). 2i 2
#)#.
S5uti5n :
697.
! $2
698.
"!2
!
699.
# =
=2
%00.
%01. %02.
= 3J K
%03.
11. 2 ;5
%0.
S5uti5n :
%0!.
#=
- i in
>
!2
=
706.
%0%.
= tan
708.
!1
709.
"!
=
=
%10. %11.
1!. ;5
%12.
S5uti5n :
%13.
=
%1.
= tan
715.
#=
=
=0 =1
%1#.
=1
%1%.
1-
%1(.
1-0=
=
#
%1). 720.
! $1
721.
"!0
%22. %23.
1%.
%2.
S5uti5n :
%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 e4unyai Re+M = 1,# dan +M
= +2,% %1.
aka e4unyai 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$5ute 8aue 56: 2 + 3i
%!0.
2%.
1− i
+ 3i 1− i
2
%!1.
An'e*:
=
=
+) 1+1
=
13 2
%!2.
%!3.
− 3i 1 + 3i 2
%!. %!!. %!#. %!%.
2). 1-2i>3
%!(.
An'e*:
1-2i>3 = 1-2i> 2 1-2i>
(
%!).
= 1-i+> 1-2i>
%#0.
= i+3> 1-2i>
%#1.
= i+(+3+#i
%#2.
= +11+12i
%#3.
=−11>2 + =−2> 2
=
%#.
=
121 +
%#!.
=
12!
%##.
=! ! ! − 2i
%#%.
31. ! + 2i
%#(.
An'e*:
! − 2i
! + 2i
! + 2i
! − 2i
=
2! + 2! +
=
2) 2)
=
1 = 1
%#). 2!
%%0.
33. 3 + i
%%1.
An'e*:
2!
2!
=
3 + i 3 − i %%2.
#2! ) + 1#
=
#2! 2!
=
2!
=!
S58e 65* a 45i$e 8aue 56 the *ea nu$e* and y in the
655'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 .... kai 2
)
%(3.
-2y-1 =0 .....kai 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 - 2iy + y 2 = 2i
%)%.
2 y2 -2iy = 2i •
2 y2 = 0
•
2y = 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 - 2iy + y 2 = 2 2iy - y 2 2+ y2 - 2iy = 2 - y2 2iy
(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!.
Tet ea;h O6 the 655'in7 e*ie 65* ;5n8e*7en;e
(2#. (2%.
3.
(2(.
S58e :
(2).
= i
(30.
n
(31.
= i
(32.
n
!1
(33.
=
=
G 1 CONVEREN>
(3. (3!.
!.
(3#.
S58e :
(3%.
= i
(3(.
n
(3).
= i
(0.
n
- i
n
- i
n
(1.
=
-
(2.
= 2 @ 1 DIVEREN>
(3. (.
%.
(!.
S58e :
(#.
= i
(%.
n
((.
= i
().
n
(!0.
= i 1>
!2
(!1.
=
=
DIVEREN>
(!2. (!3.
).
(!.
S58e :
(!!.
= i
(!#.
n
(!%.
= i
(!(.
n
(!).
=i
(#0.
=
(#1.
DIFFERENT TEST>
=
(#2. (#3.
INTERA
(#. (#!.
(##.
i= 1 /adi in = 1
(#%. (#(.
ehin77a:
(#).
(%0.
DIVEREN>
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 eua niai M
((!. ((#.
3.
((%. (((. ((). ()0. ()1. ()2. ()3.
Ca*ena G1,
().
aka k5n8e*7en 4ada eua niai 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 eu ua ni niai 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 adaah 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 adaah 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 eua niai 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$ut k5n8e*7en 4ada
102. 1025. −
1. e
PROBLEM SECTION 9
i µ
−
102#.
e
i µ
=
1 102%. = 2 1
102(. =
2
− i. 1
2
−
2
+ i
in
µ
>
2
2
i
2
2
e = - iy 1
= 2
2
1
1031.
µ
iθ
102). 1030.
;5
2
y=+2
3 µ i
1032.
3.
2
)e
1033. 3 µ i 2
;5
103.
)e
= )
103!.
= 0 - i +1>
103#.
= +i
103%.
e
iθ
=0
103).
y = +1 !.
e
2
+ i in
3µ
= - iy
103(.
100.
3 µ
! µ i
101. ! µ i
= ;5 3 µ + i in ! µ >
102.
e
103.
= ;5 µ + i in µ >
10.
= +1 - 0
10!.
= +1
#2
2 >
10#.
iθ
e
10%.
= - iy
= +1
10(.
y =0
10). 10!0.
%.
3e
2 =1+ iµ >
10!1. 10!2.
3e 2
2 =1+ iµ >
= 3e
2 + i 2 µ
i 2 µ
10!3.
= 3e . e
10!.
2 = 3e =;5 2 µ + i in 2 µ >
10!!.
= 3e 1 - 0>
10!#.
= 3e
2
2
10!%.
e
10!(.
= 3e
10!).
y=0
iθ
= - iy
2
10#0.
10#1.
).
2e
−i
µ 2
10#2.
10#3.
2e
−i
µ 2 = 2
;5
µ
2
10#.
= 2 0 i.1>
10#!.
= +2i
10##.
e
iθ
− i in
µ
2>
= - iy
10#%.
=0
10#(.
y = +2
10#). 10%0. ! i µ
10%1.
2e
11. ! i µ
10%2.
2e
2 =;5
=
! µ
+ i in !µ >
#3
10%3.
=
2 =;5 22! + i in 22!>
10%.
=
2 =− ;5 ! − i in !> 2 =−
10%!.
=
1!76.
= +1 i
10%%.
e
iθ
1 2
1
2 −i.
2>
2
= - iy
10%(.
= +1
10%).
y = +1
10(0.
=i − 3> 10(1.
10(2.
1− i
13.
=i − 3> 1− i
3
1+ i 1+ i
=
=i 2 10(3.
3
=i − 3 > 3 =1 + i > 1 − = −1>
−2
=
=2 − 2 3i >=i + i
2
−
3i + 3>=i − 3 − i 3 > 2 =
3 − i 3>
2 =2 − 2 3i >=−1 − 3 + i − i 3>
10(.
2
=
=2 − 2 3i >=−1i + 3> + i =1 − 3>
10(!.
=
10(#.
=
− 2=1 +
2
3 > + 2i =1 − 3 > + 2 3i =1 + 3 > − 2 3 =i >=1 − 3> 2
2
− 2=1 + 10(%.
2
=
−2− 2 10((.
=
3 + (i + 2 3 − # 2
=
− ( + (i 10().
3> + 2iV=1 − 3 > + = 3 + 3>U + 2 3 − #
2
= − − i #
10)0.
= +
y=
10)1.
=1 + i> 2
10)2.
1!.
10)3.
=1 + i>
10).
2
+ =1 + i>
+ =1 + i> =
=1 + i> V1 + =1 + i> U 2
2
= 1 + 2i − 1V1 + =1 + 2i − 1>U
10)!.
= 2iV1 + 2iU
10)#.
= 2i + =−1>
10)%.
=
10)(.
= +
10)).
y=2
− + 2i
1
1100.
3 + ( ) i 1 1%.
1101.
&a'a$ : 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 eiiki = 1 < y = 0 < * = 1 dan θ = #%.
3
1K 3
#(.
3
1 e4unyai * =
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 , 20 , 3#0
0
%2. %3.
3
0
1 = 1 ;5
0
0
- i in
0
> =1
0
%.
= 1 ;5120 - i in120 > = +1K2 - 1 K 2 3 i 0
%!.
0
0
= 1 ;5 20 - i in 20 > = +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 eiiki = 1 < y =0 < * = 1 dan θ = (2.
1K
(3.
1 eiiki * =
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 eiiki = 1 < y =0 < * = 1 dan θ = )%.
#
1 eiiki * = 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 , 20 , 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 ;5120 - i in120 > = +1K2 - 1 K 2 3 i
10#. 10%.
0
10(.
= 1 ;5 1(0
10).
= 1 ;5 20
0
- i in1(0
0
> = +1
- i in 20
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# eiiki = 1# < y =0 < * = 1# dan θ = ( 11#. 1# eiiki * = 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 eiiki = 1 < y =0 < * = 1 dan θ =
0
13.
!
1K !
13!.
!
1 eiiki * =
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). 10.
!
0
1 = ;5
0
0
- i in 0 0
11.
=
;5 %2
0
- i in %2
0
12. 13.
=
>= 1
0
;51 - 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.
+( eiiki = +( < y =0 < * = ( dan θ = 1(0 3
− ( eiiki * = 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.
+ eiiki = + < y =0 < * = dan θ =
1#.
− eiiki * = 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!
>= 1i
1%. 1%!. #
− #
1%#.
1!.
1%%.
&a'a$ :
1%(. 1%).
+# eiiki = +# < y =0 < * = # dan θ = 1(0 #
− # eiiki * = 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 ;51!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 eiiki = +1 < y =0 < * = 1 dan θ = !
1)!.
− 1 eiiki * = 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 ;51(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 eiiki = 0 < y =1 < * = 1 dan θ =
20).
3
210. 1K 3
211. z =
i eiiki * = 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 eiiki * = 1K 2
22#.
0
2 + 2i 3 eiiki = 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 − ( eiiki = +( < y =(
23%.
(i 3 − ( eiiki * = 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 20.
n
θ =
30
= 0
0
21.
0
0
+)0
0
θ = 30 ,120 , 210 , 300
0
0
n
0
22.
23.
(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 eiiki * = 1K !
2!2.
2 dan θ =
+ 1 i eiiki = +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 ;51()
=
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 = aa
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# eiiki = 1# < y =0 < * = 1# dan θ =
0
2%2.
1# eiiki * = 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/uahan da*i hainya = 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 $y thei* 45'e* e*ie. W*ite the 45'e* e*ie 65* eiM. By ;54a*in7 thee e*ie 5$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 655'in7 in *e;tan7ua* 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 655'in7 inte7*a e4*e the ine and ;5ine in
e45nentia 65* and then inte7*ate t5 h5' that: .
33#.
33%.
=
33(.
=
33).
30.
=
=
(1
31.
=
32.
=
33.
=
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.
E8auate
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 ;5h 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 ;5h 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 ;5h 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. ;5h2 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 ;5h 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 ;5h " % i ;5 x inh "
)0
x
2 e
iz
+ e−
iz
iz
457.
!(.
iz
%T'#b,&/
%1.
46'.
R = in x ;5h "
%0.
I = + ;5 x inh "
*=
=
%2.
=
%3.
=
%.
=
%!.
=
%#.
=
%%. 47&.
2%.
%).
in - 3i> = in ;5 3i - ;5 in 3i = in ;5h 3 - i ;5 inh 3
(0.
R = in ;5h 3
(1.
I = + ;5 inh 3
(2.
=
(3.
=
(.
=
)1
(!.
=
(#.
=
(%. ((.
31.
inh n z -
>
().
= inh n z > ;5h
- ;5h n z > inh
)0.
= inh n z > ;5h
- i ;5h n z > in
)1.
= inh n z > ;5h
> - i
)2. )3.
)#.
499.
33.
tan i =
).
=
4'5.
= i tan 1
3!.
;5h i - 2>
)%.
= ;5h i ;5h 2 inh i inh 2
)(.
= ;5 1 ;5h 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
;5h 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 655'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 eiz = u+1, and the e9uati5n $e;5e
!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 int*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. ;5h+1 > !!.
−iz
)
= 2i
+ u2 + 2 −1 = 0
a*;tan ( 2i )
!3.
i ( e iz + e
= 2i
)
u −u
!1.
−iz
et u eiz . Then eiz = u+1, and the e9uati5n $e;5e
!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 eiz = u+1, and the e9uati5n $e;5e
!%. 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 int*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 eiz = u+1, and the e9uati5n $e;5e
!#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 int*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 eiz = u+1, and the e9uati5n $e;5e
!%%. 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 int*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 eiz = u+1, and the e9uati5n $e;5e
!)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 eiz = u+1, and the e9uati5n $e;5e
#0!. u − u −1 6!6.
e iz − e − iz
5*
2i
=
! 3
#0%.
3u2 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 int int*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 eiz = u+1, and the e9uati5n $e;5e
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 eiz = u+1, and the e9uati5n $e;5e
#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 int int*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. ;5h+1 3 K2> #%.
64&.
z = = a*; ;5 3 K2> ;5 z =
u
651.
2
=
;5 z = = 3 K2>
3 2
et u eiz . Then eiz = u+1, and the e9uati5n $e;5e
#). 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 int*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. ;5h+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 eiz = u+1, and the e9uati5n $e;5e
##.
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 int*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 eiz = u+1, and the e9uati5n $e;5e
#%(.
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 eiz = u+1, and the e9uati5n $e;5e
#)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 8aue i. *int : t*y t5 58e 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 eiz = u+1, and the e9uati5n $e;5e
u − u −1 7!6.
=
tan z = i
i( u + u −
1
)
=i
= − ( u + u −1 )
−1
7!7.
u −u
7!&.
2u − 2u −1
7!'.
2u 2
%10.
u=
=0
−2=0 0 ± 0 + 1# 2
%11.
e iz = u
%12.
iz = n ( ± 2)
=
± 2
= ±2
= ±2
a*;tan i = z = i ±
%13.
+ 2nπ 2
π
714. It i int*u;ti8e n5' t5 6ind ;5 M and ee that it i i. F5* iz = n ± 2
%1!. %1#.
e iz = e n( ±2 ) e −iz
%1%.
=
1
e
tan z = 71&.
Then
iz
= ±2
=
1
e
n ( ± 2)
−iz
=
1
±2
± 2−
1
−e ±2 =i −iz = iz 1 i( e + e ) i ± 2 + ± 2 e
iz
719. 3 720. LINER EUTION ECTORS MTRICES N ETERMINNTS 721. 722.
PROBLEM SECTION 2
723.
S*' ;' <**w(- ='= *< '>,a*(= b" #'?,(- ;' a# *
#*w ';'*( <*#.
%2. %2!. %2#. %2%. %2(. %2).
1.
Penyeeaian :
2 x + " = % x − 2 " = 3 R2 - 2R1
%30. R2 - 2R1
%31.
2 1 % − 2
2 11
3
1
11
0
2 11
1 0
11
2 1
1
2 x + " = 2(1) + " =
2 x + " =
%33.
" = 2
x = 1
%32.
R3 + 2R1 2 0 1 % 2 0 0 1 ! 0 1 # − ! 1 0 #
aka , y> = 1 , 2>
%3.
1 !
−%
− 13 3. %
%3!. %3#.
%3%.
R3 + #R2 2 0 0
0
1
1
!
0
− 3%
0
1
1
!
0
1
2 0 − 3% 0 %
0
1
1
!
#
−%
− 13 %
%3(. %3).
%0. %1. %2.
2 0 0
% 2 R3 : +3% 0 1 0
0
1
1
!
0
− 3%
− 3% %
0
1
%3. u$titui
%. 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$titui M
3 x
+ + 2 = ! " +=3
%#).
− ! z = 1 − 2 z = − 3 z = 1
%%0. 3 x
%%1.
− ! =1 x = 2u$titui M
%%2. " %%3.
− 2 = −3 " = − 1
%%.
u$titui M
%%!. %%#. = 3,2,+1,1> ',,y,M> %%%.
%%(.
1 − 2 0 ! 0 1 1 2 − 1 %.
%%).
1 −R32 - R20 ! 0 1 # 2 0
%(0.
3
%
% 10
0
0
3
0
0
1
0
0
2
!
−! 1 −2 −3 1
1
%(1.
1 − 2 ! 0 # 2
%(2.
1 −R32 - R10 ! 0 1 1 0 0
1 − 2 % ! 0 10 % 0
0
1 0
0 1 0
% 1
R3 : %
%
2
%(3. 2 − "
= x − " = " = − 2 2 + z = % x + z = % z = ! x = 2
%(. %(!. %(#.
%(%. ,y,M> = 2+2,!> 788. 789. 790.
PROBLEM SECTION 3
%)1.
E8auate the dete*inant in P*5$e 1 t5 # $y *edu;in7 the a
h5'n in Ea4e and uin7 the a4a;e de8e54ent.
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
a4a;e de8e54entU
%)#.
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
a4a;e de8e54entUU
− ,1 , − 3 ,2 *5' *edu;ti5n : 3 U ,2
a4a;e de8e54entU
=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
a4a;e de8e54entU
3
%
1
#
1
1
−!
−3 −( → −% %
*5' *edu;ti5n :
,3
− 2 , U
−3 −( → − 21
−!
%
a4a;e de8e54entU
−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
%
→ a4a;e de8e54entU
−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 = 3i + !/> (2%. A B = 2+>i - 3++!>/ (2(. = +2i - (/ (2). B - 2A = i !/ - 22i - 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. (!.
Xueti5n: 6ind ;aa* 465/e;ti5n 56 A 5n B
and 4*5/e;ti5n
56 B 5n ABA> (!!. An'e*: (!#. (!%. (!(. (!).
tie 4*5/e;ti5n 56 B 5n A> (#0.
2
(#1. (#2.
tie 4*5/e;ti5n 56 A 5n B> (#3.
(#. (#!.
(##.
1!.
a> unit 8e;t5* in ae di*e;ti5n a A=2i+/-2k
(#%. (#(. (#)."nit 8e;t5* =
(%0.
$> Ve;t5* a7intude 12, ae di*e;ti5n a A.
(%1.
(%2.
;> Ve;t5* 4e*4endi;ua* t5 AVA>.
(%3.T5 6ind 8e;t5* 4e*4endi;ua* t5 A, 'e need t5 ;*5 A t5 an5the* 8e;t5*, he*e 'e ue - = (%.
(%!.
d> Vekt5* unit VA =
1).
We ha8e:
(%(.
Vekt5* 4*e4endi;ua* t5 $5th the i A B
(%#. (%%.
(%). ((0.
((1. &&2. ((3.
23.
a> I6
ea4e, B =
and A Y B = 0, B Z 0. F5*
((. ((!. ((#.
4*58ed>
((%. (((.
$> A = 2i 3/ - k. A B = 0, B Z 0, 65* 4*58eent, 'e ue B =
(().
()0. ()1. 892.
4*58ed>
PROBLEM SECTION 5
()3. 1.
E9uati5n 56 the t*ai7ht ine th*5u7h 2, +3>, 54e 3K , * = * [ - At ().
5uti5n :
()!.
A = i - 3/ , ine th*5u7h 2, +3>, 5 :
()#.
*[ = 2i 3/
()%.
5 the e9uati5n 4a*aet*i; i
()(.
* = 2i 3/> - i - 3/> t
()). 3. W*ite in 4a*aet*i; 65* a in 4*5$e 1>. The e9uati5n 56 the t*ai7ht ine that /5in 1, +2> and 3, 0 > )00.
5uti5n :
)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*aet*i; i
)0#.
* = *[ - At
)0%.
* = = i 2/> - 2i - 2/> t
)0(. !.
S5uti5n : )0).
the 4a*aet*i; i : y = y[ - $t
)10. %.
ine th*5u7h 2, 3, > and !, 1, +2> )11.
5uti5n :
)12.
the e9uati5n yet*i; i :
)13.
2
)1.
2
=
y3
M +
=
+3
+#
)1!. )1#.
).
ine th*5u7h +1, 3, %> and +1, +2, %>
)1%.
S5uti5n :
)1(.
A = +1 +1> > i - +2 + 3> / - % %> k
)1).
= +!/
)20.
*[ = + i - 3 / - % k
)21.
5 the e9uati5n 4a*aet*i; i
)22.
* = *[ - At
)23.
* = + i - 3 / - % k> - + ! /> t
)2. )2!. 11.
ine th*5u7h , +1, 3> and 4a*ae t5 i 2 k )2#.
5uti5n :
)2%.
1 )2(.
13.
=
0
y +1>
=
M +3
+2
ine th*5u7h 3, 0, +!> and 4a*ae t5 the ine * = 2, 1, +!> - 0+,
+3, 1> t )2).
S5uti5n :
)30.
N =3i!k
)31.
*[ = 2 i - / ! k
* + * [ = + 3 / - k
)32.
5 the e9uati5n 56 4ane :
)33.
N . * + *[ > = 0 \ 3 i ! k> . + 3 / - k> = 0 a. ! k = 0 $. ! M = 0
)3. 1!.
Pane th*5u7h the 5*i7in and the 45int in 4*5$e (. ine th*5u7h 0,+ 2, > and 3, + 2, + 1> )3!. )3#.
5uti5n :
)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%.
Pane th*5u7h the 45int and the 4e*4endi;ua* t5 the ine in 4*5$e 13 )3. ).
5uti5n :
)!.
* = 2 i - / + !> - +3 / - k > t
, 5 :
)#.
ine th*5u7h i 2, 1, + !>
)%.
the 4e*4endi;ua* i 8e;t5* + 3 / - k >
)(. )). )!0.
1).
Pane ;5ntainin7 the t'5 4a*ae ine in 4*5$e 13
)!1.
S5uti5n :
)!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 4ane: ;*5
4*5du;t 56 thee 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 4ane 1! ! , +2 , +1 > ;a thi 45int B, then %, 1, +2 > 1!A AB = 2, 3, +1 > F*5 the e9uti5n 56 the 4ane '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 4ane 1! 2 , 1 , 0 > ;a thi 45int B, then +2, , ! >
• •
1!A )(%. AB = +, 3, ! > F*5 the e9uti5n 56 the 4ane 'e 7et n5*a 8e;t5* N = 2i - # / 3 k We 7et $y n di8idin7 N $y
)((. )().
=
))0.
991.
1.
PROBLEMS SECTION 6
A ;5e7e kee4 a *e;5*d in the 65* 56 a at*i that i, a ta$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 thee at*i;e *e4*eent What d5e thei*
di66e*en;e *e4*eent 1001. In 4*5$e 3, 6ind -/ -/ $ -/ 5/ 3-/ 53-. 5$e*8e that 0 -. Sh5' that det -) det -) (det ) (det -)/ $ut that det $-)
Z det $det -/ det 505 det/ and det 3-03det -. in 4*5$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# • DetA-B> Z det A - det B • Det!A Z ! det A • Det 3B> Z 3 det B • Det 3B = 2% det B 1011. 1012.
!.
54ute the 4*5du;t 56 ea;h 56 the at*i;e in 4*5$e 'ith it
t*an45e in $5th 5*de*, that i, 1013. 101.
An'e*:
and
, et;.>
101!.
101#.
101%. 101(.
%.
I6 'e uti4y a ;54e nu$e*
$y
, 'e 7et
, that i a ;54e
nu$e* 'ith the ae r , $ut 'ith it an7e in;*eaed $y that yhe 8e;t5* the an7e
6*5 the 5*i7in t5 the 45int
. We ;an ay
ha $een *5tated $y
. In 6i7u*e #.1, 'e *5tated the ae th*5u7h an7e
i e9ui8aent t5 *5tatin7 the 8e;t5* # th*5u7h the an7e
ae un;han7ed. Thu i6
'hi;h
, ea8in7 the
, 'e h5ud 6ind
. Take *ea and ia7ina*y 4a*t 56 thi e9uati5n t5 5$tain e9uati5n #.3>. 101).
1020.
).
"e *ae*] *ue t5 58e the *5tati5n e9uati5n #.3> 65* x and "
in te* 56 x and " . h5' that y5u* *eut ;5**e45nd t5 a *5tati5n th*5u7h the an7e 1021. 1022.
.
An'e*:
1023. 102.
=
102!.
102#.
102%. 102(. 11. 102).
1030.
The e9uati5n
Re4*eent *5tati5n 56 ae in th*ee dieni5n. Find the at*i 56
the *eutant *5tati5n and de;*i$e 7e5et*i;ay the net *eut 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 uti4yin7 5ut the the at*i;e, that the 655'in7 e9uati5n *e4*eent an ei4e. 103!. 103#.
= 30
An'e*:
103%.
= 30
103(.
103). 100. 101. 101.
1!.
The 655'in7 655'in7 at*i 4*5du;t 4*5du;t i ued in di;uin7 di;uin7 t'5 thin ene
in ai* 102.
=
103. 103. Whe*e Whe*e
and
a*e the 65;a 65;a en7th en7th 56 the ene ene and d i i the
ditan;e $et'een the. A in P*5$e 1, eeent
i
'he*e 6 i the 65;a en7th 56 the ;5$inati5n. Find , det , and 1K6. 10. 10!.
1%.
= " and 10#.
F5* the at*i;e in #.1%> and #.22>, 8e*i6y that A A = ". Penyeeaian :
10%.
10(. 10).
10!0.
10!1.
10!2.
10!3.
10!.
Den7an et5de k56akt5* :
10!!.
10!#.
10!%.
10!(.
10!).
10#0. det
10#1.
10#2.
A
=
10#3.
=
10#.
10#!.
10##. &adi 10#%.
In 4*5$e 1( t5 21, 6ind the in8e*e 56 the 7i8en at*i a> $y *5'
*edu;ti5n, 10#(.
=
$> $y the 65*ua #.2>.
10#).
1) 10%0. Penyeeaian : a> By *5' *edu;ti5n 10%1.
10%2.
S5
$> By the 65*ua #.2> 10%3. A= 10%. 10%!. 10%#. 10%%. 10%(. 10%). 10(0.
10(1. 10(2.
21.
10(3. Penyeeaian : a. By *5' *edu;ti5n 10(.
=
10(!.
10(#.
10(%.
10((.
10().
10)0.
10)1.
10)2.
10)3.
10).
=
10)!.
10)#.
10)%.
10)(. 10)).23.
P*5$e 22$> i a 4e;ia ;ae 56 the 7ene*a the5*e that
the in8e*e 56 a 4*5du;t i the 56 at*i;e i the 4*5du;t 56 the in8e*e in *e8e*e 5*de*, that i,
. P*58e thi
the5*e . int Thi i eay? 1100. 1101.
In 4*5$e 2 t5 2%, 58e ea;h et 56 e9uati5n $y the eth5d 56
6indin7 the in8e*e 56 the ;5e66i;ient at*i. 1102.
2!.
1103. Penyeeaian: 110.
110!.
110#.
110%.
110(. 110). 1110. 1111. 1112.
1113. 111.
111!.
111#. &adi P = , +3> 111%. 111(.
2%.
111). P'("''=aa( : 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
110. A =
111. Penyeeaian : 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 655'in7 inea* 6un;ti5n P*58e y5u* ;5n;ui5n $y
h5'in7 that *> ati6ie $5th 56 e9uati5n %.1> 5* that d5e n5t ati6y at eat 5ne 56 the. 1. *> = A . * - 3, 'he*e A i a 7i8en 8e;t5*. 3. *> = * . * 11%1. a*e the 655'in7 inea* 8e;t5* 6un;ti5n P*58e y5u* ;5n;ui5n uin7 %.2>. !. F*> = A *, 'he*e A i a 7i8en 8e;t5*. 11%2.
A*e the 655'in7 54e*at5* inea*
%. De6inite inte7*a 'ith *e4e;t t5 x 6*5 0 t5 1< the 5$/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$5ute 8aue< 54e*ate 5n ;54e nu$e*. 13. A in 4*5$e 12, i D2 - 2D - 1 inea* 1!. Find the t*an45e< 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%!.
S58e the 655'in7 et 56 iutane5u e9uati5n $y *edu;in7 the
at*i t5 *5' e;he5n 65*. 1.
11%#. at*iknya : 11%%.
=
11%(. 11%). S5uti5n : 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*iknya : 11(%.
=
11((. S5uti5n : 11().
5* = + and y = 3
11)0. O* in the 8e;t5* 65* : 11)1. * = ,y> = ,+3> 11)2. !.
11)3. at*iknya : 11).
=
11)!. S5uti5n : 11)#.
11)%. Atau 11)(. - M = 11)).
=
1200.
yM =
-! +M
1201.
y ==
1202.
M
-M
=0
1203. O* in the 8e;t5* 65* : 120. * = , y, M> =
120!.
=
-
M
120#. %.
120%. at*iknya : 120(.
120).
1210. S5uti5n :
=
=
1211.
1212. O* M
=
1213.
y
=
121.
= 0
121!. * = , y, M> = 0,
,
>
121#. ).
121%. at*iknya : 121(.
121). Sehin77a en/adi : 1220. 1221.
1222. Sehin77a en/adi :
=
1223.
=
122. S5uti5n : 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*iknya : 123.
123!.
=
a. The e9uati5n ;5**e45ndin7 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 120. = +M 121. y = M ;. In the 8e;t5* 65* 122. * = , y, M> = +M, M, M> = +1, 1>M 123. 13.
12. 12!.
at*iknya :
12#.
=
=
a. The e9uati5n ;5**e45ndin7 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**e45ndin7 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**e45ndin7 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*iknya :
, ,+
M+
, M>
, 0> - +
,
, 1>M
12(0.
=
:2
12(1.
en/adi :
12(2.
12(3.
The e9uati5n ;5**e45ndin7 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 655'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 5ny t'5 inde4endent 8e;t5*
1312. 1313.
PROBLEMSECTION 9
1. Find - and - 7i8en 131.
131!.
,
O$e*8e that - i the nu at*i< i6 'e ;a it 0, then - =
0, $ut neithe* n5* - i 0. Sh5' that i in7ua*. 131#.
An'e*: AB =
131%.
&adi te*$ukti $ah'a AB = 0
131(.
3. i8en the at*i
131).
1320.
Find the t*an45e, the ad/5int, the in8e*e, the ;54ete
;5n/u7ate, and the t*an45e ;5n/u7ate 56 . Ve*i6y that +1 = +1 = the unit at*i. 1321.
An'e* : AT
1322.
A =
1323.
Ad/ A =
=
132. 132!.
Det A = 1 0> - 0 0> - !i +2i 2i 2> = 10 A+1 = ad/ A 132#. =
132). 1330.
132%.
54ete ;5n/u7ate A =
132(.
T*an45e ;i5n/u7ate 56 A =
!. a>
Sh5' that (->T = -T T, that i, the t*an45e 56 a 4*5du;t
56 t'5 at*i;e i e9ua t5 the 4*5du;t 56 the t*an45e in *e8e*e 5*de*. a> "e a> t5 h5' that the at*i T i yet*i;. ^See ).%>_ 1331. An'e*: a. AB>T = BTAT 1332. ia: A = dan
1333.
AB> =
133.
AB> T =
133!.
BT =
=
dan AT =
133#.
BTAT =
=
133%. &adi, te*$ukti $ah'a AB> T = BTAT $. AAT =
=
133(. 133).
%. Sh5' that the at*i 56 a *5tati5n #.> i 5*th575na< h5' that a unit
at*i i 5*th575na. 130. 131.
). Sh5' that the dete*inant 56 an 5*th575na at*i i -1 5* +1. int:
What i the dete*inant 56 T 'hen i 5*th575na and h5' d5 det and det T ;54a*e 132. 133.
11. Sh5' that a *ea e*itian at*i i yet*i;. Sh5' that a *ea
unita*y at*i i 5*th575na. 13.
13. Sh5' that the 655'in7 at*i i unita*y at*i.
13!.
1!. Sh5' that t'5 eanin7 56 ad/5int in #.#> a*e e*itian. 13#. 1347.
PROBLEM SECTION 10
13(. 13).
In 4*5$e 1 t5 (, the 7i8en et 56 e9uati5n $y *edu;in7 the
at*i t5 e;he5n 65*. Say 7e5et*i;ay 'hat the 5uti5n i 5ne 45int 5n a ine a 4ane,5* n5 5uti5n>.I6 the 5uti5n i a ine, '*ite it 8e;t5* e9ueti5n.
13!0.
1
3 x + 2 " − z = 1 x − " + 2 z = ! ! x + ( " − z = −3 .
3 2 − 1 1 1 − 2 ! ! ( − − 3 S5uti5n:
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 5uti5n 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!%.
S5uti5n:
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 5uti5n ,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#!.
S5uti5n:
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 *ae*] *ue t5 58e P*5$e 3.
13%1.
S5uti5n:
2 x + ! " + 2 z = 2 x + 2 " + 2 z = % 1 3 x + " = 1 → 3
13%2. 13%3.
Find
13%.
% 1 x = 2 1 3
13%!.
Find y
13%#.
2 1 3 " = 2 1 3
!
2
2
2
!
2
! 2
2 x
2 " = % 0 z 1
0
2 = −(> + !=2> + 2= 2#> − 32 + 10 + !2 0 = 2=−(> + !=#> + 2=−2> = − 1# + 30 − 2
2
%
2
1
0
!
2
2
2 2= −2> + =#> + 2= −20>
0
=
2=−(> + !=#> + 2=−2>
− + 2 − 0 − 20 = = −2 − 1# + 30 − 10
=
=
30 10
=3
13%%.
Find M
13%(.
2 1 3 z = 2 1 3
13%).
S5, *eut 6*5 thi 4*5$e i
13(0.
11. Sh5' that i6 ea;h eeent 56 5ne *5' 5* ;5un> 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*
ea4e,
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 *eut t5 8e*ity Fa;t $ 56 Se;ti5n 3.