1-amaliy mashg‘ulot. Matrisa va ular ustida amallar

1.

1-amaliy mashg‘ulot. Matrisa va ular ustida amallar
1.1. Quyidagi matritsalarning
2 A + 3B chiziqli kombinatsiyasini toping, bu
yerda
1 2 3
−2 3 0
=
A =
,B
.
−
0
1
1
2
1
1
1 2 3
−2 3 0 2 4 6
+
3
⋅
Yechish. 2 A + 3B = 2 ⋅
2 1 1 = 0 2 −2 +
0 1 −1
−6 9 0 2 − 6 4 + 9 6 + 0 −4 13 6
+ =
=
.
6 3 3 0 + 6 2 + 3 −2 + 3 6 5 1
Berilgan matritsalarning chiziqli kombinatsiyasini toping:
2 −1 0
3 1 2
−2 1 3 .
3
4
−
2
,
B
=
1.2. 4 A − 5 B, A =
−3 1 5
0 2 −4
1 −1 −3
0 3 2
=
,B
1.3.
3 A + 2 B, A =
.
−
2
1
5
1
4
1
3 4 5
1 2 3
=
1.4. A =
va B 6 0 −2 matritsalar berilgan. AB va BA
1 0 −1
7 1 8
matritsalar koʻpaytmasi (agar mumkin boʻlsa)ni toping.
3 4 5
1 2 3
=
AB
⋅ 6 =
0 −2
Yechish.
1 0 −1 7 1 8
1⋅ 3 + 2 ⋅ 6 + 3 ⋅ 7
1 ⋅ 4 + 2 ⋅ 0 + 3 ⋅1
1 ⋅ 5 + 2 ⋅ ( −2 ) + 3 ⋅ 8
=
1 ⋅ 3 + 0 ⋅ 6 + ( −1) ⋅ 7 1 ⋅ 4 + 0 ⋅ 0 + ( −1) ⋅ 1 1 ⋅ 5 + 0 ⋅ ( −2 ) + ( −1) ⋅ 8
36 7 25
=
.
−
−
4
3
3
BA koʻpaytma mavjud emas, B matritsaning ustunlari soni A matritsaning
satrlari soniga mos kelmaydi.
AB va BA matritsalar koʻpaytmasi (agar mumkin boʻlsa)ni toping:

2.

3 −2
3 4
,
B
=
1.5. A =
2 5 .
5 −4
3
1
=
1.6. A ( 4 0 −2 3 1) , B = −1 .
5
2
1 −1
2 1 −1
1.7. A = 0 1 0 , B = 0 1 .
1 0
0 0 −1
1 −1 2
3 4 1
1.8. A = 2 3 4 , B = 0 2 5 .
−4 5 1
1 −1 4
1.9.
Agar f ( x ) =
−2 x 2 + 5 x + 9,
1 2
A=
3 0
boʻlsa,
f ( A)
koʻphadning qiymatini toping.
Yechish.
1 2 1 2 1 ⋅1 + 2 ⋅ 3 1 ⋅ 2 + 2 ⋅ 0 7 2
A2 = A ⋅ A =
⋅
=
=
,
3 0 3 0 3 ⋅1 + 0 ⋅ 3 3 ⋅ 2 + 0 ⋅ 0 3 6
7 2
1 2
1 0
f ( A ) =−2 A2 + 5 A + 9 E =−2 ⋅
+
5
⋅
+
9
⋅
3 0
0 1 =
3 6
−14 −4 5 10 9 0 0 6
=
+ 15 0 + 0 =
9 −3 .
−
6
−
12
9
f ( A ) matritsali koʻphadning qiymatini toping:
1 5
1.10. f ( x ) = 3 x3 + x 2 + 2, A =
.
0 −3
1 2 0
1.11. f ( x ) = 3 x − 5=
x + 2, A 0 2 −1 .
−2 1 4
2
matritsali

3.

1 0 0
1.12. f ( x ) = x3 − 6 x 2 + 9 x =
+ 4, A 0 2 −1 .
0 1 4
1 2
1.13. f ( x ) = 2 x3 − 3 x 2 + 5, A =
.
−2 3
1.14. Satrlari ustida elementar almashtirishlar yordamida
A matritsani
pogʻonasimon koʻrinishga keltiring:
0 −1 −1 −3
A = 1 2 4 7
5 0 10 5
0 −1 −1 −3
1 2 4 7
=
Yechish. A 1 2 4 7 I ↔ II 0 −1 −1 −3 III − 5 ⋅ I
5 0 10 5
5 0 10 5
4
7
1 2
0 −1 −1 −3 III − 10 ⋅ II
0 −10 −10 −30
1 2 4 7
B= 0 −1 −1 −3 − pog ' onasimon matritsa.
0 0 0 0
Matritsalarni pogʻonasimon koʻrinishga keltiring:
2 3 −2
1.15. 3 1 1 .
1 5 −5
1 −2 1 11
3 −1 2
5
.
1.16.
2 1 −3 −18
5 0 −1 −13
2 3 −2 3
1.17. 3 1 1 2 .
1 5 −5 4

4.

1.18. Kopxona 3 xil mahsulot ishlab chiqarish uchun 2 xil xomashyodan
foydalanadi. Xomashyo xarajatlari
2 3
A = 5 2
1 4
matritsa bilan berilgan.
Mahsulot ishlab chiqarish rejasi C = (100 80 130 ) - satr-matritsa koʻrinishida
30
berilgan. Har bir xomashyo turining bir birligi bahosi (pul.birl.) B = 50
ustun-matritsa koʻrinishida berilgan. Rejani bajarish uchun sarflanadigan
xomashyo miqdorini va xomashyoning umumiy bahosini aniqlang.
Yechish. 1-usul. Har bir xomashyo sarfi
2 3
S = C ⋅ A = (100 80 130 ) ⋅ 5 2 = ( 730 980 )
1 4
boʻlsa, xomashyoning umumiy bahosi
30
Q = S ⋅ B = ( C ⋅ A ) ⋅ B = ( 730 980 ) ⋅ = ( 70900 )
50
boʻladi.
2-usul. Avval har bir mahsulot turiga sarflanuvchi xomashyo miqdori
2 3
210
30
R = A ⋅ B = 5 2 ⋅ = 250
1 4 50 230
Soʻngra, xom ashyoning umumiy bahosini aniqlaymiz
210
Q = C ⋅ R = (100 80 130 ) ⋅ 250 = ( 70900 )
230
Quyidagi iqtisodiy mazmundagi masalani yeching:
1.19. Kopxona 3 xil mahsulot ishlab chiqarish uchun 2 xil xomashyodan
foydalanadi. Xomashyo harajatlari
2 1
A = 1 3
3 4
matritsa bilan berilgan.

5.

Maxsulot
ishlab
chiqarish
rejasi
C = (100 200 150 )
–
satr-matritsa
koʻrinishida berilgan. Har bir xomashyo turining bir birligi bahosi (pul.birl.)
10
B = – ustun-matritsa koʻrinishida berilgan. Rejani bajarish uchun
15
sarflanadigan xomashyo miqdorini va xomashyoning umumiy bahosini
aniqlang.
1.20. Kopxona 4 xil mahsulot ishlab chiqarish uchun 2 xil xomashyodan
foydalanadi. Xomashyo harajatlari
2
3
A=
1
3
1
2
4
2
matritsa bilan berilgan.
Mahsulot ishlab chiqarish rejasi C = (120 80 150 130 ) – satr-matritsa
koʻrinishida berilgan. Har bir xomashyo turining bir birligi bahosi (pul.birl.)
80
B = – ustun-matritsa koʻrinishida berilgan. Rejani bajarish uchun
60
sarflanadigan xomashyo miqdorini va xomashyoning umumiy bahosini
aniqlang.
1.21.
1 2
ikkinchi tartibli determinantni hisoblang:
3 4
Yechish.
1 2
=⋅
1 4 − 2 ⋅ 3 =−2.
3 4
Ikkinchi tartibli determinantni hisoblang:
1.22.
−7 6
.
5 −4
1.23.
10 −5
.
9 −8
sin10
sin890
− cos10
cos890
x+ y
x
1.25.
y−x
x2 − y 2
2x
x− y
.
y−x
x2 − y 2
1.24.
.

6.

1
1.26.
1
5 − a2
−a
a2
1
2
5+a
1
2
.
3 2 1
1.27. Uchinchi tartibli determinantni hisoblang: 2 5 3 .
3 4 2
Yechish. Determinantni birinchi satr elementlari boʻyicha yoyib hisoblaymiz:
3 2 1
5 3
2 3
2 5
2 5 3=
3⋅
− 2⋅
+ 1⋅
=
4 2
3 2
3 4
3 4 2
= 3 ⋅ ( 5 ⋅ 2 − 3 ⋅ 4 ) − 2 ⋅ ( 2 ⋅ 2 − 3 ⋅ 3) + 1 ⋅ ( 2 ⋅ 4 − 5 ⋅ 3) =
=3 ⋅ ( −2 ) − 2 ⋅ ( −5 ) + 1 ⋅ ( −7 ) =−3.
Uchinchi tartibli determinantlarni ixtiyoriy satr (ustun) elementlari
boʻyicha yoyib hisoblang:
1 2 3
1.28. 4 5 6 .
7 8 9
2 1 3
1.29. 5 3 2 .
1 4 3
1.30. Uchinchi tartibli determinantni uchburchak qoidasidan foydalanib
1 2 3
hisoblang: −4 5 −6 .
7 8 9
1 2 3
−4 5 −6 =1 ⋅ 5 ⋅ 9 + 2 ⋅ ( −6 ) ⋅ 7 + ( −4 ) ⋅ 3 ⋅ 8 −
Yechish.
7 8 9
−3 ⋅ 5 ⋅ 7 − ( −4 ) ⋅ 2 ⋅ 9 − 1 ⋅ ( −6 ) ⋅ 7 =
45 − 84 − 96 − 105 + 72 + 42 =
−126.
Uchburchak qoidasidan foydalanib determinantlarni hisoblang:

7.

0 0 1
1.31. 0 2 0 .
3 0 0
2 −1
3
1.32. −2 2 3 .
4 2 −3
1.33.Toʻrtinchi tartibli determinantni hisoblang:
a 0 3 5
0 0 b 2
.
∆=
1 c 2 3
0 0 0 d
Yechish. Determinantni toʻrtinchi satr elementlari boʻyicha yoyib
a 0 3
det er min antni
∆ = ( +d ) ⋅ 0 0 b =
=
2 − satr bo ' yicha yoyamiz
1 c 2
hisoblaymiz: =d ⋅ ( −b ) ⋅ a 0 =−d ⋅ b ⋅ a ⋅ c.
1 c
Satr yoki ustun elementlari boʻyicha yoyish orqali determinantlarni
hisoblang:
1
0
1.34.
3
2
1
3
1.35.
1
2
1
0
3
3
2
1
0
0
3
.
2
1
2 0 −3
1 0 4
.
5 −1 7
−2 1
0
1
Tenglamani yeching:

8.

1.36.
2x −1 x +1
x+2
x −1
= −6.
6
3
1.37. 2 x
1
x −1
0 = 0.
x+2
2
4
Uchinchi tartibli determinantlarni qulay usulda hisoblang:
a 1 a
1.38. −1 a 1 .
a −1 a
−1 −2
2 5 .
3
1.39. 1
−4
1
6
a+x
x
1.40. x
b+ x
x
x
x2
1.41. y 2
z2
x 1
y 1.
z 1
x
x .
c+x
sin 3α
cos3α 1
1.42. sin 2α
cos 2α 1 .
cos α 1
sin α
a b
c
1.43. b c a .
c a b
a x x
1.44. x b x .
x x c
2 −1 3 −2 4
1.45. Matritlmsa rangini ta’rifga asosan hisoblang: A = 4 −2 5 1 7 .
2 −1 1 8 2

9.

A matritsa 3 × 5 oʻlchamli, demak uning rangi 3 dan yuqori boʻlmaydi.
Uchinchi tartibli minorlarni hisoblaymiz:
2 −1 3
−4 − 10 − 12 + 12 + 4 + 10 = 0;
M1 =
4 −2 5 =
2 −1 1
2 −1 −2
M 2 =4 −2 1 =−32 − 2 + 8 − 8 + 32 + 2 = 0;
2 −1 8
2 −1 4
M3 =
4 −2 7 =
−8 − 14 − 16 + 16 + 8 + 14 = 0;
2 −1 2
−1 3 −2
M 4 =−2 5 1 =−40 − 3 + 4 − 10 + 48 + 1 = 0;
−1 1 8
3 −2 4
M 5 = 5 1 7 = 6 − 14 + 160 − 4 + 20 − 168 = 0;...
1
8
2
Barcha uchinchi tartibli minorlar nolga teng. Ikkinchi tartibli minorlarni
hisoblaymiz:
M 11 =
−1 3
= −5 + 6 = 1        
M 11 ≠ 0,
−2 5
r ( A) =
2.
Bu usulda noldan farqli minor topilgunga qadar hisoblashlar davom etadi.
Shuning uchun 3 va undan kattaroq tartibli matritsa rangini hisoblash birmuncha
qiyinchiliklarga olib keladi.
1.46. Matritsa rangini elementar almashtirishlar yordamida nollar yigʻib
hisoblang:
25
75
A=
75
25
Yechish:
31 17 43
94 53 132
94 54 134
32 20 48

10.

25
75
A =     
75
25
31
94
94
32
17 43 25 31 17 43 25 31 17 43
53 132 0 1 2 3 0 1 2 3
.
54 134 0 1 3 5 0 0 1 2
20 48 0 1 3 5 0 0 0 0
25 31 17
Bu matritsaning rangi 0 1 2 matritsa rangiga teng.
0 0 1
25 31 17
0 1 2= 25 ≠ 0
0 0 1
25 31 17
r 0 1 2 = 3
0 0 1
Demak, berilgan matritsaning rangi ham 3 ga teng. r ( A ) = 3.
1.47. Quyidagi matritsalar rangini minorlar ajratish usuli bilan hisoblang:
3 5 7
1 2 3 6
a ) 1 2 3 ; b) 2 3 1 6 ;
1 3 5
3 1 2 6
0 2 −4
1 2 1 3
−1 −4 5
4 −1 −5 −6
c)
; d) 3 1
7
1 −3 −4 −7
0 5 −10
2 1 −1 0
2 3
0
1.48. Quyidagi matritsalar rangini elementar almashtirish usuli bilan hisoblang:
1 2 1 3 4
a ) 3 4 2 6 8 ;
1 2 1 8 4
1 7 5 8 9 2
b) 3 21 15 24 27 6 ;
2 14 10 16 18 4
1 2 3 4
c) 2 4 6 8 ;
3 6 9 12

11.

4 5
0 2
d)
4 7
8 12
2 1 −3
1 1 2
;
3 3 −1
5 3 −4
2 3 2
−1
1.49. A = 5 1 4 matritsa uchun teskari A matritsani klassik usulda
1 −2 −1
toping.
Yechish.=
Aij ( i 1,=
2, 3;   j 1, 2, 3) A matritsa elementlarining algebraik
toʻldiruvchilari.
2 3 2
A =5 1 4 =−2 + 12 − 20 − 2 + 15 + 16 =43 − 24 =19 ≠ 0
1 −2 −1
Demak A xosmas matritsa, va A−1 teskari matritsa mavjud. Algebraik
toʻldiruvchilarni hisoblaymiz:
1 4
A11 =
=−1 + 8 =7;
−2 −1
A31 =
A21 =−
3 2
=− ( −3 + 4 ) =−1;
−2 −1
5 4
3 2
=− ( −5 − 4 ) =9;
= 12 − 2 = 10; A12 =−
−
1
1
1 4
2
2 2
−
A22 =
=−2 − 2 =−4; A32 =
5
1 −1
2
=
− ( 8 − 10 ) =
2;
4
5 1
2 3
A13 =
=−10 − 1 =−11; A23 =−
=− ( −4 − 3) =7;
1 −2
1 −2
2 3
A33 =
=−
2 15 =
−13;
5 1
topilganlarni formulaga qoʻyamiz
7 −1 10
va teskari A = 1/19 9 −4 2 matritsani olamiz. Teskari matritsaning
−11 7 −13
−1
toʻgriligini tekshirish uchun quyidagi tenglikni tekshiramiz:

12.

AA−1 = A−1 A = E
2 3 2
7 −1
5 1 4 ⋅ 1 / 19 9 −4
1 −2 −1
−11 7
14 + 27 − 22 −2 − 12 + 14
= 1 / 19 ⋅ 35 + 9 − 44 −5 − 4 + 28
7 − 18 + 11
−1 + 8 − 7
19 0 0 1 0 0
= 1 / 19 ⋅ 0 =
19 0 =
0 1 0
0 0 19 0 0 1
10
2 =
−13
20 + 6 − 26
50 + 2 − 52 =
10 − 4 + 13
E
−1
Demak, A toʻgʻri topilgan.
1 2 1
1 −1 −3 matritsa uchun A−1 matritsani Gauss-Jordan usulida
1.50. A =−
4 3 −2
toping.
−16 ≠ 0 teskari matritsa mavjud. Berilgan matritsani birlik
Yechish: A =
matritsa hisobida kengaytirib, elementar almashtirishlar bajaramiz, bu usulni to
chap tomonda A matritsa oʻrnida birlik matritsa hosil boʻlguncha davom
ettiramiz, oʻng tomonda hosil boʻlgan matritsa berilgan matritsaga nisbatan
teskari matritsa boʻladi.
1 2
−1 −1
4 3
1 2
~ 0 1
0 0
1 1 0 0 1 2 1 1 0
−3 0 1 0 0 1 −2 1 1
−2 0 0 1 0 −5 −6 −4 0
1 1 0 0 1 2 1
1
−2 1 1 0 ~ 0 1 −2 1
−16 1 5 1 0 0 1 −1 / 16
0
0
1
0
0
1
0 ~
−5 / 16 −1 / 16
1 0 5 −1
−2
0 1 0 0 −11 / 16 −7 / 16 5 / 16
~ 0 1 0 14 / 16 6 / 16 −2 / 16 ~ 0 1 0 14 / 16 6 / 16 −2 / 16
0 0 1 −1 / 16 −5 / 16 −1 / 16 0 0 1 −1 / 16 −5 / 16 −1 / 16

13.

11 7 −5
A−1 = −1 / 16 −14 −6 2 teskari matritsa toʻgʻri topilganini (3) formulaga
1
5 1
qoʻyib tekshiramiz:
1 2 1 11 7 −5
AA = −1 / 16 −1 −1 −3 ⋅ −14 −6 2 =
4 3 −2 1
5 1
7 − 12 + 5
−5 + 4 + 1
11 − 28 + 1
= −1 / 16 −11 + 14 − 3 −7 + 6 − 15
5 − 2 − 3 =
44 − 42 − 2 28 − 18 − 10 −20 + 6 − 2
−1
0 1 0 0
−16 0
0 1 0
= −1 / 16 0 −16 0 =
0
0 −16 0 0 1
demak,
teskari
matritsa
toʻg’ri
topilgan.
1.51. Berilgan kvadrat matritsalar uchun teskari matritsani ikki usulda toping:
−1 1
a)
;
4 2
1 3
b)
.
2 6
1 5 7
c) 3 1 1 ;
2 3 4
2 −1 7
d ) 5 3 2 ;
1 4 3
1.52. Berilgan kvadrat matritsalar uchun teskari matritsani qulay usulda toping:
tgα
a)
2
2
1
d )
3
0
1
ctgα
3
5
1
0
4
7
1
0
0
0
;
0
1
2 1 2
b) 1 1 1 ;
2 3 2
1
1
e)
1
1
2
1
2
1
1
3
1
3
1 0 5
c) 4 −2 −1
2 1 3
0
1
1
0

14.

… 1
1 1 1
… 1
0 1 1
f ) 0 0 1
… 1 ;
… … … … …
0 0 0 … 1
1
0
g) 0
…
0
a3 … a n
a 2 … a n−1
a … a n−2 .
… … … … …
0 0
0 … 1
a a2
1 a
0 1

15.

Foydalanishga tavsiya etiladigan adabiyotlar roʻyxati
1.
Mike Rosser. Basic mathematics for economists. London and New York
1993, 2003y.
2.
M.Harrison and P.Waldron Mathematics for economics and finance. London
and New York 2011y.
3. M. Hoy, J.Livernois et.al. Mathematics for Economics. The MIT Press,
London& Cambridge, 2011.
4. Robert M. Leekley, Applied Statistics for Businiess and Economics, USA,
2010.
5. Alpha C. Chiang, Kevin Wainwright, Fundamental Methods of Mathematical
Economics, NY 2005
6. Xashimov A.R., Xujaniyazova G.S. Iqtisodchilar uchun matematika. O’quv
qo’llanma. “Iqtisod-moliya”. 2017, 386 bet.
7.
Бабаджанов Ш.Ш. Математика для экономистов. Учебное пособие.
“Iqtisod-moliya”. 2017, 746 стр.