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0ALGEBRAIK TENGLAMALAR SISTEMASINI YECHISH
To'yliyev Abbos
DTPI talabasi https://doi.org/10.5281/zenodo.11121779
Biz quyidagi tenglamalar sistemasini yechishni qaraylik.
яцхП H ^12^2 H H ^ln^n — n+1 а21*1 H а22*2 H H a2n*n — a2 n+1
,anlXl H H ■" H ^^
a
n n+1
Tenglamalar sistemasini yechishning quyidagi usulini qaraylik. Soddalik uchun 4 nomalumli 4 ta tenglamadan iborat bo'lgan sistemani qaraymiz.
йцхП H ^12^2 H %3*3 H CI14X4 — ^15 a2l*l H a22^2 H a23^3 H a24*4 — a25 a3lxl H a32x2 H a33x3 H a34x4 — a35 ^^41X1 H Я42Х2 H CI43X3 H Й44Х4 — Я45
(1) tenglamalar sistemasini barcha tenglamalarini mos ravishda хп no'malumni oldidagi koeffitsiyentga bo'lib yuboramiz. Natijada quyidagi tenglamalar sistemasi kelib chiqadi:
Г , а12 , а13 , а14 _ al5 xI H X2 H X3 H X4 —
Яц ^11 ^11 ^11
, a22 , a23 , a24 _ a25 xI H X2 H X3 H X4 —
a2l a2l a2l a2l
, a32 , a33 , a34 _ a35 xI H X2 H X3 H X4 —
a3l a3l a3l a3l
, a42 , a43 , a44 _ a45 xI H X2 H X3 H X4 —
Я41 Я41 Я41 Я41
Ushbu sistemadan
— a(l) n — 1,2,3,4; i — 2,3,4,5
anl
deb belgilash olamiz. Natijada quyidagi sistema hosil bo'ladi:
(l)
(l)
(l) (l)
xi H
(l) (l) (l) (l) H Й22 ^2 H ^23 ^3 H ^24 ^4 — ^25
m , rn _ „(1)
(l) (l) (l) Xl H Я32 X2 H Я33 X3 H ^34 X4
a
35
(l) (l) (l) (l) Xl H 042 X2 "Г Я43 X3 H ^44 X4 — ^45
Bu sistemani 1-tenglamasidan 2-3-4-tenglamalarini ayiramiz, va
a
(l) ln
(1) (1*)
aiu — am (n — 2,3,4,5 í — 2,3,4) deb belgilash olsak, quyidagi tenglamalar
sistemasini hosil qilamiz:
(l) (l) (l) (l) H ^2 H ^13 ^3 H ^14 ^4 — ^15
(l*)
(l*)
(l*) (l*)
^2 ^2 I Я23 ^3 H ^24 ^4 — 5 (l*) (l*) (l*) (l*) Я32 ^2 ! Я33 ^3 ' ^34 ^4 — ^35
(l*) (l*)
(l*)
(l*)
^42 ^2 H ^43 *3 H ^44 X4 — ^45
hosil bo'lgan tenglamalar sistemasini 2-3-4- tenglamalarini x2 no'malumini oldidagi koeffitsiyentiga bo'lib yuboramiz va
a.
a.
(i*)
(2)
— — a. (i*) unt
n2
(n — 2,3,4; í — 3,4,5)
belgilash olamiz, natijada tenglamamiz quyidagi ko'rinishga keladi:
(i)
(i)
(i) (i)
Xi +
(2) (2) (2) X2 + X3 + ^24, ^4 — ^25 (2) (2) (2) X2 + ^3 + ^34 ^4 — ^35
(2) (2) (2) X2 + Я43 X3 + ^44 X4 — ^45
Hosil bo'lgan tenglamalar sistemasida 2-tenglamasidan 3-4-tenglamalarini ayiramiz va
a
(2) 2n
hosil qilamiz:
(2) _ _(2*)
a¿n
a¿n ) (n — 3,4,5 í — 3,4) deb belgilash olib, quyidagi tenglamalar sistemasini
(i) (i) (i) (i) ^i + ^2 + ^13 ^3 + ^4 — ^15
(2)
(2) (2)
X2 + ^^3 ^3 + ^^4 ^4 — &25
(2*) (2*) _ (2*)
a33 ^3 + ^34 ^4 — ^35
(2*) (2*) _ (2*)
tt43 X3 + tt44 X4 — tt45
hosil bo'lgan sistemani 3- va 4- tenglamalarini x3 no'malumini oldidagi koeffitsiyentiga bo'lib yuboramiz va
(2*)
"" (n — 3,4; í — 4,5)
a.
a.
m _ (3) (2*) "ni
n3
belgilash olib, quyidagi sistemani hosil qilamiz:
(i) (i) (i) (i) ^i + ft^ X2 + ^^3 ^3 + ^^4 ^4 — ^15
(2)
(2) (2)
(3*) um
X2 + ^23 X3 + X4 — ^25 (3) (3)
X3 + X4 — ^35
(3) (3)
X3 + tt44 X4 — 045
Hosil bo'lgan sistemani 3-tenglamasidan 4-tenglamasini ayiramiz va a3n n — 4,5; í — 4 belgilash olsak, quyidagi sistema hosil bo'ladi:
(3)
(3)
a. —
(i)
(i)
(i) (i)
Xi +
(2) (2) (2) X2 + ^23 X3 + ^4 — ^25
(3) (3)
X3 + X4 — ^35
a
(3*) 44
X4 — tt
(3*) 45
hosil bo'lgan sistemaning 4-tenglamasidan x4 ni topsak,
(3*)
a
X4 —
a
45 _ (4) (3*) — tt45 44
ekanligini topib olamiz. Endi teskarisidan boorish usulini qo'llab xb x2, x3, x4 larni topamiz.
X4
(4)
a
45
*2
(2)
a
25
(3) (3) (4)
X3 — tt35 tt34 tt45
(2) (4) tt24 tt45
ß(2)(ß(3)
(3) (4) tt34 tt45
)
(1) (1) (4) (1) (3) (3) (4) (1) (2) (2) (4) (2) (3) (3) (4)
Л1 — u15 u14 u45 u13 (u35 u34 u45 ) u12 (u25 u24 u45 u23 (u35 u34 u45 J)
Misol. Quyidagi tenglamalar sistemasini yeching:
'6,1x1 + 6,2x2 — 6,3x3 + 6,4x4 — 6,5 1,1x1 — 1,5x2 + 2,2x3 — 3,8x4 — 4,2 5,1x1 — 5,0x2 + 4,9x3 — 4,8x4 — 4,7 Д,8х1 + 1,9x2 + 2,0x3 — 2,1x4 — 2,2 Ushbu tenglamalar sistemasini yechish uchun jadvaldan foydalanamiz:
^¿1 a¿2 a¿3 a¿4 a¿5
6,1 6,2 -6,3 6,4 6,5
1,1 -1,5 2,2 -3,8 4,2
5,1 -5,0 4,9 -4,8 4,7
a4i 1,8 1,9 2,0 -2,1 2,2
(1) a . u1¿ 1 1,0164 -1,0328 1,0492 1,0656
(1) a . U2Í 1 -1,3636 2 -3,4545 3,8182
(1) a . U3Í 1 -0,9804 0,9608 -0,9412 0,9216
(1) a) ■ 4l 1 1,0556 1,1111 -1,1667 1,2222
(1*) a. u1¿ 1 1,0164 -1,0328 1,0492 1,0656
(1*) a. U2¿ 0 2,38 -3,0328 4,5037 -2,7526
(1*) U3Í 0 1,9968 -1,9936 1,9904 0,144
(1*) 4l 0 -0,0392 -2,1439 2,2159 -0,1566
(2) a . u1¿ 1 1,0164 -1,0328 1,0492 1,0656
(2) a . u2¿ 0 1 -1,2743 1,8923 -1,1566
(2) a . U3Í 0 1 -0,9984 0,9968 0,0721
(2) ^ ■ 4l 0 1 54,6913 -56,5281 3,9949
(2*) a. u1¿ 1 1,0164 -1,0328 1,0492 1,0656
(2*) a. u2¿ 0 1 -1,2743 1,8923 -1,1566
(2*) a. U3Í 0 0 -0,2759 0,8955 -1,2287
(2*) 4l 0 0 -55,9656 58,4204 -5,1515
(3) a . u1¿ 1 1,0164 -1,0328 1,0492 1,0656
(3) a . u2¿ 0 1 -1,2743 1,8923 -1,1566
(3) a . U3Í 0 0 1 -3,2457 4,4534
(3) a: ■ 4l 0 0 1 -1,0439 0,0921
(3*) П ■ u1t 1 1,0164 -1,0328 1,0492 1,0656
(3*) п. 0 1 -1,2743 1,8923 -1,1566
(3*) п. u3t 0 0 1 -3,2457 4,4534
(3*) п. u3i 0 0 0 -2,2018 4,3613
-2,2018x4 = 4,3613 x4 = -1,9808 = a(4} x3 = 4,4534 - (-3,2457) * -1,9808 = -1,9757 x2 = -1,1566 - 1,8923 * (-1,9808) - (-1,2743) * (-1,9757) = 0,0740 x1 = 1,0656 - 1,0492 * (-1,9808) - (-1,0328) * (-1,9757) - 1,0164 * 0,0740 = 1,0281 Javob: x1 = 1,0281; x2 = 0,0740; x3 = -1,9757; x4 = -1,9808.
REFERENCES
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