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11Central Asian Journal of
Education and Innovation
IKKI O'ZGARUVCHILI TENGLAMALAR SISTEMASI Sharipova Madina Po'latovna
Osiyo Xalqaro Universiteti "Umumtexnik fanlar" kafedrasi o'qituvchisi [email protected] Latipova Shahnoza Salim qizi Osiyo Xalqaro Universiteti "Umumtexnik fanlar" kafedrasi o'qituvchisi
[email protected] https://doi.org/10.5281/zenodo.10691337
ARTICLE INFO
Qabul qilindi: 10-February 2024 yil Ma'qullandi: 15- February 2024 yil Nashr qilindi: 22- February 2024 yil
KEY WORDS
Chiziqli tenglamalar, matritsa, Kramer formulasi,determinan t.
ABSTRACT
Maqolada ikki o'zgaruvchili tenglamalar sistemasi ,uniyechishning Kramer usullari o'rganilgan..
(1)
Ikki o'zgaruvchili tenglamalar sistemasi, matematikda, texnikada va boshqa sohalar bo'yicha juda muhim ahamiyatga ega.Ikki o'zgaruvchili tenglamalar sistemasi, har bir x qiymatiga mos keladigan y qiymatlarini hisoblash uchun amalga oshiriladi. Bu, grafiklar, ko'ordinatalar va Kramer usuli orqali ham yoritilishi mumkin.
Bunday tenglamalar sistemasining umumiy ko'rinishi
¡ax+by = c\
\a2x + ^y = c2
dan iborat bo'lib, u ikki noma'lum (o'zgaruvchanli) chiziqli tenglamalar sistemasi deb
ataladi.
1-ta'rif. x va y o'zgaruvchilarning (2) sistemaning har bir tenglamasini to'g'ri sonli tenglikka aylantiruvchi a va ß qiymatlari jufti (a,ß) uning yechimi deyiladi. (2) sistemani yechish uchun uning birinchi tenglamasini b ga, ikkinchisini b ga ko'paytirib, ularni ayiramiz. U holda
x(ap2 -a2\) = cp2 -c2b hosil bo'ladi, agar a^b2 -a2b ^ 0 bo'lsa,
x = cA - c-A (2) ap2 - a2bx
ga ega bo'lamiz.
y =
ac2 - a2c!
ap2 - a2bx
(3)
ni topamiz.
Shunday qilib, aft2 - a2b ^ 0 bo'lsa, (1) sistema yagona yechimga ega bo'ladi. Ushbu
A =
al b a b
ko'rinishdagi jadval 2 sistemaning matritsasi deyiladi. A matritsaning gorizontal qatorlari uning satrlari, vertikal qatorlari esa - ustunlari deyiladi. a, b,, a2, b lar uning
elementlari deyiladi. Qaralayotgan matritsa, ikkinchi tartibli kvadrat matritsa deyiladi. Uning chap yuqori burchagidan o'ng pastki burchagiga boruvchi dioganal uning bosh dioganali deyiladi.
3 va 4 formulalardagi kasrlarning maxraji bosh dioganaldagi elementlar ko'paytmasidan, ikkinchi dioganalda turgan elementlarning ko'paytmasini ayirish natijasida tuzilganligi ko'rinib turibdi: a:b2 — a2b . Bu ifoda A matritsaning determinanti deb ataladi va quyidagicha belgilanadi: a b
a b2
A =
Demak, ta'rifga ko'ra,
A=
a b
a2 b2
= ab - ab
Quyidagicha tasdiq o'rinli: ikkinchi tartibli determinant nolga teng bo'lishi uchun, uning satrlaridagi yoki ustunlaridagi elementlar proporsional bo'lishi zarur va yetarli.
Yuqoridagi belgilashlar asosida 3 tenglikning surati quyidagi determinantdan iborat:
A, =
ci bi C2 b2
C1b2 - C2b1 .
Bu determinant A determinantdagi birinchi ustunni ozod hadlar ustuni bilan almashtirishdan hosil qilingan. Xuddi shunga o'xshash A determinantning ikkinchi ustunini ozod hadlar bilan almashtirsak, 4 tenglikning suratidagi ifoda hosil bo'ladi:
A 2 =
A,
A,
Shunday qilib, agar A * 0 bo'lsa, 2 sistemaning yechimi x = —-, y =
AA
lardan iborat va yagonadir. Bu formulalar Kramer formulalari deyiladi. Misol. Kramer formulalaridan foydalanib, 1 sistemani yechamiz.
\x + y = 20 [4 x + 2 y = 76, 20 1 76 2
A =
1 1 4 2
= 1-2 - 4-1 = -2 * 0
A =
= 20-2 - 76-1 = -36; A =
1 20 4 76
= 76-1 - 20 - 4 = -4.
U holda, x = (-36) : (-2) = 18, y = (-4) : (-2) = 2. Endi
A=
a b a b
= axb2 -a2b =0
(4)
al c1
a2 c2
a
b
holni qaraymiz. 5 tenglikni — = — ko'rinishda yozish mumkin, ya'ni bu holda
ab
noma'lumlarning koeffisentlari proporsionaldir. Bundan tashqari, agar
C b
c b
a b, = 0 ya'ni = -L
a2
ham o'rinli bo'lsa,
a
b
a
bL ÇL b2 a2
bo'lib, biz ikki noma'lumli bitta tenglamaga ega bo'lamiz. Bu holda u sistema cheksiz ko'p yechimga ega bo'ladi. Nihoyat, agar
a, b. c b
A = 1 1 =0 lekin, 1 1 * 0
a b C2 b2
ya'ni
a b a, — = — * 1
a
b
bo'lsa, u holda sistema ziddiyatli bo'ladi va yechimga ega bo'lmaydi. 1- sistemaning x = a, y = ¡3 yechimi Dekart koordinatalari sistemasida
axx + by = C va a2x + b2y = c2 to'g'ri chiziqlarning kesishish nuqtasini ifodalaydi.
Agar
a b
a2 b
* 0 bo'lsa, to'g'ri chiziqlar har xil bo'lib, yagona umumiy nuqtaga ega
b
bo'ladi. — = — =— bo'lsa, har ikkala tenglama bitta tenglamani ifodalaydi va uning har bir
a b c
nuqtasi, berilgan to'g'ri chiziqning «kesishish nuqtalari» bo'ladi, ya'ni ular ustma-ust tushadi.
Nihoyat, agar — = — ^ — bo'lsa, to'g'ri chiziqlar parallel va ular bitta ham umumiy nuqtaga
a b c
ega bo'lmaydi.
Shuni ta'kidlab o'tamizki, maktab matematika kursida 2 sistemani o'rniga qo'yish va qo'shish usullari yordamida yechilar edi. Bu yerda yana bitta usulni o'rgandik. Ushbu usulning bitta qulayligi shundaki, uning yordamida n noma'lumli n -ta chiziqli tenglamalar sistemasini ham yechish mumkin. Bunga qiziqqan o'quvchilar [1] adabiyotga murojaat qilishlari mumkin.
Umuman, f (x, y) = 0 Ba x, y) = 0 tenglamalar sistemasi deb, ularning (f (x, y) = 0) a (p( x, y) = 0) kon'yunksiyasiga, maktabda belgilangandek, jf (x, y) = 0 x y) = 0
sistemaga aytiladi. 6 sistemaning har birini to'g'ri sonli tenglikka aylantiruvchi a Ba ¡3 sonlar jufti (a, uning yechimi deyiladi.
Bizga ma'lumki, ikki predikat kon'yunksiyasining rostlik to'plami, shu predikatlar rostlik to'plamlari kesishmasidan iborat. Xuddi shunga o'xshash 6 sistema yechimlarining
(5)
to'plami, f (x, y) = 0 Ba p(x, y) = 0 tenglamalar yechimlari to'plamining kesishmasidan iborat. Geometrik yo'l bilan bu to'plam quyidagicha topiladi: f (x, y) = 0 Ba <^(x, y) = 0 tenglamalarning grafigi chiziladi, so'ngra bu grafiklarning kesishish nuqtalari topiladi. Misol. (4; 3), (-4; -3) juftliklar
Í3x - 4 y = 0 [x2 + y2 = 25
sistema yechimlari to'plamiga tegishli bo'ladi. Haqiqatdan, x = 4, y = 3 hamda x = -4, y = -3 lar tenglamalarning har birini qanoatlantiradi: 3 • 4 - 4 • 3 = 0, 42 + 32 = 25. Ba 3 • (-4)-4 • (-3) = 0, (-4)2 + (-3)2 = 25
Bunday sistemalar, statistik hisob-kitoblarda, matematik modellarda, injenerlik amaliyotlarda, kriptografiyada, hisoblash texnologiyalarida va boshqalar kabi ko'plab sohalarda foydalaniladi.
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