chiziqli algebraik tenglamalar sistemasining matritsaviy shakli (matritsaning tatbig’i)

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1746956704.pptx                                               nn nnnn n n x x x x b b b b aaa aaa aaa a     2 1 2 1 21 22221 11211 ,, belgilashlarni kiritamiz. endi ( 12) sistemani matritsalarni ko’paytirish qoidas idan foydalanib, bax (13) ko’rinishda yozish mumkin. 0det a bo’lsa, teskari matritsa 1 a mavjud va baaxa 11   hosil bo’ladi. shunday qilib, noma’lum x matritsa ba 1 matritsaga teng bo’ladi, yahni x = ba 1 . bu (12) tenglamalar sistemasini yechishning matritsaviy yozuvini bildiradi. 1-misol. matritsalar yordamida ushbu tenglamalar sistemasini yeching:         293 ,442 ,4 321 321 …
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a matritsaning determinanti 0 dan farqli bo’lganligi uchun, unga teskari yagona 1 a matritsa mavjud va tenglamalar sistemasi yagona yechimga ega bo’ladi. endi 1 a teskari matritsani topish uchun  determinant elementlarining hamma algebraik to’ldiruvchilarini hisoblaymiz: 2 3 4 1 9 1 1 1 2 1 1 3 1 1 4 1 9 2 1 9 3 1 4 2 1 1 1 1 = × × - × × - × × - × × + × × + × × = = d 1 - a 1 - a d . 2 4 4 ; ; 9 3 1 4 2 1 1 1 1 3 2 1 ÷ ÷ ÷ ø ö ç ç ç è æ = ÷ ÷ ÷ ø ö ç ç ç è æ = ÷ ÷ ÷ ø ö ç ç ç è æ = b x x x x a b …
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apdan ga ko’paytirsak, yoki bo’lib, yahni image4.wmf oleobject4.bin image5.wmf oleobject5.bin image6.wmf oleobject6.bin image1.wmf oleobject1.bin image2.wmf oleobject2.bin image3.wmf oleobject3.bin                                                    1 3 2 25,04145,0 2)5,1(4445,2 214)3(43 2 4 4 5,015,0 5,145,2 133 x tenglik hosil bo’ladi. shunday kilib,                        1 3 2 3 2 1 x x x x yoki 1,3,2 3 2 1  ххх . (topilgan yechimlarni tenglamalar sistemasiga bevosita qo’yib, yechimning to’g’riligini tekshirib ko’rish mumkin). 1 , 3 , 2 3 2 1 - = = …
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1746956704.pptx                                               nn nnnn n n x x x x b b b b aaa aaa aaa a     2 1 2 1 21 22221 11211 ,, belgilashlarni kiritamiz. endi ( 12) sistemani matritsalarni ko’paytirish qoidas idan foydalanib, bax (13) ko’rinishda yozish mumkin. 0det a bo’lsa, teskari matritsa 1 a mavjud va baaxa 11   hosil bo’ladi. shunday qilib, noma’lum x matritsa ba 1 matritsaga teng bo’ladi, yahni x = ba …

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