Matritsalar. Chiziqli tenglamalar sistemasi Mavzu rejasi

Misollar: Tenglamalar tizimini Gauss usuli bilan yeching.

Shunday qilib, tizim cheksiz ko'p echimlarga ega.
Ushbu maqolada biz chiziqli algebraik tenglamalar tizimini echishning matritsa usuli haqida gaplashamiz, uning ta'rifini topamiz va echimiga misollar keltiramiz.
Ta'rif 1
Teskari matritsa usuli noma'lumlar soni tenglamalar soniga teng bo'lgan taqdirda SLAElarni echish uchun ishlatiladigan usul.
1-misol
N noma'lum bo'lgan n chiziqli tenglamalar tizimining echimini toping:
11 x 1 + a 12 x 2 +. ... ... + a 1 n x n \u003d b 1 a n 1 x 1 + a n 2 x 2 +. ... ... + a n n x n \u003d b n
Matritsali yozuv : A × X \u003d B
bu erda A \u003d a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋯ ⋯ ⋯ a n 1 a n 2 ⋯ a n n - bu sistemaning matritsasi.
X \u003d x 1 x 2-x n - noma'lum ustun,
B \u003d b 1 b 2 ⋮ b n - erkin koeffitsientlar ustuni.
Biz olgan tenglamadan siz X ni ifodalashingiz kerak. Buning uchun chapdagi matritsa tenglamasining ikkala tomonini A - 1 ga ko'paytirish kerak:
A - 1 × A × X \u003d A - 1 × B.
A - 1 × A \u003d E bo'lgani uchun, E × X \u003d A - 1 × B yoki X \u003d A - 1 × B.
Izoh
A matritsaga teskari matritsa faqat d e t A shart nolga teng bo'lmagan taqdirda mavjud bo'lish huquqiga ega. Shuning uchun teskari matritsa usuli bilan SLAElarni echishda, avvalo, d e t A
Agar d e t A nolga teng bo'lmagan taqdirda, tizim faqat bitta echimga ega: teskari matritsa usuli yordamida. Agar d e t A \u003d 0 bo'lsa, unda tizimni bu usul bilan echib bo'lmaydi.

Teskari matritsa usuli yordamida chiziqli tenglamalar tizimini echishga misol

2-misol
Biz teskari matritsa usuli bilan SLAE ni hal qilamiz:
2 x 1 - 4 x 2 + 3 x 3 \u003d 1 x 1 - 2 x 2 + 4 x 3 \u003d 3 3 x 1 - x 2 + 5 x 3 \u003d 2
Qanday hal qilish kerak?

• 
  Biz tizimni A X \u003d B matritsa tenglamasi shaklida yozamiz, bu erda
  

A \u003d 2 - 4 3 1 - 2 4 3 - 1 5, X \u003d x 1 x 2 x 3, B \u003d 1 3 2.

• 
  Ushbu tenglamadan X ni ifodalaymiz:
  

• 
  A matritsaning determinantini toping:
  

det A \u003d 2 - 4 3 1 - 2 4 3 - 1 5 \u003d 2 × (- 2) × 5 + 3 × (- 4) × 4 + 3 × (- 1) × 1 - 3 × (- 2) × 3 - - 1 × (- 4) × 5 - 2 × 4 - (- 1) \u003d - 20 - 48 - 3 + 18 + 20 + 8 \u003d - 25
d e t A 0 ga teng emas, shuning uchun teskari matritsali eritma usuli ushbu tizim uchun mos keladi.

• 
  Birlashma matritsasi yordamida teskari A - 1 matritsasini toping. A matritsaning mos keladigan elementlariga A i j algebraik qo'shimchalarini hisoblaymiz:
  

A 11 \u003d (- 1) (1 + 1) - 2 4 - 1 5 \u003d - 10 + 4 \u003d - 6,
A 12 \u003d (- 1) 1 + 2 1 4 3 5 \u003d - (5 - 12) \u003d 7,
A 13 \u003d (- 1) 1 + 3 1 - 2 3 - 1 \u003d - 1 + 6 \u003d 5,
A 21 \u003d (- 1) 2 + 1 - 4 3 - 1 5 \u003d - (- 20 + 3) \u003d 17,
A 22 \u003d (- 1) 2 + 2 2 3 3 5 - 10 - 9 \u003d 1,
A 23 \u003d (- 1) 2 + 3 2 - 4 3 - 1 \u003d - (- 2 + 12) \u003d - 10,
A 31 \u003d (- 1) 3 + 1 - 4 3 - 2 4 \u003d - 16 + 6 \u003d - 10,
A 32 \u003d (- 1) 3 + 2 2 3 1 4 \u003d - (8 - 3) \u003d - 5,
A 33 \u003d (- 1) 3 + 3 2 - 4 1 - 2 \u003d - 4 + 4 \u003d 0.

• 
  A matritsaning algebraik qo'shimchalaridan tashkil topgan A * birlashma matritsasini yozamiz:
  

A * \u003d - 6 7 5 17 1 - 10 - 10 - 5 0

• 
  Biz teskari matritsani quyidagi formula bo'yicha yozamiz:
  

A - 1 \u003d 1 d e t A (A *) T: A - 1 \u003d - 1 25 - 6 17 - 10 7 1 - 5 5 - 10 0,

• 
  A - 1 teskari matritsani B bo'sh atamalar ustuniga ko'paytiramiz va tizimga yechim olamiz:
  

X \u003d A - 1 × B \u003d - 1 25 - 6 17 - 10 7 1 - 5 5 - 10 0 1 3 2 \u003d - 1 25 - 6 + 51 - 20 7 + 3 - 10 5 - 30 + 0 \u003d - 1 0 1
Javob : x 1 \u003d - 1; x 2 \u003d 0; x 3 \u003d 1
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