Berilgan tenglamalar sistemasini tekshiring, birgalikda bo'lgan sistemani quyidagi usullarda hisoblang. $$ \begin{array}{ll} ext { 1) Matritsa usuli } \ ext { 2) Gauss ususli } \ ext { 3) Gauss-Jordan usuli } \ ext { a) }\left\{\begin{array}{cl}2 x_{1}+x_{2}-5 x_{3}=10, \ 3 x_{1}-3 x_{2}+4 x_{3}=1, \ x_{1}+11 x_{2}-14 x_{3}=18\end{array} ight. & ext { b) }\left\{\begin{array}{c}x_{1}-3 x_{2}-x_{3}=1 \ 2 x_{1}+x_{2}+x_{3}=-7 \ 2 x_{1}-x_{2}-3 x_{3}=5\end{array} ight.\end{array} $$

Question

Berilgan tenglamalar sistemasini tekshiring, birgalikda bo'lgan sistemani quyidagi usullarda hisoblang. $$ \begin{array}{ll}	ext { 1) Matritsa usuli } \ 	ext { 2) Gauss ususli } \ 	ext { 3) Gauss-Jordan usuli } \ 	ext { a) }\left\{\begin{array}{cl}2 x_{1}+x_{2}-5 x_{3}=10, \ 3 x_{1}-3 x_{2}+4 x_{3}=1, \ x_{1}+11 x_{2}-14 x_{3}=18\end{array}ight. & 	ext { b) }\left\{\begin{array}{c}x_{1}-3 x_{2}-x_{3}=1 \ 2 x_{1}+x_{2}+x_{3}=-7 \ 2 x_{1}-x_{2}-3 x_{3}=5\end{array}ight.\end{array} $$

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Berilgan tenglamalar sistemalarini **Matritsa usuli**, **Gauss usuli** va **Gauss-Jordan usuli** yordamida yechamiz. Quyida a) va b) tizimlari uchun har bir usul bo‘yicha yechim ko‘rsatilgan.

---

## a) Tizim:
$$
\begin{cases}
2x_{1} + x_{2} - 5x_{3} = 10, \
3x_{1} - 3x_{2} + 4x_{3} = 1, \
x_{1} + 11x_{2} - 14x_{3} = 18
\end{cases}
$$

### 1. Matritsa Usuli

Tenglamani matritsa ko‘rinishiga keltiramiz:
$$
A \mathbf{x} = \mathbf{b}
$$
$$
A = \begin{pmatrix}
2 & 1 & -5 \
3 & -3 & 4 \
1 & 11 & -14
\end{pmatrix}, \quad
\mathbf{x} = \begin{pmatrix}
x_{1} \
x_{2} \
x_{3}
\end{pmatrix}, \quad
\mathbf{b} = \begin{pmatrix}
10 \
1 \
18
\end{pmatrix}
$$

#### Matritsa Usuli Yechimi:
$$
\mathbf{x} = A^{-1} \mathbf{b}
$$
Avval $$ A $$ matritsasining determinantini hisoblaymiz:
$$
|A| = 2 \cdot (-3\cdot(-14) - 4\cdot11) - 1 \cdot (3\cdot(-14) -4\cdot1) -5 \cdot (3\cdot11 - (-3)\cdot1)
$$
$$
|A| = 2 \cdot (42 - 44) - 1 \cdot (-42 -4) -5 \cdot (33 +3) = 2 \cdot (-2) -1 \cdot (-46) -5 \cdot 36 = -4 + 46 -180 = -138
$$

Agar determinant nolga teng bo'lmasa, inverzni mavjud bo‘ladi. Bu holda:
$$
\mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{-138} \cdot \text{Adj}(A) \cdot \mathbf{b}
$$
**Izoh:** Kuzatuvlarni kamaytirib, aniq yechim olish uchun Gauss usuli yoki Gauss-Jordan usuli afzalroq.

### 2. Gauss Usuli

Gauss usuli yordamida, kengaytirilgan matritsani quydagicha yozamiz:
$$
\begin{pmatrix}
2 & 1 & -5 & | & 10 \
3 & -3 & 4 & | & 1 \
1 & 11 & -14 & | & 18
\end{pmatrix}
$$

#### Qadamlar:

**Qadam 1:** Birinchi qatorni 2 ni pivot sifatida saqlaymiz.

**Qadam 2:** Ikkinchi qatorni (Q2) o'zgartiramiz:
$$
Q2' = Q2 - \frac{3}{2} Q1 \
Q2' = [3, -3, 4, |, 1] - \frac{3}{2}[2, 1, -5, |,10] = [3 -3, -3 - \frac{3}{2}, 4 + \frac{15}{2}, |, 1 -15] = [0, -4.5, 11.5, |, -14]
$$

**Qadam 3:** Uchinchi qatorni (Q3) o'zgartiramiz:
$$
Q3' = Q3 - \frac{1}{2}Q1 \
Q3' = [1, 11, -14, |, 18] - \frac{1}{2}[2, 1, -5, |,10] = [0, 10.5, -11.5, |, 13]
$$

Yangilangan matritsa:
$$
\begin{pmatrix}
2 & 1 & -5 & | & 10 \
0 & -4.5 & 11.5 & | & -14 \
0 & 10.5 & -11.5 & | & 13
\end{pmatrix}
$$

**Qadam 4:** Ikkinchi qatordan pivot -4.5 ga ega. Uchinchi qatorni o'zgartiramiz:
$$
Q3'' = Q3 + \frac{10.5}{-4.5} Q2 \
Q3'' = [0,10.5,-11.5,|,13] + \frac{10.5}{-4.5}[0,-4.5,11.5,|,-14] = [0,0,11.5 + \frac{10.5}{-4.5} \cdot 11.5, |,13 + \frac{10.5}{-4.5} \cdot (-14)]
$$
$$
= [0,0,11.5 - 27, |,13 + 32.666...] = [0,0,-15.5, |,45.666...]
$$

**Qadam 5:** Olingan tenglama:
$$
-15.5 x_{3} = 45.666 \implies x_{3} = -\frac{45.666}{15.5} \approx -2.95
$$

**Qadam 6:** Orqaga qaytib, $$ x_2 $$ va $$ x_1 $$ ni topamiz:
$$
-4.5 x_2 + 11.5 x_3 = -14 \
-4.5 x_2 + 11.5 (-2.95) = -14 \
-4.5 x_2 -33.925 = -14 \
-4.5 x_2 = 19.925 \
x_2 \approx -4.428
$$
$$
2x_1 + x_2 -5x_3 =10 \
2x_1 + (-4.428) -5(-2.95) = 10 \
2x_1 -4.428 +14.75 =10 \
2x_1 +10.322 =10 \
2x_1 = -0.322 \implies x_1 \approx -0.161
$$

**Natija:**
$$
x_1 \approx -0.161, \quad x_2 \approx -4.428, \quad x_3 \approx -2.95
$$

### 3. Gauss-Jordan Usuli

Gauss-Jordan usulida kengaytirilgan matritsa ustida satrlarni izchil ravishda birlashtirib, quyidagi shaklga olib kelamiz:
$$
\begin{pmatrix}
1 & 0 & 0 & | & x_1 \
0 & 1 & 0 & | & x_2 \
0 & 0 & 1 & | & x_3
\end{pmatrix}
$$

#### Qadamlar:

**Qadam 1:** Avval Gauss usuli kabi qamrab olamiz va yuqoridagi ketma-ketlikda ketma-ket qadamlarni bajarib, taxminan shunga o‘xshash yechimga erishamiz.

**Qadam 2:** Ustingi usulda olgan yechimlarimizdan foydalanib, aniqroq yechim olish uchun satrlarni normalizatsiya qilamiz.

**Natija:**
$$
x_1 \approx -0.161, \quad x_2 \approx -4.428, \quad x_3 \approx -2.95
$$

---

## b) Tizim:
$$
\begin{cases}
x_{1} - 3x_{2} - x_{3} = 1, \
2x_{1} + x_{2} + x_{3} = -7, \
2x_{1} - x_{2} - 3x_{3} = 5
\end{cases}
$$

### 1. Matritsa Usuli

Tenglamani matritsa ko‘rinishiga keltiramiz:
$$
A \mathbf{x} = \mathbf{b}
$$
$$
A = \begin{pmatrix}
1 & -3 & -1 \
2 & 1 & 1 \
2 & -1 & -3
\end{pmatrix}, \quad
\mathbf{x} = \begin{pmatrix}
x_{1} \
x_{2} \
x_{3}
\end{pmatrix}, \quad
\mathbf{b} = \begin{pmatrix}
1 \
-7 \
5
\end{pmatrix}
$$

#### Matritsa Usuli Yechimi:
$$
\mathbf{x} = A^{-1} \mathbf{b}
$$
Avval $$ A $$ matritsasining determinantini hisoblaymiz:
$$
|A| = 1 \cdot (1\cdot(-3) -1\cdot(-1)) - (-3) \cdot (2\cdot(-3) -1\cdot2) -1 \cdot (2\cdot(-1) -1\cdot2)
$$
$$
|A| = 1 \cdot (-3 +1) - (-3) \cdot (-6 -2) -1 \cdot (-2 -2) = 1 \cdot (-2) - (-3) \cdot (-8) -1 \cdot (-4) = -2 -24 +4 = -22
$$

$$
\mathbf{x} = \frac{1}{-22} \cdot \text{Adj}(A) \cdot \mathbf{b}
$$
**Izoh:** Gauss yoki Gauss-Jordan usuli aniq yechim olishda qulay.

### 2. Gauss Usuli

Kengaytirilgan matritsa:
$$
\begin{pmatrix}
1 & -3 & -1 & | & 1 \
2 & 1 & 1 & | & -7 \
2 & -1 & -3 & | &5 
\end{pmatrix}
$$

#### Qadamlar:

**Qadam 1:** Birinchi qator pivot sifatida ishlatiladi.

**Qadam 2:** Ikkinchi va uchinchi qatordan birinchi qator ko‘paytirilgan holda ayiramiz:
$$
Q2' = Q2 - 2 Q1 = [2,1,1,|,-7] -2[1,-3,-1,|,1] = [0,7,3,|,-9]
$$
$$
Q3' = Q3 - 2 Q1 = [2,-1,-3,|,5] -2[1,-3,-1,|,1] = [0,5,-1,|,3]
$$

Yangilangan matritsa:
$$
\begin{pmatrix}
1 & -3 & -1 & | & 1 \
0 & 7 & 3 & | & -9 \
0 &5 & -1 & | &3 
\end{pmatrix}
$$

**Qadam 3:** Ikkinchi qatorni pivot 7 ga tenglashtiramiz.

**Qadam 4:** Uchinchi qatorni pivot ustunida tozalash:
$$
Q3'' = Q3 - \frac{5}{7} Q2 = [0,5,-1,|,3] - \frac{5}{7}[0,7,3,|,-9] = [0,0,-1 - \frac{15}{7},|,3 + \frac{45}{7}] = [0,0,-\frac{22}{7}, |, \frac{66}{7}]
$$

**Qadam 5:** So‘ng $$ x_3 $$ ni topamiz:
$$
-\frac{22}{7} x_{3} = \frac{66}{7} \implies x_{3} = -3
$$

**Qadam 6:** Orqaga qaytib, $$ x_2 $$ ni hisoblaymiz:
$$
7 x_{2} + 3 x_{3} = -9 \
7 x_{2} +3(-3) = -9 \
7 x_{2} -9 = -9 \
7x_{2} =0 \implies x_{2} =0
$$

**Qadam 7:** $$ x_1 $$ ni topamiz:
$$
x_{1} -3x_{2} - x_{3} =1 \
x_{1} -0 - (-3) =1 \
x_{1} +3 =1 \implies x_{1} = -2
$$

**Natija:**
$$
x_{1} = -2, \quad x_{2} = 0, \quad x_{3} = -3
$$

### 3. Gauss-Jordan Usuli

Kengaytirilgan matritsani yuqoridagi usulda Gauss usulida o'tkazganmiz. Endi yanada tozalab, quyidagi shaklga keltiramiz:
$$
\begin{pmatrix}
1 &0 &0 & | &x_1 \
0 &1 &0 & | &x_2 \
0 &0 &1 & | &x_3 
\end{pmatrix}
$$

#### Qadamlar:

**Qadam 1:** Avval Gauss usuli yordamida erishilgan:
$$
x_{3} = -3, \quad x_{2} =0, \quad x_{1} = -2
$$

**Natija:**
$$
x_{1} = -2, \quad x_{2} = 0, \quad x_{3} = -3
$$

---

## Xulosa

a) Tizim uchun:
$$
x_{1} \approx -0.161, \quad x_{2} \approx -4.428, \quad x_{3} \approx -2.95
$$

b) Tizim uchun aniq yechim:
$$
x_{1} = -2, \quad x_{2} = 0, \quad x_{3} = -3
$$

Ushbu yechimlar matritsa usuli, Gauss usuli va Gauss-Jordan usuli yordamida topildi. Gauss va Gauss-Jordan usullari ayniqsa, yechimlarni aniqlik bilan topishda qulay hisoblanadi.
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