1. Matritsaning rangini toping: $$ A=\left(\begin{array}{cccccc} 1 & 7 & 5 & 8 & 9 & 2 \ 3 & 21 & 15 & 24 & 27 & 6 \ 2 & 14 & 10 & 16 & 18 & 4 \end{array}\right) $$ 2. Determinantni hisoblang: $$ \left|\begin{array}{rrr} 1 & 2 & -1 \ 3 & 7 & 2 \ 2 & 3 & -7 \end{array}\right| $$ 3. Chiziqli algebraik tenglamalar sistemasining ildizini toping: $$ \begin{array}{l} x+2 y+3 z=0 \ 2 x+4 y+5 z=-1 \ 3 x+5 y+6 z=1 \end{array} $$

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1. **Matritsaning rangini topish**

Matritsa $$ A $$ quyidagicha berilgan:
$$
A = \begin{pmatrix}
1 & 7 & 5 & 8 & 9 & 2 \
3 & 21 & 15 & 24 & 27 & 6 \
2 & 14 & 10 & 16 & 18 & 4
\end{pmatrix}
$$

Matritsaning rangini topish uchun uni qatorlar bo'yicha tartibga solamiz va qayerda qatorlar bir-biriga bog'liq ekanligini aniqlaymiz.

**Qadamlar:**

1. **Qator 2 ni tekshirish:**
   $$
   \text{Qator 2} = 3 \times \text{Qator 1} = 3 \times (1, 7, 5, 8, 9, 2) = (3, 21, 15, 24, 27, 6)
   $$
   Demak, Qator 2 Qator 1 ning uch baravaridir.

2. **Qator 3 ni tekshirish:**
   $$
   \text{Qator 3} = 2 \times \text{Qator 1} = 2 \times (1, 7, 5, 8, 9, 2) = (2, 14, 10, 16, 18, 4)
   $$
   Demak, Qator 3 ham Qator 1 ning ikkinchi baravaridir.

**Natija:**
Barcha qatorlar Qator 1 ning ko'paytmalari bo'lgani uchun matritsaning rangі 1 ga teng.

**Demak, matritsaning rangi:** $$ \boxed{1} $$

2. **Determinantni hisoblash**

Berilgan determinant:
$$
\left|\begin{array}{rrr}
1 & 2 & -1 \
3 & 7 & 2 \
2 & 3 & -7
\end{array}\right|
$$

3x3 determinantini hisoblash formulasi:
$$
\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
$$
Bu yerda:
$$
\begin{aligned}
a &= 1, & b &= 2, & c &= -1, \
d &= 3, & e &= 7, & f &= 2, \
g &= 2, & h &= 3, & i &= -7.
\end{aligned}
$$

**Hisoblash:**
$$
\begin{aligned}
\det(A) &= 1 \times (7 \times (-7) - 2 \times 3) - 2 \times (3 \times (-7) - 2 \times 2) + (-1) \times (3 \times 3 - 7 \times 2) \
&= 1 \times (-49 - 6) - 2 \times (-21 - 4) + (-1) \times (9 - 14) \
&= 1 \times (-55) - 2 \times (-25) + (-1) \times (-5) \
&= -55 + 50 + 5 \
&= 0
\end{aligned}
$$

**Demak, determinanti:** $$ \boxed{0} $$

3. **Chiziqli algebraik tenglamalar sistemasining ildizini topish**

Berilgan tenglamalar:
$$
\begin{cases}
x + 2y + 3z = 0 \
2x + 4y + 5z = -1 \
3x + 5y + 6z = 1
\end{cases}
$$

**Avval, kengaytirilgan matritsani tuzamiz:**
$$
\begin{pmatrix}
1 & 2 & 3 & | & 0 \
2 & 4 & 5 & | & -1 \
3 & 5 & 6 & | & 1
\end{pmatrix}
$$

**Qadamlar:**

1. **Qator 2 dan Qator 1 ni olib chiqamiz:**
   $$
   \text{Q2}' = \text{Q2} - 2 \times \text{Q1} = (2-2, 4-4, 5-6 | -1-0) = (0, 0, -1 | -1)
   $$

2. **Qator 3 dan Qator 1 ni olib chiqamiz:**
   $$
   \text{Q3}' = \text{Q3} - 3 \times \text{Q1} = (3-3, 5-6, 6-9 | 1-0) = (0, -1, -3 | 1)
   $$

Matritsa endi:
$$
\begin{pmatrix}
1 & 2 & 3 & | & 0 \
0 & 0 & -1 & | & -1 \
0 & -1 & -3 & | & 1
\end{pmatrix}
$$

3. **Qatorlarni almashtiramiz:**
   $$
   \text{Q2} \leftrightarrow \text{Q3}
   $$
   Natija:
   $$
   \begin{pmatrix}
   1 & 2 & 3 & | & 0 \
   0 & -1 & -3 & | & 1 \
   0 & 0 & -1 & | & -1
   \end{pmatrix}
   $$

4. **Orqaga almashtirish yordamida yechimni topamiz:**

- **Q3:** $$-1z = -1 $$ ⇒ $$ z = 1 $$
   - **Q2:** $$-y - 3z = 1 $$ ⇒ $$-y - 3(1) = 1 $$ ⇒ $$-y = 4 $$ ⇒ $$ y = -4 $$
   - **Q1:** $$ x + 2y + 3z = 0 $$ ⇒ $$ x + 2(-4) + 3(1) = 0 $$ ⇒ $$ x - 8 + 3 = 0 $$ ⇒ $$ x = 5 $$

**Natija:**
$$
x = 5, \quad y = -4, \quad z = 1
$$

**Demak, tenglamalar sistemasining yechimi:** $$ \boxed{(x, y, z) = (5, -4, 1)} $$
1. **Matritsaning rangi:** 1 2. **Determinantni hisoblash:** 0 3. **Tenglamalar sistemasining ildizi:** $$ x = 5 $$, $$ y = -4 $$, $$ z = 1 $$

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